7.3: Maximum Likelihood
Basic Theory
The Method
Suppose again that we have an observable random variable \(\bs{X}\) for an experiment, that takes values in a set \(S\). Suppose also that distribution of \(\bs{X}\) depends on an unknown parameter \(\theta\), taking values in a parameter space \(\Theta\). Of course, our data variable \(\bs{X}\) will almost always be vector valued. The parameter \(\theta\) may also be vector valued. We will denote the probability density function of \(\bs{X}\) on \(S\) by \(f_\theta\) for \(\theta \in \Theta\). The distribution of \( \bs{X} \) could be discrete or continuous.
The likelihood function is the function obtained by reversing the roles of \(\bs{x}\) and \(\theta\) in the probability density function; that is, we view \(\theta\) as the variable and \(\bs{x}\) as the given information (which is precisely the point of view in estimation).
The likelihood function at \( \bs{x} \in S \) is the function \( L_{\bs{x}}: \Theta \to [0, \infty) \) given by \[ L_\bs{x}(\theta) = f_\theta(\bs{x}), \quad \theta \in \Theta \]
In the method of maximum likelihood , we try to find the value of the parameter that maximizes the likelihood function for each value of the data vector.
Suppose that the maximum value of \( L_{\bs{x}} \) occurs at \( u(\bs{x}) \in \Theta \) for each \( \bs{x} \in S \). Then the statistic \( u(\bs{X}) \) is a maximum likelihood estimator of \( \theta \).
The method of maximum likelihood is intuitively appealing—we try to find the value of the parameter that would have most likely produced the data we in fact observed.
Since the natural logarithm function is strictly increasing on \( (0, \infty) \), the maximum value of the likelihood function, if it exists, will occur at the same points as the maximum value of the logarithm of the likelihood function.
The log-likelihood function at \( \bs{x} \in S \) is the function \( \ln L_{\bs{x}} \): \[ \ln L_{\bs{x}}(\theta) = \ln f_\theta(\bs{x}), \quad \theta \in \Theta \] If the maximum value of \( \ln L_{\bs{x}} \) occurs at \( u(\bs{x}) \in \Theta \) for each \( \bs{x} \in S \). Then the statistic \( u(\bs{X}) \) is a maximum likelihood estimator of \( \theta \)
The log-likelihood function is often easier to work with than the likelihood function (typically because the probability density function \(f_\theta(\bs{x})\) has a product structure).
Vector of Parameters
An important special case is when \(\bs{\theta} = (\theta_1, \theta_2, \ldots, \theta_k)\) is a vector of \(k\) real parameters, so that \(\Theta \subseteq \R^k\). In this case, the maximum likelihood problem is to maximize a function of several variables. If \(\Theta\) is a continuous set, the methods of calculus can be used. If the maximum value of \(L_\bs{x}\) occurs at a point \(\bs{\theta}\) in the interior of \(\Theta\), then \(L_\bs{x}\) has a local maximum at \(\bs{\theta}\). Therefore, assuming that the likelihood function is differentiable, we can find this point by solving \[ \frac{\partial}{\partial \theta_i} L_\bs{x}(\bs{\theta}) = 0, \quad i \in \{1, 2, \ldots, k\} \] or equivalently \[ \frac{\partial}{\partial \theta_i} \ln L_\bs{x}(\bs{\theta}) = 0, \quad i \in \{1, 2, \ldots, k\} \] On the other hand, the maximum value may occur at a boundary point of \(\Theta\), or may not exist at all.
Random Samples
The most important special case is when the data variables form a random sample from a distribution.
Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample of size \(n\) from the distribution of a random variable \(X\) taking values in \(R\), with probability density function \(g_\theta\) for \(\theta \in \Theta\). Then \(\bs{X}\) takes values in \(S = R^n\), and the likelihood and log-likelihood functions for \( \bs{x} = (x_1, x_2, \ldots, x_n) \in S \) are \begin{align*} L_\bs{x}(\theta) & = \prod_{i=1}^n g_\theta(x_i), \quad \theta \in \Theta \\ \ln L_\bs{x}(\theta) & = \sum_{i=1}^n \ln g_\theta(x_i), \quad \theta \in \Theta \end{align*}
Extending the Method and the Invariance Property
Returning to the general setting, suppose now that \(h\) is a one-to-one function from the parameter space \(\Theta\) onto a set \(\Lambda\). We can view \(\lambda = h(\theta)\) as a new parameter taking values in the space \(\Lambda\), and it is easy to re-parameterize the probability density function with the new parameter. Thus, let \( \hat{f}_\lambda(\bs{x}) = f_{h^{-1}(\lambda)}(\bs{x})\) for \( \bs{x} \in S \) and \( \lambda \in \Lambda \). The corresponding likelihood function for \( \bs{x} \in S \) is \[ \hat{L}_\bs{x}(\lambda) = L_\bs{x}\left[h^{-1}(\lambda)\right], \quad \lambda \in \Lambda \] Clearly if \(u(\bs{x}) \in \Theta\) maximizes \(L_\bs{x}\) for \(\bs{x} \in S\). Then \(h\left[u(\bs{x})\right] \in \Lambda\) maximizes \(\hat{L}_\bs{x}\) for \(\bs{x} \in S\). It follows that if \(U\) is a maximum likelihood estimator for \(\theta\), then \(V = h(U)\) is a maximum likelihood estimator for \( \lambda = h(\theta) \).
If the function \(h\) is not one-to-one, the maximum likelihood function for the new parameter \(\lambda = h(\theta)\) is not well defined, because we cannot parameterize the probability density function in terms of \(\lambda\). However, there is a natural generalization of the method.
Suppose that \( h: \Theta \to \Lambda \), and let \( \lambda = h(\theta) \) denote the new parameter. Define the likelihood function for \( \lambda \) at \( \bs{x} \in S\) by \[ \hat{L}_\bs{x}(\lambda) = \max\left\{L_\bs{x}(\theta): \theta \in h^{-1}\{\lambda\} \right\}; \quad \lambda \in \Lambda \] If \( v(\bs{x}) \in \Lambda \) maximizes \( \hat{L}_{\bs{x}} \) for each \( \bs{x} \in S \), then \( V = v(\bs{X}) \) is a maximum likelihood estimator of \( \lambda \).
This definition extends the maximum likelihood method to cases where the probability density function is not completely parameterized by the parameter of interest. The following theorem is known as the invariance property : if we can solve the maximum likelihood problem for \( \theta \) then we can solve the maximum likelihood problem for \( \lambda = h(\theta) \).
In the setting of the previous theorem, if \( U \) is a maximum likelihood estimator of \( \theta \), then \( V = h(U) \) is a maximum likelihood estimator of \( \lambda \).
Proof
As before, if \(u(\bs{x}) \in \Theta\) maximizes \(L_\bs{x}\) for \(\bs{x} \in S\). Then \(h\left[u(\bs{x})\right] \in \Lambda\) maximizes \(\hat{L}_\bs{x}\) for \(\bs{x} \in S\).
Examples and Special Cases
In the following subsections, we will study maximum likelihood estimation for a number of special parametric families of distributions. Recall that if \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from a distribution with mean \(\mu\) and variance \(\sigma^2\), then the method of moments estimators of \(\mu\) and \(\sigma^2\) are, respectively, \begin{align} M & = \frac{1}{n} \sum_{i=1}^n X_i \\ T^2 & = \frac{1}{n} \sum_{i=1}^n (X_i - M)^2 \end{align} Of course, \(M\) is the sample mean, and \(T^2 \) is the biased version of the sample variance. These statistics will also sometimes occur as maximum likelihood estimators. Another statistic that will occur in some of the examples below is \[ M_2 = \frac{1}{n} \sum_{i=1}^n X_i^2 \] the second-order sample mean. As always, be sure to try the derivations yourself before looking at the solutions.
The Bernoulli Distribution
Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample of size \(n\) from the Bernoulli distribution with success parameter \(p \in [0, 1]\). Recall that the Bernoulli probability density function is \[ g(x) = p^x (1 - p)^{1 - x}, \quad x \in \{0, 1\} \] Thus, \(\bs{X}\) is a sequence of independent indicator variables with \(\P(X_i = 1) = p\) for each \(i\). In the usual language of reliability, \(X_i\) is the outcome of trial \(i\), where 1 means success and 0 means failure. Let \(Y = \sum_{i=1}^n X_i\) denote the number of successes, so that the proportion of successes (the sample mean) is \(M = Y / n\). Recall that \(Y\) has the binomial distribution with parameters \(n\) and \(p\).
The sample mean \(M\) is the maximum likelihood estimator of \(p\) on the parameter space \( (0, 1) \).
Proof
Note that \(\ln g(x) = x \ln p + (1 - x) \ln(1 - p)\) for \( x \in \{0, 1\} \) Hence the log-likelihood function at \( \bs{x} = (x_1, x_2, \ldots, x_n) \in \{0, 1\}^n \) is \[ \ln L_{\bs{x}}(p) = \sum_{i=1}^n [x_i \ln p + (1 - x_i) \ln(1 - p)], \quad p \in (0, 1) \] Differentiating with respect to \(p\) and simplifying gives \[ \frac{d}{dp} \ln L_{\bs{x}}(p) = \frac{y}{p} - \frac{n - y}{1 - p} \] where \(y = \sum_{i=1}^n x_i\). Thus, there is a single critical point at \(p = y / n = m\). The second deriviative is \[ \frac{d^2}{d p^2} \ln L_{\bs{x}}(p) = -\frac{y}{p^2} - \frac{n - 1}{(1 - p)^2} \lt 0 \] Hence the log-likelihood function is concave downward and so the maximum occurs at the unique critical point \(m\).
Recall that \(M\) is also the method of moments estimator of \(p\). It's always nice when two different estimation procedures yield the same result. Next let's look at the same problem, but with a much restricted parameter space.
Suppose now that \(p\) takes values in \(\left\{\frac{1}{2}, 1\right\}\). Then the maximum likelihood estimator of \(p\) is the statistic \[ U = \begin{cases} 1, & Y = n\\ \frac{1}{2}, & Y \lt n \end{cases} \]
- \(\E(U) = \begin{cases} 1, & p = 1 \\ \frac{1}{2} + \left(\frac{1}{2}\right)^{n+1}, & p = \frac{1}{2} \end{cases}\)
- \(U\) is positively biased, but is asymptotically unbiased.
- \(\mse(U) = \begin{cases} 0 & p = 1 \\ \left(\frac{1}{2}\right)^{n+2}, & p = \frac{1}{2} \end{cases}\)
- \(U\) is consistent.
Proof
Note that the likelihood function at \( \bs{x} = (x_1, x_2, \ldots, x_n) \in \{0, 1\}^n \) is \(L_{\bs{x}}(p) = p^y (1 - p)^{n-y}\) for \( p \in \left\{\frac{1}{2}, 1\right\} \) where as usual, \(y = \sum_{i=1}^n x_i\). Thus \(L_{\bs{x}}\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^y\). On the other hand, \(L_{\bs{x}}(1) = 0\) if \(y \lt n\) while \(L_{\bs{x}}(1) = 1\) if \(y = n\). Thus, if \(y = n\) the maximum occurs when \(p = 1\) while if \(y \lt n\) the maximum occurs when \(p = \frac{1}{2}\).
- If \(p = 1\) then \(\P(U = 1) = \P(Y = n) = 1\), so trivially \(\E(U) = 1\). If \(p = \frac{1}{2}\), \[ \E(U) = 1 \P(Y = n) + \frac{1}{2} \P(Y \lt n) = 1 \left(\frac{1}{2}\right)^n + \frac{1}{2}\left[1 - \left(\frac{1}{2}\right)^n\right] = \frac{1}{2} + \left(\frac{1}{2}\right)^{n+1} \]
- Note that \( \E(U) \ge p \) and \(\E(U) \to p\) as \(n \to \infty\) both in the case that \(p = 1\) and \(p = \frac{1}{2}\).
