11.3: The Geometric Distribution
Basic Theory
Definitions
Suppose again that our random experiment is to perform a sequence of Bernoulli trials \(\bs{X} = (X_1, X_2, \ldots)\) with success parameter \(p \in (0, 1]\). In this section we will study the random variable \(N\) that gives the trial number of the first success and the random variable \( M \) that gives the number of failures before the first success.
Let \( N = \min\{n \in \N_+: X_n = 1\} \), the trial number of the first success, and let \( M = N - 1 \), the number of failures before the first success. The distribution of \( N \) is the geometric distribution on \( \N_+ \) and the distribution of \( M \) is the geometric distribution on \( \N \) . In both cases, \( p \) is the success parameter of the distribution.
Since \( N \) and \( M \) differ by a constant, the properties of their distributions are very similar. Nonetheless, there are applications where it more natural to use one rather than the other, and in the literature, the term geometric distribution can refer to either. In this section, we will concentrate on the distribution of \( N \), pausing occasionally to summarize the corresponding results for \( M \).
The Probability Density Function
\( N \) has probability density function \( f \) given by \(f(n) = p (1 - p)^{n-1}\) for \(n \in \N_+\).
Proof
Note first that \(\{N = n\} = \{X_1 = 0, \ldots, X_{n-1} = 0, X_n = 1\}\). By independence, the probability of this event is \((1 - p)^{n-1} p\).
Check that \( f \) is a valid PDF
By standard results for geometric series \[\sum_{n=1}^\infty \P(N = n) = \sum_{n=1}^\infty (1 - p)^{n-1} p = \frac{p}{1 - (1 - p)} = 1\]
A priori , we might have thought it possible to have \(N = \infty\) with positive probability; that is, we might have thought that we could run Bernoulli trials forever without ever seeing a success. However, we now know this cannot happen when the success parameter \(p\) is positive.
The probability density function of \(M\) is given by \(\P(M = n) = p (1 - p)^n\) for \( n \in \N\).
In the negative binomial experiment, set \(k = 1\) to get the geometric distribution on \(\N_+\). Vary \(p\) with the scroll bar and note the shape and location of the probability density function. For selected values of \(p\), run the simulation 1000 times and compare the relative frequency function to the probability density function.
Note that the probability density functions of \( N \) and \( M \) are decreasing, and hence have modes at 1 and 0, respectively. The geometric form of the probability density functions also explains the term geometric distribution .
Distribution Functions and the Memoryless Property
Suppose that \( T \) is a random variable taking values in \( \N_+ \). Recall that the ordinary distribution function of \( T \) is the function \( n \mapsto \P(T \le n) \). In this section, the complementary function \( n \mapsto \P(T \gt n) \) will play a fundamental role. We will refer to this function as the right distribution function of \( T \). Of course both functions completely determine the distribution of \( T \). Suppose again that \( N \) has the geometric distribution on \( \N_+ \) with success parameter \( p \in (0, 1] \).
\( N \) has right distribution function \( G \) given by \(G(n) = (1 - p)^n\) for \(n \in \N\).
Proof from Bernoulli trials
Note that \(\{N \gt n\} = \{X_1 = 0, \ldots, X_n = 0\}\). By independence, the probability of this event is \((1 - p)^n\).
Direct proof
Using geometric series, \[ \P(N \gt n) = \sum_{k=n+1}^\infty \P(N = k) = \sum_{k=n+1}^\infty (1 - p)^{k-1} p = \frac{p (1 - p)^n}{1 - (1 - p)} = (1 - p)^n \]
From the last result, it follows that the ordinary (left) distribution function of \(N\) is given by \[ F(n) = 1 - (1 - p)^n, \quad n \in \N \] We will now explore another characterization known as the memoryless property .
For \( m \in \N \), the conditional distribution of \(N - m\) given \(N \gt m\) is the same as the distribution of \(N\). That is, \[\P(N \gt n + m \mid N \gt m) = \P(N \gt n); \quad m, \, n \in \N\]
Proof
From the result above and the definition of conditional probability, \[ \P(N \gt n + m \mid N \gt m) = \frac{\P(N \gt n + m)}{\P(N \gt m)} = \frac{(1 - p)^{n+m}}{(1 - p)^m} = (1 - p)^n = \P(N \gt n)\]
Thus, if the first success has not occurred by trial number \(m\), then the remaining number of trials needed to achieve the first success has the same distribution as the trial number of the first success in a fresh sequence of Bernoulli trials. In short, Bernoulli trials have no memory. This fact has implications for a gambler betting on Bernoulli trials (such as in the casino games roulette or craps). No betting strategy based on observations of past outcomes of the trials can possibly help the gambler.
Conversely, if \(T\) is a random variable taking values in \(\N_+\) that satisfies the memoryless property , then \(T\) has a geometric distribution.
Proof
Let \(G(n) = \P(T \gt n)\) for \(n \in \N\). The memoryless property and the definition of conditional probability imply that \(G(m + n) = G(m) G(n)\) for \(m, \; n \in \N\). Note that this is the law of exponents for \(G\). It follows that \(G(n) = G^n(1)\) for \(n \in \N\). Hence \(T\) has the geometric distribution with parameter \(p = 1 - G(1)\).
Moments
Suppose again that \(N\) is the trial number of the first success in a sequence of Bernoulli trials, so that \(N\) has the geometric distribution on \(\N_+\) with parameter \(p \in (0, 1]\). The mean and variance of \(N\) can be computed in several different ways.
\(\E(N) = \frac{1}{p}\)
Proof from the density function
Using the derivative of the geometric series, \begin{align} \E(N) &= \sum_{n=1}^\infty n p (1 - p)^{n-1} = p \sum_{n=1}^\infty n (1 - p)^{n-1} \\ &= p \sum_{n=1}^\infty - \frac{d}{dp}(1 - p)^n = - p \frac{d}{d p} \sum_{n=0}^\infty (1 - p)^n \\ &= -p \frac{d}{dp} \frac{1}{p} = -p \left(-\frac{1}{p^2}\right) = \frac{1}{p}\end{align}
Proof from the right distribution function
Recall that since \( N \) takes positive integer values, its expected value can be computed as the sum of the right distribution function. Hence \[ \E(N) = \sum_{n=0}^\infty \P(N \gt n) = \sum_{n=0}^\infty (1 - p)^n = \frac{1}{p} \]
Proof from Bernoulli trials
We condition on the first trial \( X_1 \): If \( X_1 = 1 \) then \( N = 1 \) and hence \( \E(N \mid X_1 = 1) = 1 \). If \( X_1 = 0 \) (equivalently \( N \gt 1) \) then by the memoryless property, \( N - 1 \) has the same distribution as \( N \). Hence \( \E(N \mid X_1 = 0) = 1 + \E(N) \). In short \[ \E(N \mid X_1) = 1 + (1 - X_1) \E(N) \] It follows that \[ \E(N) = \E\left[\E(N \mid X_1)\right] = 1 + (1 - p) \E(N)\] Solving gives \( \E(N) = \frac{1}{p}\).
This result makes intuitive sense. In a sequence of Bernoulli trials with success parameter \( p \) we would
expect
to wait \( 1/p \) trials for the first success.
\( \var(N) = \frac{1 - p}{p^2} \)
Direct proof
We first compute \( \E\left[N(N - 1)\right] \). This is an example of a factorial moment, and we will compute the general factorial moments below. Using derivatives of the geometric series again, \begin{align} \E\left[N(N - 1)\right] & = \sum_{n=2}^\infty n (n - 1) p (1 - p)^{n-1} = p (1 - p) \sum_{n=2}^\infty n(n - 1) (1 - p)^{n-2} \\ & = p (1 - p) \frac{d^2}{d p^2} \sum_{n=0}^\infty (1 - p)^n = p (1 - p) \frac{d^2}{dp^2} \frac{1}{p} = p (1 - p) \frac{2}{p^3} = 2 \frac{1 - p}{p^2} \end{align} Since \( \E(N) = \frac{1}{p} \), it follows that \( \E\left(N^2\right) = \frac{2 - p}{p^2} \) and hence \( \var(N) = \frac{1 - p}{p^2} \)
Proof from Bernoulli trials
Recall that \[ \E(N \mid X_1) = 1 + (1 - X_1) \E(N) = 1 + \frac{1}{p} (1 - X_1) \] and by the same reasoning, \( \var(N \mid X_1) = (1 - X_1) \var(N) \). Hence \[ \var(N) = \var\left[\E(N \mid X_1)\right] + \E\left[\var(N \mid X_1)\right] = \frac{1}{p^2} p(1 - p) + (1 - p) \var(N) \] Solving gives \( \var(N) = \frac{1 - p}{p^2} \).
Note that \( \var(N) = 0 \) if \( p = 1 \), hardly surprising since \( N \) is deterministic (taking just the value 1) in this case. At the other extreme, \( \var(N) \uparrow \infty \) as \( p \downarrow 0 \).
In the negative binomial experiment, set \(k = 1\) to get the geometric distribution. Vary \(p\) with the scroll bar and note the location and size of the mean\(\pm\)standard deviation bar. For selected values of \(p\), run the simulation 1000 times and compare the sample mean and standard deviation to the distribution mean and standard deviation.
the probability generating function \( P \) of \(N\) is given by \[ P(t) = \E\left(t^N\right) = \frac{p t}{1 - (1 - p) t}, \quad \left|t\right| \lt \frac{1}{1 - p} \]
Proof
This result follows from yet another application of geometric series: \[ \E\left(t^N\right) = \sum_{n=1}^\infty t^n p (1 - p)^{n-1} = p t \sum_{n=1}^\infty \left[t (1 - p)\right]^{n-1} = \frac{p t}{1 - (1 - p) t}, \quad \left|(1 - p) t\right| \lt 1 \]
Recall again that for \( x \in \R \) and \( k \in \N \), the falling power of \( x \) of order \( k \) is \( x^{(k)} = x (x - 1) \cdots (x - k + 1) \). If \( X \) is a random variable, then \( \E\left[X^{(k)}\right] \) is the factorial moment of \( X \) of order \( k \).
The factorial moments of \(N\) are given by \[ \E\left[N^{(k)}\right] = k! \frac{(1 - p)^{k-1}}{p^k}, \quad k \in \N_+ \]
Proof from geometric series
Using derivatives of geometric series again, \begin{align} \E\left[N^{(k)}\right] & = \sum_{n=k}^\infty n^{(k)} p (1 - p)^{n-1} = p (1 - p)^{k-1} \sum_{n=k}^\infty n^{(k)} (1 - p)^{n-k} \\ & = p (1 - p)^{k-1} (-1)^k \frac{d^k}{dp^k} \sum_{n=0}^\infty (1 - p)^n = p (1 - p)^{k-1} (-1)^k \frac{d^k}{dp^k} \frac{1}{p} = k! \frac{(1 - p)^{k-1}}{p^k} \end{align}
Proof from the generating function
Recall that \(\E\left[N^{(k)}\right] = P^{(k)}(1)\) where \(P\) is the probability generating function of \(N\). So the result follows from standard calculus.
Suppose that \( p \in (0, 1) \). The skewness and kurtosis of \(N\) are
- \(\skw(N) = \frac{2 - p}{\sqrt{1 - p}}\)
- \(\kur(N) = \frac{p^2}{1 - p}\)
Proof
The factorial moments can be used to find the moments of \(N\) about 0. The results then follow from the standard computational formulas for skewness and kurtosis.
Note that the geometric distribution is always positively skewed. Moreover, \( \skw(N) \to \infty \) and \( \kur(N) \to \infty \) as \( p \uparrow 1 \).
Suppose now that \(M = N - 1\), so that \(M\) (the number of failures before the first success) has the geometric distribution on \(\N\). Then
- \(\E(M) = \frac{1 - p}{p}\)
- \(\var(M) = \frac{1 - p}{p^2}\)
- \(\skw(M) = \frac{2 - p}{\sqrt{1 - p}}\)
- \(\kur(M) = \frac{p^2}{1 - p}\)
- \(\E\left(t^M\right) = \frac{p}{1 - (1 - p) \, t}\) for \(\left|t\right| \lt \frac{1}{1 - p}\)
Of course, the fact that the variance, skewness, and kurtosis are unchanged follows easily, since \(N\) and \(M\) differ by a constant.
The Quantile Function
Let \(F\) denote the distribution function of \(N\), so that \(F(n) = 1 - (1 - p)^n\) for \(n \in \N\). Recall that \(F^{-1}(r) = \min \{n \in \N_+: F(n) \ge r\}\) for \(r \in (0, 1)\) is the quantile function of \(N\).
The quantile function of \(N\) is \[ F^{-1}(r) = \left\lceil \frac{\ln(1 - r)}{\ln(1 - p)}\right\rceil, \quad r \in (0, 1) \]
Of course, the quantile function, like the probability density function and the distribution function, completely determines the distribution of \(N\). Moreover, we can compute the median and quartiles to get measures of center and spread.
The first quartile, the median (or second quartile), and the third quartile are
- \(F^{-1}\left(\frac{1}{4}\right) = \left\lceil \ln(3/4) \big/ \ln(1 - p)\right\rceil \approx \left\lceil-0.2877 \big/ \ln(1 - p)\right\rceil\)
- \(F^{-1}\left(\frac{1}{2}\right) = \left\lceil \ln(1/2) \big/ \ln(1 - p)\right\rceil \approx \left\lceil-0.6931 \big/ \ln(1 - p)\right\rceil\)
- \(F^{-1}\left(\frac{3}{4}\right) = \left\lceil \ln(1/4) \big/ \ln(1 - p)\right\rceil \approx \left\lceil-1.3863 \big/ \ln(1 - p)\right\rceil\)
Open the special distribution calculator, and select the geometric distribution and CDF view. Vary \( p \) and note the shape and location of the CDF/quantile function. For various values of \( p \), compute the median and the first and third quartiles.
The Constant Rate Property
Suppose that \(T\) is a random variable taking values in \(\N_+\), which we interpret as the first time that some event of interest occurs.
The function \( h \) given by \[ h(n) = \P(T = n \mid T \ge n) = \frac{\P(T = n)}{\P(T \ge n)}, \quad n \in \N_+ \] is the rate function of \(T\).
If \(T\) is interpreted as the (discrete) lifetime of a device, then \(h\) is a discrete version of the failure rate function studied in reliability theory. However, in our usual formulation of Bernoulli trials, the event of interest is success rather than failure (or death), so we will simply use the term rate function to avoid confusion. The constant rate property characterizes the geometric distribution. As usual, let \(N\) denote the trial number of the first success in a sequence of Bernoulli trials with success parameter \(p \in (0, 1)\), so that \(N\) has the geometric distribution on \(\N_+\) with parameter \(p\).
\(N\) has constant rate \(p\).
Proof
From the results above, \( \P(N = n) = p (1 - p)^{n-1} \) and \( \P(N \ge n) = \P(N \gt n - 1) = (1 - p)^{n-1} \), so \( \P(N = n) \big/ \P(N \ge n) = p \) for \( n \in \N_+ \).
Conversely, if \(T\) has constant rate \(p \in (0, 1)\) then \(T\) has the geometric distrbution on \(\N_+\) with success parameter \(p\).
Proof
Let \(H(n) = \P(T \ge n)\) for \(n \in \N_+\). From the constant rate property, \(\P(T = n) = p \, H(n)\) for \(n \in \N_+\). Next note that \(\P(T = n) = H(n) - H(n + 1)\) for \(n \in \N_+\). Thus, \(H\) satisfies the recurrence relation \(H(n + 1) = (1 - p) \, H(n)\) for \(n \in \N_+\). Also \(H\) satisfies the initial condition \(H(1) = 1\). Solving the recurrence relation gives \(H(n) = (1 - p)^{n-1}\) for \(n \in \N_+\).
Relation to the Uniform Distribution
Suppose again that \( \bs{X} = (X_1, X_2, \ldots) \) is a sequence of Bernoulli trials with success parameter \( p \in (0, 1) \). For \( n \in \N_+ \), recall that \(Y_n = \sum_{i=1}^n X_i\), the number of successes in the first \(n\) trials, has the binomial distribution with parameters \(n\) and \(p\). As before, \( N \) denotes the trial number of the first success.
Suppose that \( n \in \N_+ \). The conditional distribution of \(N\) given \(Y_n = 1\) is uniform on \(\{1, 2, \ldots, n\}\).
Proof from sampling
We showed in the last section that given \( Y_n = k \), the trial numbers of the successes form a random sample of size \( k \) chosen without replacement from \( \{1, 2, \ldots, n\} \). This result is a simple corollary with \( k = 1 \)
Direct proof
For \( j \in \{1, 2, \ldots, n\} \) \[ \P(N = j \mid Y_n = 1) = \frac{\P(N = j, Y_n = 1)}{\P(Y_n = 1)} = \frac{\P\left(Y_{j-1} = 0, X_j = 1, Y_{n} - Y_j = 0\right)}{\P(Y_n = 1)} \] In words, the events in the numerator of the last fraction are that there are no successes in the first \( j - 1 \) trials, a success on trial \( j \), and no successes in trials \( j + 1 \) to \( n \). These events are independent so \[ \P(N = j \mid Y_n = 1) = \frac{(1 - p)^{j-1} p (1 - p)^{n-j}}{n p (1 - p)^{n - 1}} = \frac{1}{n}\]
Note that the conditional distribution does not depend on the success parameter \(p\). If we know that there is exactly one success in the first \(n\) trials, then the trial number of that success is equally likely to be any of the \(n\) possibilities.