- If \( p = 1 \) then \( U = 1 \) with probability 1, so trivially \( \mse(U) = 0 \). If \( p = \frac{1}{2} \), \[ \mse(U) = \left(1 - \frac{1}{2}\right)^2 \P(Y = n) + \left(\frac{1}{2} - \frac{1}{2}\right)^2 \P(Y \lt n) = \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^{n+2}\]
- From (c), \( \mse(U) \to 0 \) as \( n \to \infty \).
Note that the Bernoulli distribution in the last exercise would model a coin that is either fair or two-headed. The last two exercises show that the maximum likelihood estimator of a parameter, like the solution to any maximization problem, depends critically on the domain.
\(U\) is uniformly better than \(M\) on the parameter space \(\left\{\frac{1}{2}, 1\right\}\).
Proof
Recall that \( \mse(M) = \var(M) = p (1 - p) / n \). If \( p = 1 \) then \( \mse(M) = \mse(U) = 0 \) so that both estimators give the correct answer. If \( p = \frac{1}{2} \), \( \mse(U) = \left(\frac{1}{2}\right)^{n+2} \lt \frac{1}{4 n} = \mse(M) \).
Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample of size \(n\) from the Bernoulli distribution with unknown success parameter \(p \in (0, 1)\). Find the maximum likelihood estimator of \(p (1 - p)\), which is the variance of the sampling distribution.
Answer
By the invariance principle, the estimator is \(M (1 - M)\) where \(M\) is the sample mean.
The Geometric Distribution
Recall that the geometric distribution on \(\N_+\) with success parameter \(p \in (0, 1)\) has probability density function \[ g(x) = p (1 - p)^{x-1}, \quad x \in \N_+ \] The geometric distribution governs the trial number of the first success in a sequence of Bernoulli trials.
Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the geometric distribution with unknown parameter \(p \in (0, 1)\). The maximum likelihood estimator of \(p\) is \(U = 1 / M\).
Proof
Note that \( \ln g(x) = \ln p + (x - 1) \ln(1 - p) \) for \( x \in \N_+ \). Hence the log-likelihood function corresponding to the data \( \bs{x} = (x_1, x_2, \ldots, x_n) \in \N_+^n \) is \[ \ln L_\bs{x}(p) = n \ln p + (y - n) \ln(1 - p), \quad p \in (0, 1) \] where \( y = \sum_{i=1}^n x_i \). So \[ \frac{d}{dp} \ln L(p) = \frac{n}{p} - \frac{y - n}{1 - p} \] The derivative is 0 when \( p = n / y = 1 / m \). Finally, \( \frac{d^2}{dp^2} \ln L_\bs{x}(p) = -n / p^2 - (y - n) / (1 - p)^2 \lt 0 \) so the maximum occurs at the critical point.
Recall that \(U\) is also the method of moments estimator of \(p\). It's always reassuring when two different estimation procedures produce the same estimator.
The Negative Binomial Distribution
More generally, the negative binomial distribution on \( \N \) with shape parameter \( k \in (0, \infty) \) and success parameter \( p \in (0, 1) \) has probability density function \[ g(x) = \binom{x + k - 1}{k - 1} p^k (1 - p)^x, \quad x \in \N \] If \( k \) is a positive integer, then this distribution governs the number of failures before the \( k \)th success in a sequence of Bernoulli trials with success parameter \( p \). However, the distribution makes sense for general \( k \in (0, \infty) \). The negative binomial distribution is studied in more detail in the chapter on Bernoulli Trials.
Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample of size \(n\) from the negative binomial distribution on \( \N \) with known shape parameter \( k \) and unknown success parameter \( p \in (0, 1) \). The maximum likelihood estimator of \( p \) is \[ U = \frac{k}{k + M} \]
Proof
Note that \( \ln g(x) = \ln \binom{x + k - 1}{k - 1} + k \ln p + x \ln(1 - p) \) for \( x \in \N \). Hence the log-likelihood function corresponding to \( \bs{x} = (x_1, x_2, \ldots, x_n) \in \N^n \) is \[ \ln L_\bs{x}(p) = n k \ln p + y \ln(1 - p) + C, \quad p \in (0, 1) \] where \( y = \sum_{i=1}^n x_i \) and \( C = \sum_{i=1}^n \ln \binom{x_i + k - 1}{k - 1} \). Hence \[ \frac{d}{dp} \ln L_\bs{x}(p) = \frac{n k}{p} - \frac{y}{1 - p} \] The derivative is 0 when \( p = n k / (n k + y) = k / (k + m) \) where as usual, \( m = y / n \). Finally, \( \frac{d^2}{dp^2} \ln L_\bs{x}(p) = - n k / p^2 - y / (1 - p)^2 \lt 0 \), so the maximum occurs at the critical point.
Once again, this is the same as the method of moments estimator of \( p \) with \( k \) known.
The Poisson Distribution
Recall that the
Poisson distribution
with parameter \(r \gt 0\) has probability density function \[ g(x) = e^{-r} \frac{r^x}{x!}, \quad x \in \N \] The Poisson distribution is named for Simeon Poisson and is widely used to model the number of random
points
in a region of time or space. The parameter \( r\) is proportional to the size of the region. The Poisson distribution is studied in more detail in the chapter on the Poisson process.
Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the Poisson distribution with unknown parameter \(r \in (0, \infty)\). The maximum likelihood estimator of \(r\) is the sample mean \(M\).
Proof
Note that \( \ln g(x) = -r + x \ln r - \ln(x!) \) for \( x \in \N \). Hence the log-likelihood function corresponding to \( \bs{x} = (x_1, x_2, \ldots, x_n) \in \N^n\) is \[ \ln L_\bs{x}(r) = -n r + y \ln r - C, \quad r \in (0, \infty) \] where \( y = \sum_{i=1}^n x_i \) and \( C = \sum_{i=1}^n \ln(x_i!) \). Hence \( \frac{d}{dr} \ln L_\bs{x}(r) = -n + y / r \). The derivative is 0 when \( r = y / n = m \). Finally, \( \frac{d^2}{dr^2} \ln L_\bs{x}(r) = -y / r^2 \lt 0 \), so the maximum occurs at the critical point.
Recall that for the Poisson distribution, the parameter \(r\) is both the mean and the variance. Thus \(M\) is also the method of moments estimator of \(r\). We showed in the introductory section that \(M\) has smaller mean square error than \(S^2\), although both are unbiased.
Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the Poisson distribution with parameter \(r \in (0, \infty)\), and let \(p = \P(X = 0) = e^{-r}\). Find the maximum likelihood estimator of \(p\) in two ways:
- Directly, by finding the likelihood function corresponding to the parameter \(p\).
- By using the result of the last exercise and the invariance property.
Answer
\(e^{-M}\) where \(M\) is the sample mean.
The Normal Distribution
Recall that the normal distribution with mean \(\mu\) and variance \(\sigma^2\) has probability density function \[ g(x) = \frac{1}{\sqrt{2 \, \pi} \sigma} \exp \left[-\frac{1}{2} \left(\frac{x - \mu}{\sigma}\right)^2\right], \quad x \in \R \] The normal distribution is often used to model physical quantities subject to small, random errors, and is studied in more detail in the chapter on Special Distributions
Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the normal distribution with unknown mean \(\mu \in \R\) and variance \(\sigma^2 \in (0, \infty)\). The maximum likelihood estimators of \(\mu\) and \(\sigma^2\) are \(M\) and \(T^2\), respectively.
Proof
Note that \[ \ln g(x) = -\frac{1}{2} \ln(2 \pi) - \frac{1}{2} \ln(\sigma^2) - \frac{1}{2 \sigma^2} (x - \mu)^2, \quad x \in \R \] Hence the log-likelihood function corresponding to the data \( \bs{x} = (x_1, x_2, \ldots, x_n) \in \R^n \) is \[ \ln L_\bs{x}(\mu, \sigma^2) = -\frac{n}{2} \ln(2 \pi) - \frac{n}{2} \ln(\sigma^2) - \frac{1}{2 \sigma^2} \sum_{i=1}^n (x_i - \mu)^2, \quad (\mu, \sigma^2) \in \R \times (0, \infty) \] Taking partial derivatives gives \begin{align*} \frac{\partial}{\partial \mu} \ln L_\bs{x}(\mu, \sigma^2) &= \frac{1}{\sigma^2} \sum_{i=1}^n (x_i - \mu) = \frac{1}{\sigma^2}\left(\sum_{i=1}^n x_i - n \mu\right) \\ \frac{\partial}{\partial \sigma^2} \ln L_\bs{x}(\mu, \sigma^2) &= -\frac{n}{2 \sigma^2} + \frac{1}{2 \sigma^4} \sum_{i=1}^n (x_i - \mu)^2 \end{align*} The partial derivatives are 0 when \( \mu = \frac{1}{n} \sum_{i=1}^n x_i\) and \( \sigma^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \mu)^2 \). Hence the unique critical point is \( (m, t^2) \). Finally, with a bit more calculus, the second partial derivatives evaluated at the critical point are \[ \frac{\partial^2}{\partial \mu^2} \ln L_\bs{x}(m, t^2) = -n / t^2, \; \frac{\partial^2}{\partial \mu \partial \sigma^2} \ln L_\bs{x}(m, t^2) = 0, \; \frac{\partial^2}{\partial (\sigma^2)^2} \ln L_\bs{x}(m, t^2) = -n / t^4\] Hence the second derivative matrix at the critical point is negative definite and so the maximum occurs at the critical point.
Of course, \(M\) and \(T^2\) are also the method of moments estimators of \(\mu\) and \(\sigma^2\), respectively.
Run the Normal estimation experiment 1000 times for several values of the sample size \(n\), the mean \(\mu\), and the variance \(\sigma^2\). For the parameter \(\sigma^2\), compare the maximum likelihood estimator \(T^2\) with the standard sample variance \(S^2\). Which estimator seems to work better in terms of mean square error?
Suppose again that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the normal distribution with unknown mean \(\mu \in \R\) and unknown variance \(\sigma^2 \in (0, \infty)\). Find the maximum likelihood estimator of \(\mu^2 + \sigma^2\), which is the second moment about 0 for the sampling distribution.
Answer
By the invariance principle, the estimator is \(M^2 + T^2\) where \(M\) is the sample mean and \(T^2\) is the (biased version of the) sample variance.
The Gamma Distribution
Recall that the gamma distribution with shape parameter \(k \gt 0\) and scale parameter \(b \gt 0\) has probability density function \[ g(x) = \frac{1}{\Gamma(k) \, b^k} x^{k-1} e^{-x / b}, \quad 0 \lt x \lt \infty \] The gamma distribution is often used to model random times and certain other types of positive random variables, and is studied in more detail in the chapter on Special Distributions
Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the gamma distribution with known shape parameter \(k\) and unknown scale parameter \(b \in (0, \infty)\). The maximum likelihood estimator of \(b\) is \(V_k = \frac{1}{k} M\).
Proof
Note that for \( x \in (0, \infty) \), \[ \ln g(x) = -\ln \Gamma(k) - k \ln b + (k - 1) \ln x - \frac{x}{b} \] and hence the log-likelihood function corresponding to the data \( \bs{x} = (x_1, x_2, \ldots, x_n) \in (0, \infty)^n \) is \[ \ln L_\bs{x}(b) = - n k \ln b - \frac{y}{b} + C, \quad b \in (0, \infty)\] where \( y = \sum_{i=1}^n x_i \) and \( C = -n \ln \Gamma(k) + (k - 1) \sum_{i=1}^n \ln x_i \). It follows that \[ \frac{d}{d b} \ln L_\bs{x}(b) = -\frac{n k}{b} + \frac{y}{b^2} \] The derivative is 0 when \( b = y / n k = 1 / k m \). Finally, \( \frac{d^2}{db^2} \ln L_\bs{x}(b) = n k / b^2 - 2 y / b^3 \). At the critical point \( b = y / n k \), the second derivative is \(-(n k)^3 / y^2 \lt 0\) so the maximum occurs at the critical point.