Another connection between the geometric distribution and the uniform distribution is given below in the alternating coin tossing game : the conditional distribution of \( N \) given \( N \le n \) converges to the uniform distribution on \( \{1, 2, \ldots, n\} \) as \( p \downarrow 0 \).
Relation to the Exponential Distribution
The
Poisson process
on \( [0, \infty) \), named for Simeon Poisson, is a model for random points in continuous time. There are many deep and interesting connections between the Bernoulli trials process (which can be thought of as a model for random points in
discrete
time) and the Poisson process. These connections are explored in detail in the chapter on the Poisson process. In this section we just give the most famous and important result—the convergence of the geometric distribution to the exponential distribution. The geometric distribution, as we know, governs the time of the first
random point
in the Bernoulli trials process, while the exponential distribution governs the time of the first random point in the Poisson process.
For reference, the exponential distribution with rate parameter \( r \in (0, \infty) \) has distribution function \( F(x) = 1 - e^{-r x} \) for \( x \in [0, \infty) \). The mean of the exponential distribution is \( 1 / r \) and the variance is \( 1 / r^2 \). In addition, the moment generating function is \( s \mapsto \frac{1}{s - r} \) for \( s \gt r \).
For \( n \in \N_+ \), suppose that \( U_n \) has the geometric distribution on \( \N_+ \) with success parameter \( p_n \in (0, 1) \), where \( n p_n \to r \gt 0 \) as \( n \to \infty \). Then the distribution of \( U_n / n \) converges to the exponential distribution with parameter \( r \) as \( n \to \infty \).
Proof
Let \( F_n \) denote the CDF of \( U_n / n \). Then for \( x \in [0, \infty) \) \[ F_n(x) = \P\left(\frac{U_n}{n} \le x\right) = \P(U_n \le n x) = \P\left(U_n \le \lfloor n x \rfloor\right) = 1 - \left(1 - p_n\right)^{\lfloor n x \rfloor} \] But by a famous limit from calculus, \( \left(1 - p_n\right)^n = \left(1 - \frac{n p_n}{n}\right)^n \to e^{-r} \) as \( n \to \infty \), and hence \( \left(1 - p_n\right)^{n x} \to e^{-r x} \) as \( n \to \infty \). But by definition, \( \lfloor n x \rfloor \le n x \lt \lfloor n x \rfloor + 1\) or equivalently, \( n x - 1 \lt \lfloor n x \rfloor \le n x \) so it follows that \( \left(1 - p_n \right)^{\lfloor n x \rfloor} \to e^{- r x} \) as \( n \to \infty \). Hence \( F_n(x) \to 1 - e^{-r x} \) as \( n \to \infty \), which is the CDF of the exponential distribution.
Note that the condition \( n p_n \to r \) as \( n \to \infty \) is the same condition required for the convergence of the binomial distribution to the Poisson that we studied in the last section.
Special Families
The geometric distribution on \( \N \) is an infinitely divisible distribution and is a compound Poisson distribution. For the details, visit these individual sections and see the next section on the negative binomial distribution.
Examples and Applications
Simple Exercises
A standard, fair die is thrown until an ace occurs. Let \(N\) denote the number of throws. Find each of the following:
- The probability density function of \(N\).
- The mean of \(N\).
- The variance of \(N\).
- The probability that the die will have to be thrown at least 5 times.
- The quantile function of \(N\).
- The median and the first and third quartiles.
Answwer
- \(\P(N = n) = \left(\frac{5}{6}\right)^{n-1} \frac{1}{6}\) for \( n \in \N_+\)
- \(\E(N) = 6\)
- \(\var(N) = 30\)
- \(\P(N \ge 5) = 525 / 1296\)
- \(F^{-1}(r) = \lceil \ln(1 - r) / \ln(5 / 6)\rceil\) for \( r \in (0, 1)\)
- Quartiles \(q_1 = 2\), \(q_2 = 4\), \(q_3 = 8\)
A type of missile has failure probability 0.02. Let \(N\) denote the number of launches before the first failure. Find each of the following:
- The probability density function of \(N\).
- The mean of \(N\).
- The variance of \(N\).
- The probability of 20 consecutive successful launches.
- The quantile function of \(N\).
- The median and the first and third quartiles.
Answer
- \(\P(N = n) \ \left(\frac{49}{50}\right)^{n-1} \frac{1}{50}\) for \( n \in \N_+\)
- \(\E(N) = 50\)
- \(\var(N) = 2450\)
- \(\P(N \gt 20) = 0.6676\)
- \(F^{-1}(r) = \lceil \ln(1 - r) / \ln(0.98)\rceil\) for \( r \in (0, 1)\)
- Quartiles \(q_1 = 15\), \(q_2 = 35\), \(q_3 = 69\)
A student takes a multiple choice test with 10 questions, each with 5 choices (only one correct). The student blindly guesses and gets one question correct. Find the probability that the correct question was one of the first 4.
Answer
0.4
Recall that an American roulette wheel has 38 slots: 18 are red, 18 are black, and 2 are green. Suppose that you observe red or green on 10 consecutive spins. Give the conditional distribution of the number of additional spins needed for black to occur.
Answer
Geometric with \(p = \frac{18}{38}\)
The game of roulette is studied in more detail in the chapter on Games of Chance.
In the negative binomial experiment, set \(k = 1\) to get the geometric distribution and set \(p = 0.3\). Run the experiment 1000 times. Compute the appropriate relative frequencies and empirically investigate the memoryless property \[ \P(V \gt 5 \mid V \gt 2) = \P(V \gt 3) \]
The Petersburg Problem
We will now explore a gambling situation, known as the Petersburg problem , which leads to some famous and surprising results. Suppose that we are betting on a sequence of Bernoulli trials with success parameter \(p \in (0, 1)\). We can bet any amount of money on a trial at even stakes : if the trial results in success, we receive that amount, and if the trial results in failure, we must pay that amount. We will use the following strategy, known as a martingale strategy :
- We bet \(c\) units on the first trial.
- Whenever we lose a trial, we double the bet for the next trial.
- We stop as soon as we win a trial.
Let \( N \) denote the number of trials played, so that \( N \) has the geometric distribution with parameter \( p \), and let \( W \) denote our net winnings when we stop.
\(W = c\)
Proof
The first win occurs on trial \(N\), so the initial bet was doubled \(N - 1\) times. The net winnings are \[W = -c \sum_{i=0}^{N-2} 2^i + c 2^{N-1} = c\left(1 - 2^{N-1} + 2^{N-1}\right) = c\]
Thus, \(W\) is not random and \(W\) is independent of \(p\)! Since \(c\) is an arbitrary constant, it would appear that we have an ideal strategy. However, let us study the amount of money \(Z\) needed to play the strategy.
\(Z = c (2^N - 1)\)
The expected amount of money needed for the martingale strategy is \[ \E(Z) = \begin{cases} \frac{c}{2 p - 1}, & p \gt \frac{1}{2} \\ \infty, & p \le \frac{1}{2} \end{cases} \]
Thus, the strategy is fatally flawed when the trials are unfavorable and even when they are fair, since we need infinite expected capital to make the strategy work in these cases.
Compute \(\E(Z)\) explicitly if \(c = 100\) and \(p = 0.55\).
Answer
$1000
In the negative binomial experiment, set \(k = 1\). For each of the following values of \(p\), run the experiment 100 times. For each run compute \(Z\) (with \(c = 1\)). Find the average value of \(Z\) over the 100 runs:
- \(p = 0.2\)
- \(p = 0.5\)
- \(p = 0.8\)
For more information about gambling strategies, see the section on Red and Black. Martingales are studied in detail in a separate chapter.
The Alternating Coin-Tossing Game
A coin has probability of heads \(p \in (0, 1]\). There are \(n\) players who take turns tossing the coin in round-robin style: player 1 first, then player 2, continuing until player \(n\), then player 1 again, and so forth. The first player to toss heads wins the game.
Let \(N\) denote the number of the first toss that results in heads. Of course, \(N\) has the geometric distribution on \(\N_+\) with parameter \(p\). Additionally, let \(W\) denote the winner of the game; \(W\) takes values in the set \(\{1, 2, \ldots, n\}\). We are interested in the probability distribution of \(W\).
For \(i \in \{1, 2, \ldots, n\}\), \(W = i\) if and only if \(N = i + k n\) for some \(k \in \N\). That is, using modular arithmetic, \[ W = [(N - 1) \mod n] + 1 \]
The winning player \(W\) has probability density function \[ \P(W = i) = \frac{p (1 - p)^{i-1}}{1 - (1 - p)^n}, \quad i \in \{1, 2, \ldots, n\} \]
Proof
This follows from the previous exercise and the geometric distribution of \(N\).
\(\P(W = i) = (1 - p)^{i-1} \P(W = 1)\) for \(i \in \{1, 2, \ldots, n\}\).
Proof
This result can be argued directly, using the memoryless property of the geometric distribution. In order for player \(i\) to win, the previous \(i - 1\) players must first all toss tails. Then, player \(i\) effectively becomes the first player in a new sequence of tosses. This result can be used to give another derivation of the probability density function in the previous exercise.
Note that \(\P(W = i)\) is a decreasing function of \(i \in \{1, 2, \ldots, n\}\). Not surprisingly, the lower the toss order the better for the player.
Explicitly compute the probability density function of \(W\) when the coin is fair (\(p = 1 / 2\)).
Answer
\(\P(W = i) = 2^{n-1} \big/ (2^n - 1), \quad i \in \{1, 2, \ldots, n\}\)
Note from the result above that \(W\) itself has a truncated geometric distribution .
The distribution of \(W\) is the same as the conditional distribution of \(N\) given \(N \le n\): \[ \P(W = i) = \P(N = i \mid N \le n), \quad i \in \{1, 2, \ldots, n\} \]
The following problems explore some limiting distributions related to the alternating coin-tossing game.
For fixed \(p \in (0, 1]\), the distribution of \(W\) converges to the geometric distribution with parameter \(p\) as \(n \uparrow \infty\).
For fixed \(n\), the distribution of \(W\) converges to the uniform distribution on \(\{1, 2, \ldots, n\}\) as \(p \downarrow 0\).
Players at the end of the tossing order should hope for a coin biased towards tails.
Odd Man Out
In the game of
odd man out
, we start with a specified number of players, each with a coin that has the same probability of heads. The players toss their coins at the same time. If there is an
odd man
, that is a player with an outcome different than all of the other players, then the odd player is eliminated; otherwise no player is eliminated. In any event, the remaining players continue the game in the same manner. A slight technical problem arises with just two players, since different outcomes would make both players
odd
. So in this case, we might (arbitrarily) make the player with tails the odd man.
Suppose there are \(k \in \{2, 3, \ldots\}\) players and \(p \in [0, 1]\). In a single round, the probability of an odd man is \[r_k(p) = \begin{cases} 2 p (1 - p), & k = 2 \\ k p (1 - p)^{k-1} + k p^{k-1} (1 - p), & k \in \{3, 4, \ldots\} \end{cases}\]
Proof
Let \(Y\) denote the number of heads. If \(k = 2\), the event that there is an odd man is \(\{Y = 1\}\). If \(k \ge 3\), the event that there is an odd man is \(\{Y \in \{1, k - 1\}\}\). The result now follows since \(Y\) has a binomial distribution with parameters \( k \) and \( p \).
The graph of \(r_k\) is more interesting than you might think.
For \( k \in \{2, 3, \ldots\} \), \(r_k\) has the following properties:
- \(r_k(0) = r_k(1) = 0\)
- \(r_k\) is symmetric about \(p = \frac{1}{2}\)
- For fixed \(p \in [0, 1]\), \(r_k(p) \to 0\) as \(k \to \infty\).
Proof
These properties are clear from the functional form of \( r_k(p) \). Note that \( r_k(p) = r_k(1 - p) \).
For \( k \in \{2, 3, 4\} \), \(r_k\) has the following properties:
- \( r_k \) increases and then decreases, with maximum at \(p = \frac{1}{2}\).
- \( r_k \) is concave downward
Proof
This follows by computing the first derivatives: \(r_2^\prime(p) = 2 (1 - 2 p)\), \(r_3^\prime(p) = 3 (1 - 2 p)\), \(r_4^\prime(p) = 4 (1 - 2 p)^3\), and the second derivatives: \( r_2^{\prime\prime}(p) = -4 \), \( r_3^{\prime\prime}(p) = - 6 \), \( r_4^{\prime\prime}(p) = -24 (1 - 2 p)^2 \).
For \( k \in \{5, 6, \ldots\} \), \(r_k\) has the following properties:
- The maximum occurs at two points of the form \(p_k\) and \(1 - p_k\) where \( p_k \in \left(0, \frac{1}{2}\right) \) and \(p_k \to 0\) as \(k \to \infty\).
- The maximum value \(r_k(p_k) \to 1/e \approx 0.3679\) as \(k \to \infty\).
- The graph has a local minimum at \(p = \frac{1}{2}\).
Proof sketch
Note that \(r_k(p) = s_k(p) + s_k(1 - p)\) where \(s_k(t) = k t^{k-1}(1 - t)\) for \(t \in [0, 1]\). Also, \(p \mapsto s_k(p)\) is the dominant term when \(p \gt \frac{1}{2}\) while \(p \mapsto s_k(1 - p)\) is the dominant term when \(p \lt \frac{1}{2}\). A simple analysis of the derivative shows that \(s_k\) increases and then decreases, reaching its maximum at \((k - 1) / k\). Moreover, the maximum value is \(s_k\left[(k - 1) / k\right] = (1 - 1 / k)^{k-1} \to e^{-1}\) as \(k \to \infty\). Also, \( s_k \) is concave upward and then downward, wit inflection point at \( (k - 2) / k \).
Suppose \(p \in (0, 1)\), and let \(N_k\) denote the number of rounds until an odd man is eliminated, starting with \(k\) players. Then \(N_k\) has the geometric distribution on \(\N_+\) with parameter \(r_k(p)\). The mean and variance are
- \(\mu_k(p) = 1 \big/ r_k(p)\)
- \(\sigma_k^2(p) = \left[1 - r_k(p)\right] \big/ r_k^2(p)\)
As we might expect, \(\mu_k(p) \to \infty\) and \(\sigma_k^2(p) \to \infty\) as \(k \to \infty\) for fixed \(p \in (0, 1)\). On the other hand, from the result above , \(\mu_k(p_k) \to e\) and \(\sigma_k^2(p_k) \to e^2 - e\) as \(k \to \infty\).
Suppose we start with \(k \in \{2, 3, \ldots\}\) players and \(p \in (0, 1)\). The number of rounds until a single player remains is \(M_k = \sum_{j = 2}^k N_j\) where \((N_2, N_3, \ldots, N_k)\) are independent and \(N_j\) has the geometric distribution on \(\N_+\) with parameter \(r_j(p)\). The mean and variance are
- \(\E(M_k) = \sum_{j=2}^k 1 \big/ r_j(p)\)
- \(\var(M_k) = \sum_{j=2}^k \left[1 - r_j(p)\right] \big/ r_j^2(p)\)
Proof
The form of \(M_k\) follows from the previous result: \(N_k\) is the number of rounds until the first player is eliminated. Then the game continues independently with \(k - 1\) players, so \(N_{k-1}\) is the number of additional rounds until the second player is eliminated, and so forth. Parts (a) and (b) follow from the previous result and standard properties of expected value and variance.
Starting with \(k\) players and probability of heads \(p \in (0, 1)\), the total number of coin tosses is \(T_k = \sum_{j=2}^k j N_j\). The mean and variance are
- \(\E(T_k) = \sum_{j=2}^k j \big/ r_j(p)\)
- \(\var(T_k) = \sum_{j=2}^k j^2 \left[1 - r_j(p)\right] \big/ r_j^2(p)\)
Proof
As before, the form of \(M_k\) follows from result above : \(N_k\) is the number of rounds until the first player is eliminated, and each these rounds has \(k\) tosses. Then the game continues independently with \(k - 1\) players, so \(N_{k-1}\) is the number of additional rounds until the second player is eliminated with each round having \(k - 1\) tosses, and so forth. Parts (a) and (b) also follow from the result above and standard properties of expected value and variance.
Number of Trials Before a Pattern
Consider again a sequence of Bernoulli trials \( \bs X = (X_1, X_2, \ldots) \) with success parameter \( p \in (0, 1) \). Recall that the number of trials \( M \) before the first success (outcome 1) occurs has the geometric distribution on \( \N \) with parameter \( p \). A natural generalization is the random variable that gives the number of trials before a specific finite sequence of outcomes occurs for the first time. (Such a sequence is sometimes referred to as a word from the alphabet \( \{0, 1\} \) or simply a bit string ). In general, finding the distribution of this variable is a difficult problem, with the difficulty depending very much on the nature of the word. The problem of finding just the expected number of trials before a word occurs can be solved using powerful tools from the theory of renewal processes and from the theory of martingalges.