Recall that \(V_k\) is also the method of moments estimator of \(b\) when \(k\) is known. But when \(k\) is unknown, the method of moments estimator of \(b\) is \(V = \frac{T^2}{M}\).
Run the gamma estimation experiment 1000 times for several values of the sample size \(n\), shape parameter \(k\), and scale parameter \(b\). In each case, compare the method of moments estimator \(V\) of \(b\) when \(k\) is unknown with the method of moments and maximum likelihood estimator \(V_k\) of \(b\) when \(k\) is known. Which estimator seems to work better in terms of mean square error?
The Beta Distribution
Recall that the beta distribution with left parameter \(a \in (0, \infty)\) and right parameter \(b = 1\) has probability density function \[ g(x) = a x^{a-1}, \quad x \in (0, 1) \] The beta distribution is often used to model random proportions and other random variables that take values in bounded intervals. It is studied in more detail in the chapter on Special Distribution
Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the beta distribution with unknown left parameter \(a \in (0, \infty)\) and right parameter \(b = 1\). The maximum likelihood estimator of \(a\) is \[ W = - \frac{n}{\sum_{i=1}^n \ln X_i} = -\frac{n}{\ln(X_1 X_2 \cdots X_n)} \]
Proof
Note that \( \ln g(x) = \ln a + (a - 1) \ln x \) for \( x \in (0, \infty) \) Hence the log-likelihood function corresponding to the data \( \bs{x} = (x_1, x_2, \ldots, x_n) \in (0, \infty)^n \) is \[ \ln L_\bs{x}(a) = n \ln a + (a - 1) \sum_{i=1}^n \ln x_i, \quad a \in (0, \infty) \] Therefore \( \frac{d}{da} \ln L_\bs{x}(a) = n / a + \sum_{i=1}^n \ln x_i \). The derivative is 0 when \( a = -n \big/ \sum_{i=1}^n \ln x_i \). Finally, \( \frac{d^2}{da^2} \ln L_\bs{x}(a) = -n / a^2 \lt 0 \), so the maximum occurs at the critical point.
Recall that when \(b = 1\), the method of moments estimator of \(a\) is \(U_1 = M \big/ (1 - M)\), but when \(b \in (0, \infty)\) is also unknown, the method of moments estimator of \(a\) is \(U = M (M - M_2) \big/ (M_2 - M^2)\). When \(b = 1\), which estimator is better, the method of moments estimator or the maximum likelihood estimator?
In the beta estimation experiment, set \(b = 1\). Run the experiment 1000 times for several values of the sample size \(n\) and the parameter \(a\). In each case, compare the estimators \(U\), \(U_1\) and \(W\). Which estimator seems to work better in terms of mean square error?
Finally, note that \( 1 / W \) is the sample mean for a random sample of size \( n \) from the distribution of \( -\ln X \). This distribution is the exponential distribution with rate \( a \).
The Pareto Distribution
Recall that the Pareto distribution with shape parameter \(a \gt 0\) and scale parameter \(b \gt 0\) has probability density function \[ g(x) = \frac{a b^a}{x^{a+1}}, \quad b \le x \lt \infty \] The Pareto distribution, named for Vilfredo Pareto, is a heavy-tailed distribution often used to model income and certain other types of random variables. It is studied in more detail in the chapter on Special Distribution.
Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the Pareto distribution with unknown shape parameter \(a \in (0, \infty)\) and scale parameter \(b \in (0, \infty)\). The maximum likelihood estimator of \( b \) is \( X_{(1)} = \min\{X_1, X_2, \ldots, X_n\} \), the first order statistic. The maximum likelihood estimator of \( a \) is \[ U = \frac{n}{\sum_{i=1}^n \ln X_i - n \ln X_{(1)}} = \frac{n}{\sum_{i=1}^n \left(\ln X_i - \ln X_{(1)}\right)}\]
Proof
Note that \( \ln g(x) = \ln a + a \ln b - (a + 1) \ln x \) for \( x \in [b, \infty) \). Hence the log-likelihood function corresponding to the data \( \bs{x} = (x_1, x_2, \ldots, x_n) \) is \[ \ln L_\bs{x}(a, b) = n \ln a + n a \ln b - (a + 1) \sum_{i=1}^n \ln x_i; \quad 0 \lt a \lt \infty, \, 0 \lt b \le x_i \text{ for each } i \in \{1, 2, \ldots, n\} \] Equivalently, the domain is \( 0 \lt a \lt \infty \) and \( 0 \lt b \le x_{(1)} \). Note that \( \ln L_{\bs{x}}(a, b) \) is increasing in \( b \) for each \( a \), and hence is maximized when \( b = x_{(1)} \) for each \( a \). Next, \[ \frac{d}{d a} \ln L_{\bs{x}}\left(a, x_{(1)}\right) = \frac{n}{a} + n \ln x_{(1)} - \sum_{i=1}^n \ln x_i \] The derivative is 0 when \( a = n \big/ \left(\sum_{i=1}^n \ln x_i - n \ln x_{(1)}\right) \). Finally, \( \frac{d^2}{da^2} \ln L_\bs{x}\left(a, x_{(1)}\right) = -n / a^2 \lt 0 \), so the maximum occurs at the critical point.
Recall that if \(a \gt 2\), the method of moments estimators of \(a\) and \( b \) are \[ 1 + \sqrt{\frac{M_2}{M_2 - M^2}}, \; \frac{M_2}{M} \left(1 - \sqrt{\frac{M_2 - M^2}{M_2}}\right)\]
Open the the Pareto estimation experiment. Run the experiment 1000 times for several values of the sample size \(n\) and the parameters \(a\) and \( b \). Compare the method of moments and maximum likelihood estimators. Which estimators seem to work better in terms of bias and mean square error?
Often the scale parameter in the Pareto distribution is known.
Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the Pareto distribution with unknown shape parameter \(a \in (0, \infty)\) and known scale parameter \(b \in (0, \infty)\). The maximum likelihood estimator of \( a \) is \[ U = \frac{n}{\sum_{i=1}^n \ln X_i - n \ln b} = \frac{n}{\sum_{i=1}^n \left(\ln X_i - \ln b \right)}\]
Proof
Modifying the previous proof, the log-likelihood function corresponding to the data \( \bs{x} = (x_1, x_2, \ldots, x_n) \) is \[ \ln L_\bs{x}(a) = n \ln a + n a \ln b - (a + 1) \sum_{i=1}^n \ln x_i, \quad 0 \lt a \lt \infty \] The derivative is \[ \frac{d}{d a} \ln L_{\bs{x}}(a) = \frac{n}{a} + n \ln b - \sum_{i=1}^n \ln x_i \] The derivative is 0 when \( a = n \big/ \left(\sum_{i=1}^n \ln x_i - n \ln b\right) \). Finally, \( \frac{d^2}{da^2} \ln L_\bs{x}(a) = -n / a^2 \lt 0 \), so the maximum occurs at the critical point.
Uniform Distributions
In this section we will study estimation problems related to the uniform distribution that are a good source of insight and counterexamples. In a sense, our first estimation problem is the continuous analogue of an estimation problem studied in the section on Order Statistics in the chapter Finite Sampling Models. Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the uniform distribution on the interval \([0, h]\), where \(h \in (0, \infty)\) is an unknown parameter. Thus, the sampling distribution has probability density function \[ g(x) = \frac{1}{h}, \quad x \in [0, h] \] First let's review results from the last section.
The method of moments estimator of \(h\) is \(U = 2 M\). The estimator \(U\) satisfies the following properties:
- \(U\) is unbiased.
- \(\var(U) = \frac{h^2}{3 n}\) so \(U\) is consistent.
Now let's find the maximum likelihood estimator
The maximum likelihood estimator of \(h\) is \(X_{(n)} = \max\{X_1, X_2, \ldots, X_n\}\), the \(n\)th order statistic. The estimator \(X_{(n)}\) satisfies the following properties:
- \(\E\left(X_{(n)}\right) = \frac{n}{n + 1} h\)
- \(\bias\left(X_{(n)}\right) = -\frac{h}{n+1}\) so that \(X_{(n)}\) is negatively biased but asymptotically unbiased.
- \(\var\left(X_{(n)}\right) = \frac{n}{(n+2)(n+1)^2} h^2\)
- \(\mse\left(X_{(n)}\right) = \frac{2}{(n+1)(n+2)}h^2\) so that \(X_{(n)}\) is consistent.
Proof
The likelihood function corresponding to the data \( \bs{x} = (x_1, x_2, \ldots, x_n) \) is \( L_\bs{x}(h) = 1 / h^n \) for \( h \ge x_i \) for each \( i \in \{1, 2, \ldots n\} \). The domain is equivalent to \( h \ge x_{(n)} \). The function \( h \mapsto 1 / h^n \) is decreasing, and so the maximum occurs at the smallest value, namely \( x_{(n)} \). Parts (a) and (c) are restatements of results from the section on order statistics. Parts (b) and (d) follow from (a) and (c).
Since the expected value of \(X_{(n)}\) is a known multiple of the parameter \(h\), we can easily construct an unbiased estimator.
Let \(V = \frac{n+1}{n} X_{(n)}\). The estimator \(V\) satisfies the following properties:
- \(V\) is unbiased.
- \(\var(V) = \frac{h^2}{n(n + 2)}\) so that \(V\) is consistent.
- The asymptotic relative efficiency of \(V\) to \(U\) is infinite.
Proof
Parts (a) and (b) follow from the previous result and basic properties of the expected value and variance. For part (c), \[ \frac{\var(U)}{\var(V)} = \frac{h^2 / 3 n}{h^2 / n (n + 2)} = \frac{n + 2}{3} \to \infty \text{ as } n \to \infty \]
The last part shows that the unbiased version \(V\) of the maximum likelihood estimator is a much better estimator than the method of moments estimator \(U\). In fact, an estimator such as \(V\), whose mean square error decreases on the order of \(\frac{1}{n^2}\), is called super efficient . Now, having found a really good estimator, let's see if we can find a really bad one. A natural candidate is an estimator based on \(X_{(1)} = \min\{X_1, X_2, \ldots, X_n\}\), the first order statistic. The next result will make the computations very easy.
The sample \(\bs{X} = (X_1, X_2, \ldots, X_n)\) satisfies the following properties:
- \(h - X_i\) is uniformly distributed on \([0, h]\) for each \(i\).
- \((h - X_1, h - X_2, \ldots, h - X_n)\) is also a random sample from the uniform distribution on \([0, h]\).
- \(X_{(1)}\) has the same distribution as \(h - X_{(n)}\).
Proof
- This is a simple consequence of the fact that uniform distributions are preserved under linear transformations on the random variable.
- This follows from (a) and that the fact that if \( \bs{X} \) is a sequence of independent variables, then so is \( (h - X_1, h - X_2, \ldots, h - X_n) \).
- From part (b), \( X_{(1)} = \min\{X_1, X_2, \ldots, X_n\} \) has the same distribution as \( \min\{h - X_1, h - X_2, \ldots, h - X_n\} = h - \max\{X_1, X_2, \ldots, X_n\} = h - X_{(n)} \).
Now we can construct our really bad estimator.
Let \(W = (n + 1)X_{(1)}\). Then
- \( W \) is an unbiased estimator of \( h \).
- \(\var(W) = \frac{n}{n+2} h^2\), so \(W\) is not even consistent.
Proof
These results follow from the ones above:
- \( \E(X_{(1)}) = h - \E(X_{(n)}) = h - \frac{n}{n + 1} h = \frac{1}{n + 1} h \) and hence \( \E(W) = h \).
- \( \var(W) = (n + 1)^2 \var(X_{(1)}) = (n + 1)^2 \var(h - X_{(n)}) = (n + 1)^2 \frac{n}{(n + 1)^2 (n + 2)} h^2 = \frac{n}{n + 2} h^2\).
Run the uniform estimation experiment 1000 times for several values of the sample size \(n\) and the parameter \(a\). In each case, compare the empirical bias and mean square error of the estimators with their theoretical values. Rank the estimators in terms of empirical mean square error.
Our next series of exercises will show that the maximum likelihood estimator is not necessarily unique. Suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a random sample from the uniform distribution on the interval \([a, a + 1]\), where \(a \in \R\) is an unknown parameter. Thus, the sampling distribution has probability density function \[ g(x) = 1, \quad a \le x \le a + 1 \] As usual, let's first review the method of moments estimator.
The method of moments estimator of \(a\) is \(U = M - \frac{1}{2}\). The estimator \(U\) satisfies the following properties:
- \(U\) is unbiased.
- \(\var(U) = \frac{1}{12 n}\) so \(U\) is consistent.
However, as promised, there is not a unique maximum likelihood estimatr.
Any statistic \(V \in \left[X_{(n)} - 1, X_{(1)}\right]\) is a maximum likelihood estimator of \(a\).
Proof
The likelihood function corresponding to the data \( \bs{x} = (x_1, x_2, \ldots, x_n\} \) is \( L_\bs{x}(a) = 1 \) for \( a \le x_i \le a + 1 \) and \( i \in \{1, 2, \ldots, n\} \). The domain is equivalent to \( a \le x_{(1)} \) and \( a \ge x_{(n)} - 1 \). Since the likelihood function is constant on this domain, the result follows.
For completeness, let's consider the full estimation problem. Suppose that \( \bs{X} = (X_1, X_2, \ldots, X_n) \) is a random sample of size \( n \) from the uniform distribution on \( [a, a + h] \) where \( a \in \R \) and \( h \in (0, \infty) \) are both unknown. Here's the result from the last section:
Let \( U \) and \( V \) denote the method of moments estimators of \( a \) and \( h \), respectively. Then \[ U = 2 M - \sqrt{3} T, \quad V = 2 \sqrt{3} T \] where \( M = \frac{1}{n} \sum_{i=1}^n X_i \) is the sample mean, and \( T = \frac{1}{n} \sum_{i=1}^n (X_i - M)^2 \) is the biased version of the sample variance.
It should come as no surprise at this point that the maximum likelihood estimators are functions of the largest and smallest order statistics.
The maximum likelihood estimators or \( a \) and \( h \) are \( U = X_{(1)} \) and \( V = X_{(n)} - X_{(1)} \), respectively.
- \( E(U) = a + \frac{h}{n + 1} \) so \( U \) is positively biased and asymptotically unbiased.
- \( E(V) = h \frac{n - 1}{n + 1} \) so \( V \) is negatively biased and asymptotically unbiased.
- \( \var(U) = h^2 \frac{n}{(n + 1)^2 (n + 2)} \) so \( U \) is consistent.
- \( \var(V) = h^2 \frac{2(n - 1)}{(n + 1)^2(n + 2)} \) so \( V \) is consistent.
Proof
The likelihood function corresponding to the data \( \bs{x} = (x_1, x_2, \ldots, x_n) \) is \( L_\bs{x}(a, h) = \frac{1}{h^n} \) for \( a \le x_i \le a + h \) and \( i \in \{1, 2, \ldots, n\} \). The domain is equivalent to \( a \le x_{(1)} \) and \( a + h \ge x_{(n)} \). Since the likelihood function depends only on \( h \) in this domain and is decreasing, the maximum occurs when \( a = x_{(1)} \) and \( h = x_{(n)} - x_{(1)} \). Parts (a)–(d) follow from standard results for the order statistics from the uniform distribution.
The Hypergeometric Model
In all of our previous examples, the sequence of observed random variables \( \bs{X} = (X_1, X_2, \ldots, X_n) \) is a random sample from a distribution. However, maximum likelihood is a very general method that does not require the observation variables to be independent or identically distributed. In the hypergeometric model , we have a population of \( N \) objects with \( r \) of the objects type 1 and the remaining \( N - r \) objects type 0 . The population size \( N \), is a positive integer. The type 1 size \( r \), is a nonnegative integer with \( r \le N \). These are the basic parameters, and typically one or both is unknown. Here are some typical examples:
- The objects are devices, classified as good or defective .
- The objects are persons, classified as female or male .
- The objects are voters, classified as for or against a particular candidate.
- The objects are wildlife or a particular type, either tagged or untagged .
We sample \( n \) objects from the population at random, without replacement. Let \( X_i \) be the type of the \( i \)th object selected, so that our sequence of observed variables is \( \bs{X} = (X_1, X_2, \ldots, X_n) \). The variables are identically distributed indicator variables, with \( P(X_i = 1) = r / N \) for each \( i \in \{1, 2, \ldots, n\} \), but are dependent since the sampling is without replacement. The number of type 1 objects in the sample is \( Y = \sum_{i=1}^n X_i \). This statistic has the hypergeometric distribution with parameter \( N \), \( r \), and \( n \), and has probability density function given by \[ P(Y = y) = \frac{\binom{r}{y} \binom{N - r}{n - y}}{\binom{N}{n}} = \binom{n}{y} \frac{r^{(y)} (N - r)^{(n - y)}}{N^{(n)}}, \quad y \in \{\max\{0, N - n + r\}, \ldots, \min\{n, r\}\} \] Recall the falling power notation: \( x^{(k)} = x (x - 1) \cdots (x - k + 1) \) for \( x \in \R \) and \( k \in \N \). The hypergeometric model is studied in more detail in the chapter on Finite Sampling Models.
As above, let \( \bs{X} = (X_1, X_2, \ldots, X_n) \) be the observed variables in the hypergeometric model with parameters \( N \) and \( r \). Then
- The maximum likelihood estimator of \( r \) with \( N \) known is \( U = \lfloor N M \rfloor = \lfloor N Y / n \rfloor \).
- The maximum likelihood estimator of \( N \) with \( r \) known is \( V = \lfloor r / M \rfloor = \lfloor r n / Y \rfloor \) if \( Y \gt 0 \).
Proof
By a simple application of the multiplication rule, the PDF \( f \) of \( \bs{X} \) is \[ f(\bs{x}) = \frac{r^{(y)} (N - r)^{(n - y)}}{N^{(n)}}, \quad \bs{x} = (x_1, x_2, \ldots, x_n) \in \{0, 1\}^n \] where \( y = \sum_{i=1}^n x_i \).
- With \( N \) known, the likelihood function corresponding to the data \(\bs{x} = (x_1, x_2, \ldots, x_n) \in \{0, 1\}^n\) is \[ L_{\bs{x}}(r) = \frac{r^{(y)} (N - r)^{(n - y)}}{N^{(n)}}, \quad r \in \{y, \ldots, \min\{n, y + N - n\}\} \] After some algebra, \( L_{\bs{x}}(r - 1) \lt L_{\bs{x}}(r) \) if and only if \((r - y)(N - r + 1) \lt r (N - r - n + y + 1)\) if and only if \( r \lt N y / n \). So the maximum of \( L_{\bs{x}}(r) \) occurs when \( r = \lfloor N y / n \rfloor \).
- Similarly, with \( r \) known, the likelihood function corresponding to the data \(\bs{x} = (x_1, x_2, \ldots, x_n) \in \{0, 1\}^n\) is \[ L_{\bs{x}}(N) = \frac{r^{(y)} (N - r)^{(n - y)}}{N^{(n)}}, \quad N \in \{\max\{r, n\}, \ldots\} \] After some algebra, \( L_{\bs{x}}(N - 1) \lt L_{\bs{x}}(N) \) if and only if \((N - r - n + y) / (N - n) \lt (N - r) / N\) if and only if \( N \lt r n / y \) (assuming \( y \gt 0 \)). So the maximum of \( L_{\bs{x}}(r) \) occurs when \( N = \lfloor r n / y \rfloor \).
In the reliability example (1), we might typically know \( N \) and would be interested in estimating \( r \). In the wildlife example (4), we would typically know \( r \) and would be interested in estimating \( N \). This example is known as the capture-recapture model.
Clearly there is a close relationship between the hypergeometric model and the Bernoulli trials model above. In fact, if the sampling is with replacement, the Bernoulli trials model with \( p = r / N \) would apply rather than the hypergeometric model. In addition, if the population size \( N \) is large compared to the sample size \( n \), the hypergeometric model is well approximated by the Bernoulli trials model, again with \( p = r / N \).


11.3: Freud and the Psychodynamic Perspective
Learning Objectives
- Describe the assumptions of the psychodynamic perspective on personality development
- Define and describe the nature and function of the id, ego, and superego
- Define and describe the defense mechanisms
- Define and describe the psychosexual stages of personality development
Sigmund Freud (1856–1939) is probably the most controversial and misunderstood psychological theorist. When reading Freud’s theories, it is important to remember that he was a medical doctor, not a psychologist. There was no such thing as a degree in psychology at the time that he received his education, which can help us understand some of the controversy over his theories today. However, Freud was the first to systematically study and theorize the workings of the unconscious mind in the manner that we associate with modern psychology.
In the early years of his career, Freud worked with Josef Breuer, a Viennese physician. During this time, Freud became intrigued by the story of one of Breuer’s patients, Bertha Pappenheim, who was referred to by the pseudonym Anna O. (Launer, 2005). Anna O. had been caring for her dying father when she began to experience symptoms such as partial paralysis, headaches, blurred vision, amnesia, and hallucinations (Launer, 2005). In Freud’s day, these symptoms were commonly referred to as hysteria. Anna O. turned to Breuer for help. He spent \(2\) years (1880–1882) treating Anna O. and discovered that allowing her to talk about her experiences seemed to bring some relief of her symptoms. Anna O. called his treatment the “talking cure” (Launer, 2005). Despite the fact the Freud never met Anna O., her story served as the basis for the 1895 book, Studies on Hysteria , which he co-authored with Breuer. Based on Breuer’s description of Anna O.’s treatment, Freud concluded that hysteria was the result of sexual abuse in childhood and that these traumatic experiences had been hidden from consciousness. Breuer disagreed with Freud, which soon ended their work together. However, Freud continued to work to refine talk therapy and build his theory on personality.
Levels of Consciousness
To explain the concept of conscious versus unconscious experience, Freud compared the mind to an iceberg (See figure \(\PageIndex{1}\)). He said that only about one-tenth of our mind is conscious , and the rest of our mind is unconscious . Our unconscious refers to that mental activity of which we are unaware and are unable to access (Freud, 1923). According to Freud, unacceptable urges and desires are kept in our unconscious through a process called repression. For example, we sometimes say things that we don’t intend to say by unintentionally substituting another word for the one we meant. You’ve probably heard of a Freudian slip, the term used to describe this. Freud suggested that slips of the tongue are actually sexual or aggressive urges, accidentally slipping out of our unconscious. Speech errors such as this are quite common. Seeing them as a reflection of unconscious desires, linguists today have found that slips of the tongue tend to occur when we are tired, nervous, or not at our optimal level of cognitive functioning (Motley, 2002).
According to Freud, our personality develops from a conflict between two forces: our biological aggressive and pleasure-seeking drives versus our internal (socialized) control over these drives. Our personality is the result of our efforts to balance these two competing forces. Freud suggested that we can understand this by imagining three interacting systems within our minds. He called them the id, ego, and superego. See figure below:
The unconscious id contains our most primitive drives or urges, and is present from birth. It directs impulses for hunger, thirst, and sex. Freud believed that the id operates on what he called the “pleasure principle,” in which the id seeks immediate gratification. Through social interactions with parents and others in a child’s environment, the ego and superego develop to help control the id. The superego develops as a child interacts with others, learning the social rules for right and wrong. The superego acts as our conscience; it is our moral compass that tells us how we should behave. It strives for perfection and judges our behavior, leading to feelings of pride or—when we fall short of the ideal—feelings of guilt. In contrast to the instinctual id and the rule-based superego, the ego is the rational part of our personality. It’s what Freud considered to be the self, and it is the part of our personality that is seen by others. Its job is to balance the demands of the id and superego in the context of reality; thus, it operates on what Freud called the “reality principle.” The ego helps the id satisfy its desires in a realistic way.