To set up the notation, let \( \bs x \) denote a finite bit string and let \( M_{\bs x} \) denote the number of trials before \( \bs x \) occurs for the first time. Finally, let \( q = 1 - p \). Note that \( M_{\bs x} \) takes values in \( \N \). In the following exercises, we will consider \( \bs x = 10 \), a success followed by a failure. As always, try to derive the results yourself before looking at the proofs.
The probability density function \( f_{10} \) of \( M_{10} \) is given as follows:
- If \( p \ne \frac{1}{2} \) then \[ f_{10}(n) = p q \frac{p^{n+1} - q^{n+1}}{p - q}, \quad n \in \N \]
- If \(p = \frac{1}{2}\) then \( f_{10}(n) = (n + 1) \left(\frac{1}{2}\right)^{n+2} \) for \( n \in \N \).
Proof
For \( n \in \N \), the event \(\{M_{10} = n\}\) can only occur if there is an initial string of 0s of length \( k \in \{0, 1, \ldots, n\} \) followed by a string of 1s of length \( n - k \) and then 1 on trial \( n + 1 \) and 0 on trial \( n + 2 \). Hence \[ f_{10}(n) = \P(M_{10} = n) = \sum_{k=0}^n q^k p^{n - k} p q, \quad n \in \N \] The stated result then follows from standard results on geometric series.
It's interesting to note that \( f \) is symmetric in \( p \) and \( q \), that is, symmetric about \( p = \frac{1}{2} \). It follows that the distribution function, probability generating function, expected value, and variance, which we consider below, are all also symmetric about \( p = \frac{1}{2} \). It's also interesting to note that \( f_{10}(0) = f_{10}(1) = p q \), and this is the largest value. So regardless of \( p \in (0, 1) \) the distribution is bimodal with modes 0 and 1.
The distribution function \( F_{10} \) of \( M_{10} \) is given as follows:
- If \( p \ne \frac{1}{2} \) then \[ F_{10}(n) = 1 - \frac{p^{n+3} - q^{n+3}}{p - q}, \quad n \in \N \]
- If \( p = \frac{1}{2} \) then \( F_{10} = 1 - (n + 3) \left(\frac{1}{2}\right)^{n+2} \) for \( n \in \N \).
Proof
By definition, \(F_{10}(n) = \sum_{k=0}^n f_{10}(k)\) for \( n \in \N \). The stated result then follows from the previous theorem, standard results on geometric series, and some algebra.
The probability generating function \( P_{10} \) of \( M_{10} \) is given as follows:
- If \( p \ne \frac{1}{2} \) then \[ P_{10}(t) = \frac{p q}{p - q} \left(\frac{p}{1 - t p} - \frac{q}{1 - t q}\right), \quad |t| \lt \min \{1 / p, 1 / q\} \]
- If \( p = \frac{1}{2} \) then \(P_{10}(t) = 1 / (t - 2)^2\) for \( |t| \lt 2 \)
Proof
By definition, \[ P_{10}(t) = \E\left(t^{M_{10}}\right) = \sum_{n=0}^\infty f_{10}(n) t^n \] for all \( t \in \R \) for which the series converges absolutely. The stated result then follows from the theorem above, and once again, standard results on geometric series.
The mean of \( M_{10} \) is given as follows:
- If \( p \ne \frac{1}{2} \) then \[ \E(M_{10}) = \frac{p^4 - q^4}{p q (p - q)} \]
- If \( p = \frac{1}{2} \) then \( \E(M_{10}) = 2 \).
Proof
Recall that \( \E(M_{10}) = P^\prime_{10}(1) \) so the stated result follows from calculus, using the previous theorem on the probability generating function. The mean can also be computed from the definition \( \E(M_{10}) = \sum_{n=0}^\infty n f_{10}(n) \) using standard results from geometric series, but this method is more tedious.
The graph of \( \E(M_{10}) \) as a function of \( p \in (0, 1) \) is given below. It's not surprising that \( \E(M_{10}) \to \infty \) as \( p \downarrow 0 \) and as \( p \uparrow 1 \), and that the minimum value occurs when \( p = \frac{1}{2} \).
The variance of \( M_{10} \) is given as follows:
- If \( p \ne \frac{1}{2} \) then \[ \var(M_{10}) = \frac{2}{p^2 q^2} \left(\frac{p^6 - q^6}{p - q}\right) + \frac{1}{p q} \left(\frac{p^4 - q^4}{p - q}\right) - \frac{1}{p^2 q^2}\left(\frac{p^4 - q^4}{p - q}\right)^2 \]
- If \( p = \frac{1}{2} \) then \( \var(M_{10}) = 4 \).
Proof
Recall that \( P^{\prime \prime}_{10}(1) = \E[M_{10}(M_{10} - 1)] \), the second factorial moment, and so \[ \var(M_{10}) = P^{\prime \prime}_{10}(1) + P^\prime_{10}(1) - [P^\prime_{10}(1)]^2 \] The stated result then follows from calculus and the theorem above giving the probability generating function.


12.2: Variation in Diversification Rates across Clades
We know from analyses of tree balance that the tree of life is more imbalanced than birth-death models predict. We can explore this variation in diversification rates by allowing birth and death models to vary along branches in phylogenetic trees. The simplest scenario is when one has a particular prediction about diversification rates to test. For example, we might wonder if diversification rates in one clade (clade A in Figure 12.2) are higher than in the rest of the phylogenetic tree. We can test this hypothesis by fitting a multiple-rate birth-death model.
The simplest method to carry out this test is by using model selection in a ML framework (Rabosky et al. 2007) . To do this, we first fit a constant-rates birth-death model to the entire tree, and calculate a likelihood. We can then fit variable-rates birth-death models to the data, comparing the fit of these models using either likelihood ratio tests or A I C C . The simplest way to fit a variable-rates model is to adapt the likelihood formula in equation 11.18 (or eq. 11.24 if species are unsampled). We calculate the likelihood in two parts, one for the background part of the tree (with rates λ B and μ B ) and one for the focal clade that may have different diversification dynamics (with rates λ A and μ A ). We can then compare this model to one where speciation and extinction rates are constant through time.
Consider the example in Figure 12.2. We would like to know whether clade A has speciation and extinction rates, λ A and μ A , that differ from the background rates, λ B and μ B – we will call this a “variable rates” model. The alternative is a “constant rates” model where the entire clade has constant rate parameters λ T and μ T . These two models are nested, since the constant-rates model is a special case of the variable rates model where λ T = λ A = λ B and μ T = μ A = μ B . Calculating the likelihood for these two models is reasonably straightforward - we simply calculate the likelihood for each section of the tree using the relevant equation from Chapter 11, and then multiply the likelihoods from the two parts of the tree (or add the log-likelihoods) to get the overall likelihood.
For a real example, let’s look at the phylogenetic tree of amphibians and evaluate the hypothesis that the tailed and New Zealand frogs, sister clade to the rest of frogs, diversified at a slower rate than other amphibians (Figure 12.3). We can use the phylogenetic "backbone" tree from Jetz and Pyron (Jetz and Pyron 2018) , assigning diversities based on the classification associated with that publication. We can then calculate likelihoods based on Equation 11.24.
We can calculate the likelihood of the constant rates model, with two parameters λ T and μ T , to a variable rates model with four parameters λ l i o p , μ l i o p , λ o t h e r , and μ o t h e r . For this example, we obtain the following results.
| Model | Parameter estimates | ln-Likelihood | A I C c |
|---|---|---|---|
| Constant rates | λ T = 0.30 | -1053.9 | 2111.8 |
| μ T = 0.28 | |||
| Variable rates | l a m b d a l i o p = 0.010 | -1045.4 | 2101.1 |
| μ l i o p = 0.007 | |||
| l a m b d a o t h e r = 0.29 | |||
| μ o t h e r = 0.27 |
With a difference in A I C c of more than 10, we see from these results that there is good reason to think that there is a difference in diversification rates in these "oddball" frogs compared to the rest of the amphibians.
Of course, more elaborate comparisons are possible. For example, one could compare the fit of four models, as follows: Model 1, constant rates; Model 2, speciation rate in clade A differs from the background; Model 3, extinction rate in clade A differs from the background; and Model 4, both speciation and extinction rates in clade A differ from the background. In this case, some of the pairs of models are nested – for example, Model 1 is a special case of Model 2, which is, in turn, a special case of Model 4 – but all four do not make a nested series. Here we benefit from using a model selection approach based on A I C C . We can fit all four models and use their relative number of parameters to calculate A I C C scores. We can then calculate A I C C weights to evaluate the relative support for each of these four models. (As an aside, it might be difficult to differentiate among these four possibilities without a lot of data!)