The id and superego are in constant conflict, because the id wants instant gratification regardless of the consequences, but the superego tells us that we must behave in socially acceptable ways. Thus, the ego’s job is to find the middle ground. It helps satisfy the id’s desires in a rational way that will not lead us to feelings of guilt. According to Freud, a person who has a strong ego, which can balance the demands of the id and the superego, has a healthy personality. Freud maintained that imbalances in the system can lead to neurosis (a tendency to experience negative emotions), anxiety disorders, or unhealthy behaviors. For example, a person who is dominated by their id might be narcissistic and impulsive. A person with a dominant superego might be controlled by feelings of guilt and deny themselves even socially acceptable pleasures; conversely, if the superego is weak or absent, a person might become a psychopath. An overly dominant superego might be seen in an over-controlled individual whose rational grasp on reality is so strong that they are unaware of their emotional needs, or, in a neurotic who is overly defensive (overusing ego defense mechanisms).
Defense Mechanisms
Freud believed that feelings of anxiety result from the ego’s inability to mediate the conflict between the id and superego. When this happens, Freud believed that the ego seeks to restore balance through various protective measures known as defense mechanisms (See figure \(\PageIndex{3}\)). When certain events, feelings, or yearnings cause an individual anxiety, the individual wishes to reduce that anxiety. To do that, the individual’s unconscious mind uses ego defense mechanisms , unconscious protective behaviors that aim to reduce anxiety. The ego, usually conscious, resorts to unconscious strivings to protect the ego from being overwhelmed by anxiety. When we use defense mechanisms, we are unaware that we are using them. Further, they operate in various ways that distort reality. According to Freud, we all use ego defense mechanisms.
While everyone uses defense mechanisms, Freud believed that overuse of them may be problematic. For example, let’s say Joe Smith is a high school football player. Deep down, Joe feels sexually attracted to males. His conscious belief is that being gay is immoral and that if he were gay, his family would disown him and he would be ostracized by his peers. Therefore, there is a conflict between his conscious beliefs (being gay is wrong and will result in being ostracized) and his unconscious urges (attraction to males). The idea that he might be gay causes Joe to have feelings of anxiety. How can he decrease his anxiety?  This way, Joe’s unconscious impulses are further submerged.
There are several different types of defense mechanisms. For instance, in repression, anxiety-causing memories from consciousness are blocked. As an analogy, let’s say your car is making a strange noise, but because you do not have the money to get it fixed, you just turn up the radio so that you no longer hear the strange noise. Eventually you forget about it. Similarly, in the human psyche, if a memory is too overwhelming to deal with, it might be repressed and thus removed from conscious awareness (Freud, 1920). This repressed memory might cause symptoms in other areas.
Another defense mechanism is reaction formation , in which someone expresses feelings, thoughts, and behaviors opposite to their inclinations. In the above example, Joe made fun of a homosexual peer while himself being attracted to males. In regression , an individual acts much younger than their age. For example, a four-year-old child who resents the arrival of a newborn sibling may act like a baby and revert to drinking out of a bottle. In projection , a person refuses to acknowledge her own unconscious feelings and instead sees those feelings in someone else. Other defense mechanisms include rationalization , displacement , and sublimation .
Stages of Psychosexual Development
Freud believed that personality develops during early childhood: Childhood experiences shape our personalities as well as our behavior as adults. He asserted that we develop via a series of stages during childhood. Each of us must pass through these childhood stages, and if we do not have the proper nurturing and parenting during a stage, we will be stuck, or fixated, in that stage, even as adults.
In each psychosexual stage of development , the child’s pleasure-seeking urges, coming from the id, are focused on a different area of the body, called an erogenous zone. The stages are oral, anal, phallic, latency, and genital (Table \(\PageIndex{1}\)).
Freud’s psychosexual development theory is quite controversial. To understand the origins of the theory, it is helpful to be familiar with the political, social, and cultural influences of Freud’s day in Vienna at the turn of the 20th century.  Given that sex was a taboo topic, Freud assumed that negative emotional states (neuroses) stemmed from suppression of unconscious sexual and aggressive urges. For Freud, his own recollections and interpretations of patients’ experiences and dreams were sufficient proof that psychosexual stages were universal events in early childhood.
| Stage | Age (years) | Erogenous Zone | Major Conflict | Adult Fixation Example |
|---|---|---|---|---|
| Oral | 0–1 | Mouth | Weaning off breast or bottle | Smoking, overeating |
| Anal | 1–3 | Anus | Toilet training | Neatness, messiness |
| Phallic | 3–6 | Genitals | Oedipus/Electra complex | Vanity, overambition |
| Latency | 6–12 | None | None | None |
| Genital | 12+ | Genitals | None | None |
Oral Stage
In the oral stage (birth to \(1\) year), pleasure is focused on the mouth. Eating and the pleasure derived from sucking (nipples, pacifiers, and thumbs) play a large part in a baby’s first year of life. At around \(1\) year of age, babies are weaned from the bottle or breast, and this process can create conflict if not handled properly by caregivers. According to Freud, an adult who smokes, drinks, overeats, or bites her nails is fixated in the oral stage of her psychosexual development; she may have been weaned too early or too late, resulting in these fixation tendencies, all of which seek to ease anxiety.
Anal Stage
After passing through the oral stage, children enter what Freud termed the anal stage (\(1-3\) years). In this stage, children experience pleasure in their bowel and bladder movements, so it makes sense that the conflict in this stage is over toilet training. Freud suggested that success at the anal stage depended on how parents handled toilet training. Parents who offer praise and rewards encourage positive results and can help children feel competent. Parents who are harsh in toilet training can cause a child to become fixated at the anal stage, leading to the development of an anal-retentive personality. The anal-retentive personality is stingy and stubborn, has a compulsive need for order and neatness, and might be considered a perfectionist. If parents are too lenient in toilet training, the child might also become fixated and display an anal-expulsive personality. The anal-expulsive personality is messy, careless, disorganized, and prone to emotional outbursts.
Phallic Stage
Freud’s third stage of psychosexual development is the phallic stage (\(3-6\) years), corresponding to the age when children become aware of their bodies and recognize the differences between boys and girls. The erogenous zone in this stage is the genitals. Conflict arises when the child feels a desire for the opposite-sex parent, and jealousy and hatred toward the same-sex parent. For boys, this is called the Oedipus complex, involving a boy's desire for his mother and his urge to replace his father who is seen as a rival for the mother’s attention. At the same time, the boy is afraid his father will punish him for his feelings, so he experiences castration anxiety . The Oedipus complex is successfully resolved when the boy begins to identify with his father as an indirect way to have the mother. Failure to resolve the Oedipus complex may result in fixation and development of a personality that might be described as vain and overly ambitious.
Girls experience a comparable conflict in the phallic stage—the Electra complex. The Electra complex, while often attributed to Freud, was actually proposed by Freud’s protégé, Carl Jung (Jung & Kerenyi, 1963). A girl desires the attention of her father and wishes to take her mother’s place.  While Freud initially embraced the Electra complex as a parallel to the Oedipus complex, he later rejected it, yet it remains as a cornerstone of Freudian theory, thanks in part to academics in the field (Freud, 1931/1968; Scott, 2005).
Latency Period
Following the phallic stage of psychosexual development is a period known as the latency period (\(6\) years to puberty). This period is not considered a stage, because sexual feelings are dormant as children focus on other pursuits, such as school, friendships, hobbies, and sports. Children generally engage in activities with peers of the same sex, which serves to consolidate a child’s gender-role identity.
Genital Stage
The final stage is the genital stage (from puberty on). In this stage, there is a sexual reawakening as the incestuous urges resurface. The young person redirects these urges to other, more socially acceptable partners (who often resemble the other-sex parent). People in this stage have mature sexual interests, which for Freud meant a strong desire for the opposite sex. Individuals who successfully completed the previous stages, reaching the genital stage with no fixations, are said to be well-balanced, healthy adults.
While most of Freud’s ideas have not found support in modern research, we cannot discount the contributions that Freud has made to the field of psychology. It was Freud who pointed out that a large part of our mental life is influenced by the experiences of early childhood and takes place outside of our conscious awareness; his theories paved the way for others.
Summary
Sigmund Freud presented the first comprehensive theory of personality. He was also the first to recognize that much of our mental life takes place outside of our conscious awareness. Freud also proposed three components to our personality: the id, ego, and superego. The job of the ego is to balance the sexual and aggressive drives of the id with the moral ideal of the superego. Freud also said that personality develops through a series of psychosexual stages. In each stage, pleasure focuses on a specific erogenous zone. Failure to resolve a stage can lead one to become fixated in that stage, leading to unhealthy personality traits. Successful resolution of the stages leads to a healthy adult.
Glossary
- anal stage
- psychosexual stage in which children experience pleasure in their bowel and bladder movements
- conscious
- mental activity (thoughts, feelings, and memories) that we can access at any time
- defense mechanism
- unconscious protective behaviors designed to reduce ego anxiety
- displacement
- ego defense mechanism in which a person transfers inappropriate urges or behaviors toward a more acceptable or less threatening target
- ego
- aspect of personality that represents the self, or the part of one’s personality that is visible to others
- genital stage
- psychosexual stage in which the focus is on mature sexual interests
- id
- aspect of personality that consists of our most primitive drives or urges, including impulses for hunger, thirst, and sex
- latency period
- psychosexual stage in which sexual feelings are dormant
- neurosis
- tendency to experience negative emotions
- oral stage
- psychosexual stage in which an infant’s pleasure is focused on the mouth
- phallic stage
- psychosexual stage in which the focus is on the genitals
- projection
- ego defense mechanism in which a person confronted with anxiety disguises their unacceptable urges or behaviors by attributing them to other people
- psychosexual stages of development
- stages of child development in which a child’s pleasure-seeking urges are focused on specific areas of the body called erogenous zones
- rationalization
- ego defense mechanism in which a person confronted with anxiety makes excuses to justify behavior
- reaction formation
- ego defense mechanism in which a person confronted with anxiety swaps unacceptable urges or behaviors for their opposites
- regression
- ego defense mechanism in which a person confronted with anxiety returns to a more immature behavioral state
- repression
- ego defense mechanism in which anxiety-related thoughts and memories are kept in the unconscious
- sublimation
- ego defense mechanism in which unacceptable urges are channeled into more appropriate activities
- superego
- aspect of the personality that serves as one’s moral compass, or conscience
- unconscious
- mental activity of which we are unaware and unable to access
Contributors and Attributions
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Rose M. Spielman with many significant contributors. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the creative commons license and may not be reproduced without the prior and express written consent of Rice University. For questions regarding this license, please contact<EMAIL_ADDRESS>.Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>.


8.1: Vocabulario. Las partes del cuerpo
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- Romano Sánchez Domínguez
- Imperial Valley College
Objetivos
Recognize vocabulary related to parts of the body
El torso y las extremidades
La cara y la cabeza
- la frente (forehead)
- la piel (skin)
- los ojos (eyes)
- la nariz (nose)
- la boca (mouth)
- los dientes (teeth)
- los labios (lips)
- la lengua (tongue)
- el oído (ear [inside])
- la oreja (ear [outside])
- el cerebro (brain)
- el cuello (neck)
- el pelo o el cabello (hair)
El torso y las extremidades
- la espalda (back)
- el pecho (chest)
- los hombros (shoulders)
- el estómago (stomach)
- las piernas (legs)
- los pies (feet)
- las manos (hands)
- los dedos (de las manos) (fingers)
- los dedos de los pies (toes)
- los brazos (arms)
- la muñeca (wrist)
- los huesos (bones)
- los músculos (muscles)
A Practicar
Contribute!
Did you have an idea for improving this content? We’d love your input.
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2.1: Los Angeles Aqueduct
Learning Objectives
After reading this section, you should be able to:
- Identify driving factors in the construction of the Los Angeles Aqueduct
- Classify stakeholders in the construction of the Los Angeles Aqueduct
- Evaluate the impacts of Decision 1631
|
Year |
Population |
|---|---|
|
1850 |
1,610 |
|
1860 |
4,385 |
|
1870 |
5,728 |
|
1880 |
11,183 |
|
1890 |
50,395 |
|
1900 |
102,479 |
To understand current day Los Angeles, you need to understand the population explosion in the last two hundred years in the Los Angeles basin. Los Angeles was initially a Spanish pueblo when founded in 1781 and it had a small population. A combination of surface water from the Los Angeles River and groundwater from artesian springs remained sufficient supply for a long time. As you know, Los Angeles was entitled to these supplies of surface water and groundwater because of pueblo water rights. While the supplies were sufficient for the existing population, the introduction of cattle ranching and citrus cultivation in the 1880s coupled with a drought resulted in strained water supplies.
Now, imagine you are the superintendent of this water system within California, which has been reliable under current conditions, but is strained by more and more people.
Ack! Look at those numbers! You're going to run out of water!
In 50 years, the population increased from just over fifteen hundred to over one hundred thousand people. This is the fix William Mulholland found himself in as superintendent of the water system: a burgeoning population and a dwindling water supply. His friend and colleague, former mayor of Los Angeles, Fred Eaton, suggested the Owens River as a potential supply. The Owens River relied on snowmelt from the Sierras. In order to obtain the rights for the water, the land surrounding the river was purchased to obtain water rights.
Both Eaton and Mulholland were engineers. They were intrigued by the idea that the water could be conveyed entirely by gravity with a slight slope in the aqueduct from more than 3800 feet above sea level to 1400 feet above sea level. The energy from the water was even enough to generate electricity at a number of power plants that were built along the way. In many ways, their fascination with the engineering aspects of the project may have kept them from fully considering other ethical concerns as they orchestrated purchases of land in the Owens Valley.
|
Year |
Population |
|---|---|
|
1910 |
319,198 |
|
1920 |
576,673 |
|
1930 |
1,238,048 |
|
1940 |
1,504,277 |
|
1950 |
1,970,358 |
|
1960 |
2,479,015 |
|
1970 |
2,811,801 |
|
1980 |
2,968,579 |
|
1990 |
3,485,398 |
|
2000 |
3,694,742 |
|
2010 |
3,792,621 |
Eaton and Mulholland certainly seemed to disregard stakeholders in this potential project. For example, the residents of the Owens Valley, including farmers, ranchers and the indigenous Paiute people, were certainly going to be impacted by directing the water to Southern California. These stakeholders, whose lives were dependent on the water, had very little power to mount a protest at the time. And, in fact, they generally were not asked for permission. The purchases of land in the Owens Valley were conducted in ways that seemed underhanded and non-transparent, including agents for the City of Los Angeles representing themselves as from the Bureau of Reclamation. At the point when actual stakeholders in Southern California were brought in to fund and approve the project, the land was already purchased. This is a classic example of stakeholders not being consulted until the project is almost a fait accompli .
Even though Mulholland had found enough water for the existing population in Los Angeles, the population continued to grow. By 1923, Mulholland had explored the possibility of bringing water from the Colorado River as well as the second portion of the Los Angeles Aqueduct to Mono Lake. Both projects moved ahead because it was clear that the original Los Angeles Aqueduct was not enough.
And, in the end, even the second addition to the Los Angeles Aqueduct was not enough. The Los Angeles Aqueduct was completed in three parts:
- Part One: 1913 - Los Angeles voted 10 to 1 to authorize $23 million for the first Los Angeles Aqueduct and it is subsequently built.
- Part Two: 1940 - Los Angeles votes to extend the Los Angeles Aqueduct to the Mono Lake watershed for $40 million and the extension is built.
- Part Three: 1963 - Los Angeles begins the construction of the second Los Angeles Aqueduct project
In sum, using only gravity, the Los Angeles Aqueduct encompasses canals from the Mono Lake watershed to Los Angeles stretching over 340 miles in three separate projects.
While the Los Angeles Aqueduct provided a great benefit to the residents in Southern California, it left the Owens Valley a dust bowl, including carcinogenic dust containing cadmium and nickel. After numerous years of protest and litigation, the State Water Resources Control Board in Decision 1631 drastically reduced the amount of water that could be removed from the Owens Valley from 90,000 acre-feet per year (AFY) to 16,000 AFY. How did this happen? This decision applied the public trust doctrine in a new way. The public trust doctrine addresses rights to things that are owned collectively for public use by the government, such as water and air. This interpretation allowed the water in the Owens Valley to be seen as a public resource and not something that could solely benefit Los Angeles.
As a water supply, the Los Angeles Aqueduct provides water that is dependent on local hydrology within Owens Valley. Because of the variability and Decision 1631, Los Angeles has been forced to focus on demand reduction, rely more on alternate sources of water, such as those from Metropolitan Water District of Southern California, and remediation of existing groundwater supplies in the San Fernando Groundwater Basin.
Misconception Alert!
Many people think that all of the water supply in Los Angeles comes from the Owens Valley. As you saw in this section, only 16,000 AFY currently come through the Los Angeles Aqueduct. The rest of the supply is a matter of other imported water, including the Colorado River, as well as groundwater in the San Fernando Groundwater Basin.
Try It!
- What were some of the drawbacks of removing water from the Owens Valley?
- What were the effects of Decision 1631?
- What stakeholders were not consulted in the development of the Los Angeles Aqueduct?


10.2B: Neuroglia of the Peripheral Nervous System
The two kinds of glia cells in the PNS, schwann cells and satellite cells; each have unique functions.
- Identify the neuroglia of the peripheral nervous system
Key Points
- There are two kinds of neuroglia in the peripheral nervous system (PNS): Schwann cells and satellite cells.
- Schwann cells provide myelination to peripheral neurons. Functionally, the schwann cells are similar to oligodendrocytes of the central nervous system (CNS).
- Satellite cells play an important role in modulating the PNS following injury and inflammation. These resemble the astrocytes of the CNS and assist in regulating the external chemical environment.
Key Terms
- glia : Non-neuronal cells that maintain homeostasis, form myelin, and provide support and protection for neurons in the brain and in other parts of the nervous system.
- Schwann cells : The principal glia of the PNS, these cells are involved in the conduction of nervous impulses along axons, nerve development and regeneration, trophic support for neurons, production of the nerve extracellular matrix, modulation of neuromuscular synaptic activity, and presentation of antigens to T-lymphocytes.
- Satellite glial cells : These cells line the exterior surface of neurons in the PNS and neuron cell bodies within ganglia.
The PNS has two kinds of neuroglia: schwann cells and satellite cells. Schwann cells provide myelination to peripheral neurons. They also perform phagocytic functions and clear cellular debris, allowing for the regrowth of PNS neurons. Functionally, the schwann cells are similar to oligodendrocytes of the CNS.
Satellite cells are small glia that surround neurons’ sensory ganglia in the ANS. These resemble the astrocytes of the CNS and assist in regulating the external chemical environment. PNS satellite glia are very sensitive to injury and may exacerbate pathological pain.
Schwann Cell : A Schwann cell in culture
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2: Knowing Yourself as a Learner
-
- 2.4: Learning Styles
- Several decades ago, a new way of thinking about learning became very prominent in education. It was based on the concept that each person has a preferred way to learn. It was thought that these preferences had to do with each person’s natural tendencies toward one of their senses. The idea was that learning might be easier if a student sought out content that was specifically oriented to their favored sense.
-
- 2.5: Personality Types and Learning
- Much like learning styles, there have been a number of theories surrounding the idea that different personality types may prefer different kinds of learning. Again, this builds on the original learning style concept that people may have a single preference toward how they learn, and then adds to it that certain personality traits may determine which learning style a person prefers.


1.2.4: Polarizability
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- Kelly Cox & Dana Reusser
- University of Illinois Springfield
The term polarizability refers to the tendency of molecules to generate induced electric dipole moments when subjected to an electric field. It originates from the fact that nuclei and electrons are generally not fixed in space and that when molecules are subject to an electric field, the negatively charged electrons and positively charged atomic nuclei are subject to opposite forces and undergo charge separation. Polarizability allows us to better understand the interactions between nonpolar atoms and molecules and other electrically charged species, such as ions or polar molecules with dipole moments.
Introduction
Neutral nonpolar species have spherically symmetric arrangements of electrons in their electron clouds. When in the presence of an electric field, their electron clouds can be distorted (Figure \(\PageIndex{1}\)). The ease of this distortion is the polarizability of the atom or molecule. The created distortion of the electron cloud causes the originally nonpolar molecule or atom to acquire a dipole moment. This induced dipole moment is related to the polarizability of the molecule or atom and the strength of the electric field by the following equation:
\[μ_{ind} = \alpha E \label{1}\]
where \(E\) denotes the strength of the electric field and \(\alpha\) is the polarizability of the atom or molecule with units of C m 2 V -1 .
In general, polarizability correlates with the interaction between electrons and the nucleus. The amount of electrons in a molecule affects how tight the nuclear charge can control the overall charge distribution. Atoms with fewer electrons will have smaller, denser electron clouds, as there is a strong interaction between the few electrons in the atoms’ orbitals and the positively charged nucleus. There is also less shielding in atoms with fewer electrons contributing to the stronger interaction of the outer electrons and the nucleus. With the electrons held tightly in place in these smaller atoms, these atoms are typically not easily polarized by external electric fields. In contrast, large atoms with many electrons, such as negative ions with excess electrons, are easily polarized. These atoms typically have very diffuse electron clouds and large atomic radii that limit the interaction of their external electrons and the nucleus.
Factors that Influence Polarizability
The relationship between polarizability and the factors of electron density, atomic radii, and molecular orientation is as follows:
- The greater the number of electrons, the less control the nuclear charge has on charge distribution, and thus the increased polarizability of the atom.
- The greater the distance of electrons from nuclear charge, the less control the nuclear charge has on the charge distribution, and thus the increased polarizability of the atom.
- Molecular orientation with respect to an electric field can affect polarizability (labeled Orientation-dependent), except for molecules that are: tetrahedral, octahedral or icosahedral (labeled Orientation-independent). This factor is more important for unsaturated molecules that contain areas of electron-dense regions, such as 2,4-hexadiene. Greatest polarizability in these molecules is achieved when the electric field is applied parallel to the molecule rather than perpendicular to the molecule.
Polarizability Influences Dispersion Forces
The dispersion force is the weakest intermolecular force. It is an attractive force that arises from surrounding temporary dipole moments in nonpolar molecules or species. These temporary dipole moments arise when there are instantaneous deviations in the electron clouds of the nonpolar species. Surrounding molecules are influenced by these temporary dipole moments and a sort of chain reaction results in which subsequent weak, dipole-induced dipole interactions are created (Figure \(\PageIndex{2}\)).
These cumulative dipole-induced dipole interactions create attractive dispersion forces. Dispersion forces are the forces that make nonpolar substances condense to liquids and freeze into solids when the temperature is low enough.
Polarizability affects dispersion forces in the following ways:
- As polarizability increases , the dispersion forces also become stronger . Thus, molecules attract one another more strongly and melting and boiling points of covalent substances increase with larger molecular mass.
- Polarizability also affects dispersion forces through the molecular shape of the affected molecules. Elongated molecules with electrons that are delocalized (e.g., with double or triple bonds) will have some of those electrons easily shifted over a greater distance. This is manifested in an greater polarizability and thus greater dispersion forces.