But what if you do not have an a priori reason to predict differential diversification rates across clades? Or, what if the only reason you think one clade might have a different diversification rate than another is that it has more species? (Such reasoning is circular, and will wreak havoc with your analyses!) In such cases, we can use methods that allow us to fit general models where diversification rates are allowed to vary across clades in the tree. Available methods use stepwise AIC (MEDUSA, Alfaro et al. 2009; but see May and Moore 2016) , or reversible-jump Bayesian MCMC (Rabosky 2014, 2017; but see Moore et al. 2016) .
For example, running a stepwise-AIC algorithm on the amphibian data (Alfaro et al. 2009) results in a model with 11 different speciation and extinction regimes (Figure 12.4). This is good evidence that diversification rates have varied wildly through the history of amphibians.
One note: all current approaches fit a model where birth and death rates change at discrete times in the phylogenetic tree - that is, along certain branches in the tree leading to extant taxa. One might wish for an approach, then, that models such changes - using, for example, a Poisson process - and then locates the changes on the tree. However, we still lack the mathematics to solve for E ( t ) (e.g. Equation 11.19) under such a model (Moore et al. 2016) . Given that, we can view current implementations of models where rates vary across clades as an approximation to the likelihood, and one that discounts the possibility of shifts in speciation and/or extinction rates among any clades that did not happen to survive until the present day (Rabosky 2017) - and we are stuck with that until a better alternative is developed!


6.2: Test-Taking Strategy Specifics
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Dave Dillon
“By failing to prepare, you are preparing to fail.”
– Benjamin Franklin
Discipline, Preparation and Execution
From what I have seen in my own educational experiences, along with coaching and teaching at the college and university level for over 20 years, test taking (with few exceptions) comes down to discipline, preparation and execution. Students wanting to be successful have to have the self-discipline to schedule time to study well in advance of the exam. They have to actually do the work: the preparation needed in order to have the best opportunity for success on the exam. Then they must execute: they have to be able to apply their preparation accordingly and perform well on the exam.
Preparation for an exam is not glamorous. It’s easy to find other things to do that are more interesting and fun. Students need to keep themselves motivated with their “eyes on the prize.” Think of it like this: if the most important event of your life was coming up and you wanted to perform to the best of your ability in that event, you would likely spend some time preparing for it, rehearsing for it, practicing it, etc. A student may argue that an exam they will be taking would not be the most important event of their life, but think back to the chapter on Passion. If you’re already spending the time, effort, energy and money to attend college, why not do it to the best of your ability?
It would be beneficial to spread this preparation and practice out over time and prepare periodically rather than to wait until the last minute and binge study or cram. Your preparation would not be the same and this may affect your test score. Binge studying and cramming also are not healthy. Staying up late puts stress on our brain and body, and not getting adequate sleep places our bodies at risk for getting sick.
“One of the most important keys to success is having the discipline to do what you know you should do, even when you don’t feel like doing it.”
– Unknown
“The will to succeed is important, but what’s more important is the will to prepare.”
– Bobby Knight
Everyone wants to be successful. When the exam is passed out, everyone wants to perform well. But what often separates successful students and less successful students is the preparation time put in.
Studying the right thing is a process and a skill. As you gain more experience, you will learn how to become better at knowing what to study. It can be very frustrating to spend a lot of time preparing and studying and then finding out that what you studied was not on the exam. You will see a lot of variance with exams due to different instructors, classes and types of tests. The better you become at predicting what will be on the exam and study accordingly, the better you will perform on your exams. Try placing yourself in your instructor’s shoes and design questions you think your instructor would ask. It’s often an eye-opening experience for students and a great study strategy.
Preparation for Exam Strategies
Find out as much about the exam in advance as you can.
Some professors will tell you how many questions there will be, what format the exam will be in, how much time you will have, etc., and others will not. I encourage students to ask questions about the exam if there is not information given. I also encourage students to ask those questions before class, after class, in professors’ office hours or via e-mail rather than during class.
Know the test
If you know how many questions, what the format is, and/or how much time you will have, you can start to mentally prepare for the exam much more so than if you are coming in with no information. There are two more important aspects that you may or may not know: a) what will be covered or asked on the exam; b) how the exam will be scored. Obviously, the more you know about what will be covered, the easier it is for you to be able to prepare for the exam. Most exam scoring is standardized, but not always.
Look for opportunities where some areas of the exam are worth more points than others. For example: An exam consists of 21 questions, with 10 being True/False, 10 being multiple choice, and one essay question. The T/F questions are worth 1 point each (10 points), the multiple-choice questions are worth 2 points each (20 points), and the essay question is worth 30 points. We know that the essay question is the most valuable (it is worth half of the value of the exam). And we should allocate our time for it accordingly. I would advise starting with the essay question. Do a quick analysis of time to be able to spend your time on the exam wisely. You want to spend some time with the exam question since it is so valuable, without sacrificing adequate time to ensure the T/F and multiple-choice questions are answered.
Often, the order of the exam in this scenario will be: T/F first, multiple choice second and essay third. Most students will go in the chronological order of the exam, but a savvy student would start with the essay. If an exam were to last for 30 minutes with this format of questions, I might recommend a student spent 15 minutes on the essay question, ten minutes on the multiple choice, three minutes on the T/F and two minutes reviewing their answer.
Also, look for situations where exams penalize students for incorrectly answering a question. This does not occur very often, but is the case with some exams. With the SAT for example, students are awarded one point for a correct answer and ¼ of a point subtracted for an incorrect answer. Points are not awarded nor subtracted for leaving a question blank. Thus, the strategy for a multiple-choice question is: if you can narrow down the potentially correct answer to two rather than four or five, it is statistically advantageous to answer the question and guess between the two answers; however, if a student had no idea if any of the answers were correct or incorrect, it would be best to leave the answer blank. Remember, this is rare, but it is important to understand the strategy when students take these exams.
In conclusion, the more information you have about the exam, the better you can prepare for content, allocation of time spent on aspects of the exam, and the more confident you will be in knowing how and when to attempt to answer questions.
Take care of your body
Before the exam, it is important to prepare your brain and body for optimal performance for your exam. Do not cram the night before. Get a good night’s sleep.  I recommend exercising the day before and if possible a few hours before the exam.
Strategies for Specific Exam Formats
True or False Questions
Look for qualifiers. A qualifier is a word that is absolute. Examples are: all, never, no, always, none, every, only, entirely. They are often seen in false statements. This is because it is more difficult to create a true statement using a qualifier like never, no, always, etc. For example, “All cats chase mice.” Cats may be known for chasing mice, but not all of them do so. The answer here is false and the qualifier “all” gave us a tip. Qualifiers such as: sometimes, many, some, most, often, and usually are commonly found in true statements. For example: “Most cats chase mice.” This is true and the qualifier “most” gave us a tip.
Make sure to read the entire statement. All parts of a sentence must be true if the whole statement is to be true. If one part of it is false, the whole sentence is false. Long sentences are often false for this reason.
Students should guess on True or False questions they do not know the answer to unless there is a penalty for an incorrect answer.
Multiple Choice Questions
Think of multiple choice questions as four (or five) true or false statements in one. One of the statements is true (the correct answer) and the others will be false. Apply the same strategy toward qualifiers. If you see an absolute qualifier in one of the answer choices, it is probably false and not the correct answer. Try to identify the true statement. If you can do this, you have the answer as there is only one. If you cannot do this at first, try eliminating answers you know to be false.
If there is no penalty for incorrect answers, my suggestion is to guess if you are not certain of the answer. If there is a penalty for incorrect answers, common logic is to guess if you can eliminate two of the answers as incorrect (pending what the penalty is). If there’s a penalty and you cannot narrow down the answers, it’s best to leave it blank. You may wish to ask your instructor for clarification.
Answers that are strange and unrelated to the question are usually false. If two answers have a word that looks or sounds similar, one of those is usually correct. For example: abductor/ adductor. If you see these as two of the four or five choices, one of them is usually correct. Also look for answers that are grammatically incorrect. These are usually incorrect answers. If you have to completely guess, choose B or C. It is statistically proven to be correct more than 25 percent of the time. If there are four answers for each question, and an exam had standardized the answers, each answer on the exam A, B, C and D would be equal. But most instructors do not standardize their answers, and more correct answers are found in the middle (B and C then the extremes A and D or E). “People writing isolated four-choice questions hide the correct answer in the two middle positions about 70% of the time.” [1] This is 20 percent more correct answers found in B or C than a standardized exam with equal correct answers for each letter.