Structure Dependence to Dispersion Forces
The influence of polarizability on the strength of the dispersion forces is also modulated by the 3-D structure of the molecules. For example, neo-pentane and n-pentane (Figure \(\PageIndex{3}\)) exhibit comparable polarizabilities, but have different boiling points. The lack of polar bonding in both indicates that dispersion is the dominant interaction responsible in both substances. The higher boiling point of n-pentane indicates there are stronger intermolecular interactions.
| molecule | Boiling Points (ºC) | \(\alpha\) (Å 3 ) |
|---|---|---|
| neopentane | 9.5 | 10.24 |
| n-pentane | 36.0 | 9.879 |
Both neopentane and n-pentane would be expected to exhibits comparable dispersion forces since they have comparable polarizabilities. However, the structure of the n-pentane molecule facilitates greater dispersion forces due to more contacts between n-pentane molecules. In contrast, neopentane has a smaller dispersion force since the interactions between neopentane are reduced since neopentane is more compact and symmetrical.
The relationship between polarizability and dispersion forces can be seen in the following equation, which can be used to quantify the interaction between two like nonpolar atoms or molecules (e.g., \(\ce{A}\) with \(\ce{A}\)):
\[ V(r) = \dfrac{-3}{4} \dfrac{\alpha^2 I}{r^6} \label{2}\]
where \(r\) is the distance between the atoms or molecules, \(I\) is the first ionization energy of the atom or molecule, and \(\alpha\) is the polarizability constant expressed in units of m 3 .
This expression of \(\alpha\) is related to \(\alpha'\) by the following equation:
\[\alpha' = \dfrac{\alpha}{4 \pi \epsilon_o} \label{3}\]
To quantify the interaction between unlike atoms or molecules (\(\ce{A}\) and \(\ce{B}\)), Equation \(\ref{2}\) becomes:
\[ V(r) = \dfrac{-3}{2}\dfrac{I_AI_B}{I_A+I_B} \dfrac{\alpha_A \alpha_B}{r^6} \label{4}\]
References
- Petrucci, Ralph H., et al. General Chemistry: Principles and Modern Applications.Upper Saddle River, NJ: Prentice Hall, 2007
- Chang, Raymond. "Chapter 13: Intermolecular Forces/ Ion-Induced Dipole and Dipole-Induced Dipole Interactions/ Dispersion, or London, Interactions." Physical Chemistry for the Biosciences . Sansalito, CA: University Science, 2005. 495-98. Print.
- Housecroft, Catherine E., and Alan G. Sharpe. "Chapter 6: Structures and Energetics of Metallic and Ionic Solids." Inorganic Chemistry . Harlow [etc.: Prentice Hall, 2008. 173-74. Print.
- Miessler, Gary L., and Donald A. Tarr. "Chapter 6: Acid-Base and Donor-Acceptor Chemistry." Inorganic Chemistry . Upper Saddle River, NJ: Pearson Education, 2004. 183-86. Print


2.3: Magnitude, Direction, and Components of a Vector
The Magnitude of a Vector
It is productive to represent the horizontal and vertical components of a vector \(\vec{v}\) as \(v_x\) and \(v_y\) , respectively.
The magnitude (the length) of a vector \(\vec{v}=\left\langle v_x, v_y\right\rangle\) is \[\|\vec{v}\|=\sqrt{v_x^2+v_y^2} \nonumber \]
Magnitude of vector \(\vec{v}=\left\langle 5,\left.-8\right\rangle \right.\):
Solution
The vector \(\vec{v}=\left\langle 5,\left.-8\right\rangle \right.\) has magnitude:
\(\left\|\overrightarrow{v}\right\|=\sqrt{{v_x}^2+{v_y}^2}\) = \(\sqrt{5^2+{\left(-8\right)}^2}=\ \sqrt{25+64}\) = \(\sqrt{89}\)
Interpret this as the length of the vector \(\overrightarrow{v}=\left\langle 5,\left.-8\right\rangle \right.\) is \(\sqrt{89}\) units.
The Direction of a Vector
The direction of a vector \(\vec{v}\) is the angle the vector makes with the positive x-axis.
It is typically represented with the uppercase Greek letter theta \(\theta\) . We use some trigonometry to determine the angle \(\theta\) .
\(\tan \theta=\dfrac{y}{x}\) or \(\theta=\tan ^{-1} \dfrac{y}{x}\)
The angle \(\theta \) is always between \(0^{\circ}\) and \(360^{\circ}\).
To approximate the direction of the vector \(\vec{v}=\langle 5,8\rangle\), use \(\theta=\tan ^{-1} \frac{y}{x}\), with \(x=5\) and \(y=8\).
Solution
\(\begin{aligned}
\vec{v} & =\langle 5,8\rangle \\
\theta & =\tan ^{-1} \frac{y}{x} \\
\theta & =\tan ^{-1} \frac{8}{5}
\end{aligned}\)
Using a calculator, we get
\(\begin{aligned}
& \theta=57.99^{\circ} \\
& \theta=58^{\circ}
\end{aligned}\)
To approximate the direction of the vector \(\vec{v}=\langle 5,-8\rangle\), use \(\theta=\tan ^{-1} \frac{y}{x}\), with \(x=5\) and \(y=-8\).
Solution
\(\begin{gathered}
\vec{v}=\langle 5,-8\rangle \\
\theta=\tan ^{-1} \frac{y}{x} \\
\theta=\tan ^{-1} \frac{-8}{5}
\end{gathered}\)
Using a calculator, we get
\(\theta=-57.99^{\circ}\)
Vertical component is in Quadrant IV and \(\theta\) must be in the interval \(\left[0,\left.360\right)\right.\) , therefore we calculate \(\theta\) by
\[\theta =360{}^\circ -57.99{}^\circ =302.005{}^\circ\nonumber \] \[\theta =302{}^\circ \nonumber \]
The Components of a Vector
The lengths of the \(x\) - and \(y\) - components of a vector \(\left\|\vec{v}\right\|=\sqrt{{v_x}^2+{v_y}^2}\) in two dimensions can be found using trigonometric ratios.
\(\vec{v}_x=\|\vec{v}\|_{\cos \theta}\)and \(\vec{v}_y=\|\vec{v}\|_{\sin \theta}\)
\(\vec{v}_x\) is the horizontal component of \(\vec{v}\) and \(\vec{v}_y\) is the vertical component.
The angle \(\theta\) is always between \(0^{\circ}\) and \(360^{\circ}\).
Suppose the magnitude of a vector \(\vec{v}\boldsymbol{=}\left\langle v_x,\left.v_y\right\rangle \right.\) is 20 units, and that \(\vec{v}\) makes a 60 \(\mathrm{{}^\circ}\) angle with the horizontal. Then, the components of \(\vec{v}\) are
\[\begin{aligned}
\vec{v}_x & =\|\vec{v}\| \cos \theta \\
= & 20 \cos 60^{\circ} \\
& =20 \cdot \frac{1}{2} \\
& =10
\end{aligned} \nonumber \]
and
\[\begin{aligned}
\vec{v}_y & =\|\vec{v}\|_{\sin } \theta \\
= & 20 \sin 60^{\circ} \\
& =20 \cdot \frac{\sqrt{3}}{2} \\
& =10 \sqrt{3}
\end{aligned} \nonumber \]
So, we could write \(\vec{v}\boldsymbol{=}\left\langle v_x,\left.v_y\right\rangle \right.\) as \(\vec{v}\boldsymbol{=}\left\langle 10,\left.10\sqrt{3}\right\rangle \right.\)
Using Technology
We can use technology to determine the magnitude of a vector.
Go to www.wolframalpha.com.
To find the magnitude of the vector \(\vec{v}\boldsymbol{=}\left\langle 2,\left.4\right\rangle ,\right.\) enter magnitude of the vector \(\mathrm{<}\) 2,4 \(\mathrm{>}\) in the entry field. Wolframalpha tells you what it thinks you entered, then tells you its answer. In this case, \(\left\|\vec{v}\right\|=2\sqrt{5}\) .
. The result is 2 times the square root of 5."<EMAIL_ADDRESS>To find the direction of the vector \(\vec{v}\boldsymbol{=}\left\langle 5,\left.8\right\rangle ,\right.\) enter direction of the vector \(\mathrm{<}\) 5,8 \(\mathrm{>}\) in the entry field. Wolframalpha answers \(57.9946{}^\circ \approx 58{}^\circ\) .
. The result is 9.43398 radians (r) and 57.9946 degrees from the x-axis (theta)."<EMAIL_ADDRESS>Try These
Find the magnitude of the vector \(\vec{v}=\langle 3,-4\rangle\).
- Answer
-
\(\left\|\overrightarrow{v}\right\|=5\)
Find the magnitude of the vector \(\vec{v}=\langle -3,-3\rangle\).
- Answer
-
\(\left\|\overrightarrow{v}\right\|=3\sqrt{2}\)
Find the components of the vector \(\vec{v}\) if the magnitude of \(\vec{v}\) is 6 and it makes a \(30^{\circ}\) angle with the horizontal.
- Answer
-
\({\overrightarrow{v}}_x=\ 3\sqrt{3}\) and \({\overrightarrow{v}}_y\mathrm{=3}\)
Approximate the direction of the vector \(\vec{v}=\langle 3,10\rangle\).
- Answer
-
\(\theta \approx 73.3008{}^\circ\)


2.1.12: Variable Assignments
A variable assignment \(s\) provides a value for every variable—and there are infinitely many of them. This is of course not necessary. We require variable assignments to assign values to all variables simply because it makes things a lot easier. The value of a term \(t\) , and whether or not a formula \(A\) is satisfied in a structure with respect to \(s\) , only depend on the assignments \(s\) makes to the variables in \(t\) and the free variables of \(A\) . This is the content of the next two propositions. To make the idea of “depends on” precise, we show that any two variable assignments that agree on all the variables in \(t\) give the same value, and that \(A\) is satisfied relative to one iff it is satisfied relative to the other if two variable assignments agree on all free variables of \(A\) .
If the variables in a term \(t\) are among \(x_1\) , …, \(x_n\) , and \(s_1(x_i) = s_2(x_i)\) for \(i = 1\) , …, \(n\) , then \(\Value[s_1]{t}{M} = \Value[s_2]{t}{M}\) .
Proof. By induction on the complexity of \(t\) . For the base case, \(t\) can be a constant symbol or one of the variables \(x_1\) , …, \(x_n\) . If \(t = c\) , then \(\Value[s_1]{t}{M} = \Assign{c}{M} = \Value[s_2]{t}{M}\) . If \(t = x_i\) , \(s_1(x_i) = s_2(x_i)\) by the hypothesis of the proposition, and so \(\Value[s_1]{t}{M} = s_1(x_i) = s_2(x_i) = \Value[s_2]{t}{M}\) .
For the inductive step, assume that \(t = \Atom{f}{t_1, \dots, t_k}\) and that the claim holds for \(t_1\) , …, \(t_k\) . Then
\[\begin{aligned}
\Value[s_1]{t}{M}
& = \Value[s_1]{\Atom{f}{t_1, \dots, t_k}}{M} =\\
& = \Assign{f}{M}(\Value[s_1]{t_1}{M}, \dots, \Value[s_1]{t_k}{M})
\end{aligned}\]
For \(j = 1, \dots, k\), the variables of \(t_j\) are among \(x_1, \dots,x_n\). So by induction hypothesis, \(\Value[s_1]{t_j}{M} = \Value[s_2]{t_j}{M}\). So,
\[\begin{aligned}
\Value[s_1]{t}{M} & = \Value[s_2]{\Atom{f}{t_1, \dots, t_k}}{M} =\\ & = \Assign{f}{M}(\Value[s_1]{t_1}{M}, \dots, \Value[s_1]{t_k}{M}) =\\ & = \Assign{f}{M}(\Value[s_2]{t_1}{M}, \dots, \Value[s_2]{t_k}{M}) =\\ & = \Value[s_2]{\Atom{f}{t_1, \dots, t_k}}{M} = \Value[s_2]{t}{M}.\end{aligned}\] ◻
If the free variables in \(A\) are among \(x_1\) , …, \(x_n\) , and \(s_1(x_i) = s_2(x_i)\) for \(i = 1\) , …, \(n\) , then \(\Sat[,s_1]{M}{A}\) iff \(\Sat[,s_2]{M}{A}\) .