Matching Questions
Although less common than the other types of exams, you will likely see some matching exams during your time in college. First, read the instructions and take a look at both lists to determine what the items are and their relationship. It is especially important to determine if both lists have the same number of items and if all items are to be used, and used only once.
Matching exams become much more difficult if one list has more items than the other or if items either might not be used or could be used more than once. If your exam instructions do not discern this, you may wish to ask your instructor for further clarification. I advise students to take a look at the whole list before selecting an answer because a more correct answer may be found further into the list. Mark items when you are sure you have a match (pending the number of items in the list this may eliminate answers for the future). Guessing (if needed) should take place once you have selected answers you are certain about.
Short-Answer Questions
Read all of the instructions first. Budget your time and then read all of the questions. Answer the ones you know best or feel the most confident with. Then go back to the other ones. If you do not know the answer and there is no penalty for incorrect answers, guess. Use common sense. Sometimes instructors will award partial credit for a logical answer that is related even if it is not the correct answer.
Essay Questions
Keep in mind that knowing the format of the exam can help you determine how to study. If I know that I am taking a True-False exam, I know that I will need to discern whether a statement is True or False. I will need to know subject content for the course. But if I am studying for short answer and especially for essay questions, I must know a lot more. For essay questions, I must have much greater content knowledge and be able to make a coherent argument that answers the question using information from textbooks, lectures or other course materials. I have to place a lot more time and thought into studying for an essay exam than for True-False or Multiple-Choice exams.
Read the essay question(s) and the instructions first. Plan your time wisely and organize your answer before you start to write. Address the answer to the question in your first or second sentence. It may help to restate the original question. Write clearly and legibly. Instructors have difficulty grading essays that they cannot read. Save some time for review when you have finished writing to check spelling, grammar and coherent thought in your answer. Make sure you have addressed all parts of the essay question.
During the Exam:
Always read the directions first. Read them thoroughly.
Preview the exam to help you allocate proper time for each area.
Skip questions if you do not know the answer but make a mark somewhere to ensure you are able to go back to those questions (you may need to reallocate your prepared time for this depending on how many there are).
Allocate some time to review your answers before submitting your exam or the exam time expiring.
One of the biggest mistakes that students make after they take an exam in a course is that they do not use the exam for the future. The exam contains a lot of information that can be helpful in studying for future exams. Students that perform well on an exam often put it away thinking they do not need it anymore. Students who do poorly on an exam often put it away, not wanting to think about it any further.
In both cases, students are missing out on the value of reviewing their exams. It is wise to review exams for three reasons: 1) students should review the answers that were correct because they may see those questions on future exams and it is important to reinforce learning; 2) students should review the answers that were incorrect in order to learn what the correct answer was and why. These questions also may appear on a future exam. In addition, occasionally an answer is marked incorrect, when it should have been marked correct. The student would never know this if they didn’t review their exam; and 3) there is value in reviewing the exam to try to predict what questions or what format will be used by a professor for a future exam in the same course.
Author’s Story
What does it mean to be a “poor test taker?” Think about that. Does it mean that the student has put effort into studying but has difficulty under pressure? Does it mean that a student studies the wrong material? Is the student prepared but does not execute well? Could the student have a learning disability? Are they missing key strategies for taking tests? It could mean any of those things. And while I believe that it may be true that a student may be a “poor test taker,” it does not by any means mean it is permanent. Students willing to work hard and learn can improve their test taking skills and raise their confidence.
In high school chemistry, I earned a D grade. In microeconomics at UC Santa Cruz, I struggled and barely passed. Both of these classes had grading systems that were heavily weighted by exams. I was an above average student in my other classes in both high school and college, and so I was told (and believed for a long time) that I was not a “good test taker.” I no longer believe this. I will not make excuses. The responsibility for the grade I earned rests solely with me. I mention this because many students come to my office and complain that their professor grades in a way that is unfair to them because so much emphasis is on exams and that they understand the material but are “not good test takers.” It may be the case that my grade or another students’ grade would have been better had the course grade been determined with less weight levied to exams. That to me is irrelevant. It does not absolve me from being responsible for knowing what the grading metrics were at the beginning of the course and choosing to continue in the course after learning this knowledge. In the end, I realized I was not a poor test taker and that I needed to spend more time studying and preparing to perform well on exams. I changed my attitude from being afraid of exams to one of believing I was going to do well regardless of the class or the format of the exam. Once I had confidence that I had the necessary tools and was willing to work hard, it changed my entire perspective and experience with exams.
Note: if you think you may have a learning disability and would like to get assessed, contact your college to see what steps to follow to begin the assessment process.
Thinking about excuses reminds me of a story about Mike Krzyzewski, the successful Duke University men’s basketball coach. Before Krzyzewski coached, he was a student at the United States Military Academy at West Point. One day in his first year, Krzyzweski and his roommate were walking and his roommate stepped in a puddle. Mud splashed onto Krzyzewski’s uniform. Immediately, an upperclassman screamed at Krzyzewski about not knowing the rules of wearing a clean uniform. Plebes (first-year students at military academies) are allowed three answers when asked a question: “Yes sir!” “No sir!” And “No excuse sir!” Krzyzewski repeated himself, “No excuse sir!” Despite wanting to explain what had happened and that it was not his fault, he realized that there was no excuse. It was his responsibility to keep a clean uniform.
Licenses and Attributions:
Content previously copyrighted, published in Blueprint for Success in College: Indispensable Study Skills and Time Management Strategies (by Dave Dillon), now licensed as CC BY Attribution.
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Yigal Attali and Maya Bar-Hillel, “Guess Where: The Position of Correct Answers in Multiple-Choice Test Items as a Psychometric Variable,”
Journal of Educational Measurement
40 no. 2 (2003):109-128. ↵


9.4B: The Conflict Perspective- Class Conflict and Scarce Resources
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Conflict theory of stratification holds that inequality is harmful to society because it creates a fixed system of winners and losers.
Learning Objectives
- Compare the conflict theory of inequality to the funcionalist theory of inequality
Key Points
- According to conflict theory, capitalist economic competition unfairly privileges the rich, who have the power to perpetrate an unfair system that works to their advantage.
- “Losers” who are at the bottom of the social stratification have little opportunity to improve their situation, since those at the top tend to have far more political and economic power.
- Functionalists argue against the conflict theory approach by contending that people don’t always act out of economic self-interest, and that people who want to succeed can do so through hard work.
Key Terms
- conflict theory : A social science perspective that holds that stratification is dysfunctional and harmful in society, with inequality perpetuated because it benefits the rich and powerful at the expense of the poor.
- social stratification : The hierarchical arrangement of social classes, or castes, within a society.
- tax break : A deduction in tax that is given in order to encourage a certain economic activity or a social objective.
Conflict theorists argue that stratification is dysfunctional and harmful in society. According to conflict theory, social stratification benefits the rich and powerful at the expense of the poor. Thus, it creates a system of winners and losers that is maintained by those who are on the top. The people who are losers do not get a fair chance to compete, and thus are stuck on the bottom. For example, many wealthy families pay low wages to nannies to care for their children, to gardeners to attend to their rose gardens, and to maids to pick up their dirty socks. These low wage workers do not make enough to move beyond a paycheck-to-paycheck lifestyle, and have no means to move ahead. Therefore, conflict theorists believe that this competitive system, together with the way the game is “fixed”, ends up creating and perpetuating stratification systems.
According to conflict theory, capitalism, an economic system based on free-market competition, particularly benefits the rich by assuming that the “trickle down” mechanism is the best way to spread the benefits of wealth across society. Governments that promote capitalism often establish corporate welfare through direct subsidies, tax breaks, and other support that benefit big businesses. They assume that the market will allow these benefits to the rich to make their way to the poor through competition. For example, the Walton family, the owners of Wal-Mart, receives enormous tax breaks. Whether the benefits of these tax breaks have made their way down to ordinary people through better business practices or better working conditions for Wal-Mart employees is questionable. Conflict theorists would argue that they haven’t, but rather have been used by the Walton family to solidify the patterns of stratification that keep the family rich.
Functionalists criticize this approach by arguing that people do not always act largely out of economic self-interest. For example, Chuck Feeney, the creator of Duty Free Shoppers, has given $4 billion to charities. Bill Gates has given 58% of his wealth to charity. Functionalists also argue that conflict theorists underestimate people’s ability to move upward in society. They argue that if people really want to succeed, they can do so through hard work.