Proof. We use induction on the complexity of \(A\) . For the base case, where \(A\) is atomic, \(A\) can be: \(\lfalse\) , \(\Atom{R}{t_1, \dots, t_k}\) for a \(k\) -place predicate \(R\) and terms \(t_1\) , …, \(t_k\) , or \(\eq[t_1][t_2]\) for terms \(t_1\) and \(t_2\) .
- \(\indcase{A}{\lfalse}\) both \(\SatN[,s_1]{M}{A}\) and \(\SatN[,s_2]{M}{A}\) .
- \(\indcase{A}{\Atom{R}{t_1, \ldots, t_k}}\) let \(\Sat[,s_1]{M}{\indfrm}\) . Then \[\langle \Value[s_1]{t_1}{M}, \ldots, \Value[s_1]{t_k}{M} \rangle \in \Assign{R}{M}.\nonumber\] For \(i = 1\) , …, \(k\) , \(\Value[s_1]{t_i}{M} = \Value[s_2]{t_i}{M}\) by Proposition \(\PageIndex{1}\) . So we also have \(\langle \Value[s_2]{t_i}{M}, \ldots, \Value[s_2]{t_k}{M} \rangle \in \Assign{R}{M}\) .
- \(\indcase{A}{\eq[t_1][t_2]}\) suppose \(\Sat[,s_1]{M}{\indfrm}\) . Then \(\Value[s_1]{t_1}{M} = \Value[s_1]{t_2}{M}\) . So, \[\begin{aligned} \Value[s_2]{t_1}{M} & = \Value[s_1]{t_1}{M} & \text{(by Proposition $\PageIndex{1}$)} \\ & = \Value[s_1]{t_2}{M} & \text{(since $\Sat[,s_1]{M}{\eq[t_1][t_2]}$)}\\ &= \Value[s_2]{t_2}{M} & \text{(by Proposition $\PageIndex{1}$),} \end{aligned}\] so \(\Sat[,s_2]{M}{\eq[t_1][t_2]}\) .
Now assume \(\Sat[,s_1]{M}{B}\) iff \(\Sat[,s_2]{M}{B}\) for all formulas \(B\) less complex than \(A\) . The induction step proceeds by cases determined by the main operator of \(A\) . In each case, we only demonstrate the forward direction of the biconditional; the proof of the reverse direction is symmetrical. In all cases except those for the quantifiers, we apply the induction hypothesis to sub-formulas \(B\) of \(A\) . The free variables of \(B\) are among those of \(A\) . Thus, if \(s_1\) and \(s_2\) agree on the free variables of \(A\) , they also agree on those of \(B\) , and the induction hypothesis applies to \(B\) .
- \(\indcase{A}{\lnot B}\) if \(\Sat[,s_1]{M}{\indfrm}\) , then \(\SatN[,s_1]{M}{B}\) , so by the induction hypothesis, \(\SatN[,s_2]{M}{B}\) , hence \(\Sat[,s_2]{M}{\indfrm}\) .
- \(\indcase{A}{B \land C}\) exercise.
- \(\indcase{A}{B \lor C}\) if \(\Sat[,s_1]{M}{\indfrm}\) , then \(\Sat[,s_1]{M}{B}\) or \(\Sat[,s_1]{M}{C}\) . By induction hypothesis, \(\Sat[,s_2]{M}{B}\) or \(\Sat[,s_2]{M}{C}\) , so \(\Sat[,s_2]{M}{\indfrm}\) .
- \(\indcase{A}{B \lif C}\) exercise.
- \(\indcase{A}{\lexists{x}{B}}\) if \(\Sat[,s_1]{M}{\indfrm}\) , there is an \(x\) -variant \(s_1'\) of \(s_1\) so that \(\Sat[,s_1']{M}{B}\) . Let \(s_2'\) be the \(x\) -variant of \(s_2\) that assigns the same thing to \(x\) as does \(s_1'\) . The free variables of \(B\) are among \(x_1\) , …, \(x_n\) , and \(x\) . \(s_1'(x_i) = s_2'(x_i)\) , since \(s_1'\) and \(s_2'\) are \(x\) -variants of \(s_1\) and \(s_2\) , respectively, and by hypothesis \(s_1(x_i) = s_2(x_i)\) . \(s_1'(x) = s_2'(x)\) by the way we have defined \(s_2'\) . Then the induction hypothesis applies to \(B\) and \(s_1'\) , \(s_2'\) , so \(\Sat[,s_2']{M}{B}\) . Hence, there is an \(x\) -variant of \(s_2\) that satisfies \(B\) , and so \(\Sat[,s_2]{M}{\indfrm}\) .
- \(\indcase{A}{\lforall{x}{B}}\) exercise.
By induction, we get that \(\Sat[,s_1]{M}{A}\) iff \(\Sat[,s_2]{M}{A}\) whenever the free variables in \(A\) are among \(x_1\) , …, \(x_n\) and \(s_1(x_i)=s_2(x_i)\) for \(i = 1\) , …, \(n\) . ◻
Complete the proof of Proposition \(\PageIndex{2}\) .
Sentences have no free variables, so any two variable assignments assign the same things to all the (zero) free variables of any sentence. The proposition just proved then means that whether or not a sentence is satisfied in a structure relative to a variable assignment is completely independent of the assignment. We’ll record this fact. It justifies the definition of satisfaction of a sentence in a structure (without mentioning a variable assignment) that follows.
If \(A\) is a sentence and \(s\) a variable assignment, then \(\Sat[,s]{M}{A}\) iff \(\Sat[,s']{M}{A}\) for every variable assignment \(s'\) .
Proof. Let \(s'\) be any variable assignment. Since \(A\) is a sentence, it has no free variables, and so every variable assignment \(s'\) trivially assigns the same things to all free variables of \(A\) as does \(s\) . So the condition of Proposition \(\PageIndex{2}\) is satisfied, and we have \(\Sat[,s]{M}{A}\) iff \(\Sat[,s']{M}{A}\) . ◻
If \(A\) is a sentence, we say that a structure \(\Struct M\) satisfies \(A\) , \(\Sat{M}{A}\) , iff \(\Sat[,s]{M}{A}\) for all variable assignments \(s\) .
If \(\Sat{M}{A}\) , we also simply say that \(A\) is true in \(\Struct{M}\) .
Let \(\Struct{M}\) be a structure, \(A\) be a sentence, and \(s\) a variable assignment. \(\Sat{M}{A}\) iff \(\Sat[,s]{M}{A}\) .
Proof. Exercise. ◻
Prove Proposition \(\PageIndex{3}\).
Suppose \(A(x)\) only contains \(x\) free, and \(\Struct M\) is a structure. Then:
- \(\Sat{M}{\lexists{x}{A(x)}}\) iff \(\Sat[,s]{M}{A(x)}\) for at least one variable assignment \(s\) .
- \(\Sat{M}{\lforall{x}{A(x)}}\) iff \(\Sat[,s]{M}{A(x)}\) for all variable assignments \(s\) .
Proof. Exercise. ◻
Prove Proposition \(\PageIndex{4}\) .
Suppose \(\Lang L\) is a language without function symbols. Given a structure \(\Struct M\) , \(c\) a constant symbol and \(a \in \Domain M\) , define \(\Struct M[a/c]\) to be the structure that is just like \(\Struct M\) , except that \(\Assign{c}{M[a/c]} = a\) . Define \(\Struct M \mathrel{||}\joinrel\Relbar A\) for sentences \(A\) by:
- \(\indcase{A}{\lfalse}\) not \(\Struct{M} \mathrel{||}\joinrel\Relbar\indfrm\) .
- \(\indcase{A}{\Atom{R}{d_1, \dots, d_n}}\) \(\Struct M \mathrel{||}\joinrel\Relbar\indfrm\) iff \(\langle \Assign{d_1}{M}, \dots, \Assign{d_n}{M} \rangle \in \Assign{R}{M}\) .
- \(\indcase{A}{\eq[d_1][d_2]}\) \(\Struct M \mathrel{||}\joinrel\Relbar\indfrm\) iff \(\Assign{d_1}{M} = \Assign{d_2}{M}\) .
- \(\indcase{A}{\lnot B}\) \(\Struct{M} \mathrel{||}\joinrel\Relbar\indfrm\) iff not \(\Struct{M} \mathrel{||}\joinrel\Relbar{B}\) .
- \(\indcase{A}{(B \land C)}\) \(\Struct{M} \mathrel{||}\joinrel\Relbar \indfrm\) iff \(\Struct{M} \mathrel{||}\joinrel\Relbar B\) and \(\Struct{M} \mathrel{||}\joinrel\Relbar C\) .
- \(\indcase{A}{(B \lor C)}\) \(\Struct{M} \mathrel{||}\joinrel\Relbar\indfrm\) iff \(\Struct{M} \mathrel{||}\joinrel\Relbar B\) or \(\Struct{M} \mathrel{||}\joinrel\Relbar C\) (or both).
- \(\indcase{A}{(B \lif C)}\) \(\Struct{M} \mathrel{||}\joinrel\Relbar\indfrm\) iff not \(\Struct {M} \mathrel{||}\joinrel\Relbar B\) or \(\Struct M \mathrel{||}\joinrel\Relbar C\) (or both).
- \(\indcase{A}{\lforall{x}{B}}\) \(\Struct{M} \mathrel{||}\joinrel\Relbar{\indfrm}\) iff for all \(a \in \Domain{M}\) , \(\Struct{M[a/c]} \mathrel{||}\joinrel\Relbar\Subst{B}{c}{x}\) , if \(c\) does not occur in \(B\) .
- \(\indcase{A}{\lexists{x}{B}}\) \(\Struct{M} \mathrel{||}\joinrel\Relbar {\indfrm}\) iff there is an \(a \in \Domain M\) such that \(\Struct{M[a/c]} \mathrel{||}\joinrel\Relbar\Subst{B}{c}{x}\) , if \(c\) does not occur in \(B\) .
Let \(x_1\) , …, \(x_n\) be all free variables in \(A\) , \(c_1\) , …, \(c_n\) constant symbols not in \(A\) , \(a_1\) , …, \(a_n \in \Domain M\) , and \(s(x_i) = a_i\) .
Show that \(\Sat[,s]{M}{A}\) iff \(\Struct M[a_1/c_1,\dots,a_n/c_n] \mathrel{||}\joinrel\Relbar\Subst{\Subst{A}{c_1}{x_1}\dots}{c_n}{x_n}\) .
(This problem shows that it is possible to give a semantics for first-order logic that makes do without variable assignments.)
Suppose that \(f\) is a function symbol not in \(A(x,y)\) . Show that there is a structure \(\Struct{M}\) such that \(\Sat{M}{\lforall{x}{\lexists{y}{A(x,y)}}}\) iff there is an \(\Struct M'\) such that \(\Sat{M'}{\lforall{x}{A(x,f(x))}}\) .
(This problem is a special case of what’s known as Skolem’s Theorem; \(\lforall{x}{A(x,f(x))}\) is called a Skolem normal form of \(\lforall{x}{\lexists{y}{A(x,y)}}\) .)


4.1: Introduction to How Environment Affects Strategy
What you’ll learn to do: explain the key aspects of the environment that can affect strategy
Managers develop strategies to guide their organizations to success. They analyze external and internal conditions to make sure that their assumptions are based as much as possible on solid data. The truth is, however, that the environment today is changing faster than it ever did in the past. And the rate of change is increasing, as well. There are some attributes in the modern environment that make strategic planning a risky business.
Contributors and Attributions
CC licensed content, Original
- Introduction to How Environment Affects Strategy. Authored by : John/Lynn Bruton and Lumen Learning. License : CC BY: Attribution