2.1: Hello World
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Overview
A “ Hello, world !” program is a computer program that outputs or displays “Hello, world!” to a user. Being a very simple program in most programming languages, it is often used to illustrate the basic syntax of a programming language for a working program, and as such is often the very first program people write. [1]
Discussion
A “Hello, world!” program is traditionally used to introduce novice programmers to a programming language. “Hello, world!” is also traditionally used in a sanity test to make sure that a computer language is correctly installed, and that the operator understands how to use it. [2]
The tradition of using the phrase “Hello, world!” as a test message was influenced by an example program in the seminal book The C Programming Language . The example program from that book prints “hello, world” (without capital letters or exclamation mark), and was inherited from a 1974 Bell Laboratories internal memorandum by Brian Kernighan. [3]
In addition to displaying “Hello, world!”, a “Hello, world!” program might include comments. A comment is a programmer-readable explanation or annotation in the source code of a computer program. They are added with the purpose of making the source code easier for humans to understand, and are generally ignored by compilers and interpreters. The syntax of comments in various programming languages varies considerably. [4]
Pseudocode
Function Main ... This program displays "Hello world!" Output "Hello world!" End
Output
Hello world!
Each code element represents: [5]
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Function Main
begins the main function -
...
begins a comment -
Output
indicates the following value(s) will be displayed or printed -
"Hello world!"
is the literal string to be displayed -
End
ends a block of code
Flowchart
Examples
The following pages provide examples of “Hello, world!” programs in different programming languages. Each page includes an explanation of the code elements that comprise the program and links to IDEs you may use to test the program.
Key Terms
- comment
- A programmer-readable explanation or annotation in the source code of a computer program.
References
- Wikiversity: Computer Programming
- Flowgorithm – Flowchart Programming Language
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Wikipedia: "Hello, World!" program
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Wikipedia: "Hello, World!" program
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Wikipedia: "Hello, World!" program
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Wikipedia: Comment (computer programming)
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Wikibooks: Programming Fundamentals/Hello World
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2.4: Introduction to Quantification
In this section and the next, we introduce first-order logic —also referred to as predicate logic , quantificational logic , and first-order predicate calculus . The sentence “x > 0” is not itself a proposition because its truth value depends on x. In this case, we say that x is a free variable . A sentence with at least one free variable is called a predicate (or open sentence ). To turn a predicate into a proposition, we must either substitute values for each free variable or “quantify” the free variables. We will use notation such as \(P (x)\) and \(Q(a,b)\) to represent predicates with free variables x and a,b, respectively. The letters “P ” and “Q” that we used in the previous sentence are not special; we can use any letter or symbol we want. For example, each of the following represents a predicate with the indicated free variables.
- S(x) B “x2 − 4 = 0”
- L(a,b) B “a < b”
- F(x,y) B “x is friends with y”
Note that we used quotation marks above to remove some ambiguity. What would S(x) = x 2 − 4 = 0 mean? It looks like S(x) equals 0, but actually we want S(x) to represent the whole sentence “x 2 − 4 = 0”. Also, notice that the order in which we utilize the free variables might matter. For example, compare L(a,b) with L(b,a).
One way we can make propositions out of predicates is by assigning specific values to the free variables. That is, if P (x) is a predicate and x0 is specific value for x, then P (x0 ) is now a proposition that is either true or false.
Problem 2.58. Consider S(x) and L(a,b) as defined above. Determine the truth values of S(0), S(−2), L(2,1), and L(−3,−2). Is L(2,b) a proposition or a predicate?
Besides substituting specific values for free variables in a predicate, we can also make a claim about which values of the free variables apply to the predicate.
Problem 2.59. Both of the following sentences are propositions. Decide whether each is true or false. What would it take to justify your answers?
(a) For all x ∈ R , x 2 − 4 = 0.
(b) There exists x ∈ R such that x 2 − 4 = 0.
Definition 2.60. “For all” is the universal quantifier and “there exists. . . such that” is the existential quantifier .
In mathematics, the phrases “for all”, “for any”, “for every”, and “for each” can be used interchangeably (even though they might convey slightly different meanings in colloquial language). We can replace “there exists. . . such that” with phrases like “for some” (possibly with some tweaking of the wording of the sentence). It is important to note that the existential quantifier is making a claim about “at least one”, not “exactly one.”
Variables that are quantified with a universal or existential quantifier are said to be bound . To be a proposition, all variables of a predicate must be bound.
We must take care to specify the collection of acceptable values for the free variables. Consider the sentence “For all x, x > 0.” Is this sentence true or false? The answer depends on what set the universal quantifier applies to. Certainly, the sentence is false if we apply it for all x ∈ Z. However, the sentence is true for all x ∈ N . Context may resolve ambiguities, but otherwise, we must write clearly: “For all x ∈ Z , x > 0” or “For all x ∈ N , x > 0.” The collection of intended values for a variable is called the universe of discourse .
Problem 2.61. Suppose our universe of discourse is the set of integers.
(a) Provide an example of a predicate P (x) such that “For all x, P (x)” is true.
(b) Provide an example of a predicate Q(x) such that “For all x, Q(x)” is false while “There exists x such that Q(x)” is true.
If a predicate has more than one free variable, then we can build propositions by quantifying each variable. However, the order of the quantifiers is extremely important !
Problem 2.62. Let P (x,y) be a predicate with free variables x and y in a universe of discourse U. One way to quantify the variables is “For all x ∈ U, there exists y ∈ U such that P (x,y).” How else can the variables be quantified?
The next problem illustrates that at least some of the possibilities you discovered in the previous problem are not equivalent to each other.
Problem 2.63. Suppose the universe of discourse is the set of people and consider the predicate M(x,y) B “x is married to y”. We can interpret the formal statement “For all x, there exists y such that M(x,y)” as meaning “Everybody is married to somebody.” Interpret the meaning of each of the following statements in a similar way.
(a) For all x, there exists y such that M(x,y).
(b) There exists y such that for all x, M(x,y).
(c) For all x, for all y, M(x,y).
(d) There exists x such that there exists y such that M(x,y).
Problem 2.64. Suppose the universe of discourse is the set of real numbers and consider the predicate F(x,y) B “x = y 2 ”. Interpret the meaning of each of the following statements.
(a) There exists x such that there exists y such that F(x,y).
(b) There exists y such that there exists x such that F(x,y).
(c) For all y, for all x, F(x,y).
There are a couple of key points to keep in mind about quantification. To be a proposition, all variables must be quantified. This can happen in at least two ways:
- The variables are explicitly bound by quantifiers in the same sentence.
- The variables are implicitly bound by preceding sentences or by context. Statements of the form “Let x = ...” and “Assume x ∈ ...” bind the variable x and remove ambiguity.
Also, the order of the quantification is important. Reversing the order of the quantifiers can substantially change the meaning of a proposition.
Quantification and logical connectives (“and,” “or,” “If. . . , then. . . ,” and “not”) enable complex mathematical statements. For example, if f is a function while c and L are real numbers, then the formal definition of lim x →c f (x) = L, which you may have encountered in calculus, is:
For all ε > 0, there exists δ > 0 such that for all x, if 0 < |x − c| < δ, then |f (x) − L| < ε.
In order to study the abstract nature of complicated mathematical statements, it is useful to adopt some notation.
Definition 2.65. The universal quantifier “for all” is denoted \(∀\), and the existential quantifier “there exists. . . such that” is denoted \(∃\).
Using our abbreviations for the logical connectives and quantifiers, we can symbolically represent mathematical propositions.
Problem 2.66. Convert the following propositions into statements using only logical and mathematical symbols. Assume that the universe of discourse is the set of real numbers.
(a) There exists x such that x 2 + 1 is greater than zero.
(b) There exists a natural number n such that n 2 = 36.
(c) For every x, x 2 is greater than or equal to zero.
Problem 2.67. Express the formal definition of a limit (given above Definition 2.65) in logical and mathematical symbols.
If you look closely, many of the theorems that we have encountered up until this point were of the form A(x) =⇒ B(x), where A(x) and B(x) are predicates. For example, consider
Theorem 2.2, which states, “If n is an even integer, then n2 is an even integer.” In this case, “n is an even integer” and “n2 is an even integer” are both predicates. So, it would be reasonable to assume that the entire theorem statement is a predicate. However, it is standard practice to interpret the sentence A(x) =⇒ B(x) to mean (∀x)(A(x) =⇒ B(x)) (where the universe of discourse for x needs to be made clear). We can also retool such statements to “hide” the implication. In particular, (∀x)(A(x) =⇒ B(x)) has the same meaning as (∀x ∈ U0 )B(x), where U0 is the collection of items from the universe of discourse U that makes A(x) true. For example, we could rewrite the statement of Theorem 2.2 as “For every even integer n, n 2 is even.”
Problem 2.68. Reword Theorem 2.7 so that it explicitly reads as a universally quantified statement. Compare with Problem 2.47.
Problem 2.69. Find at least two other instances of theorem statements that appeared earlier in the book and are written in the form A(x) =⇒ B(x). Rewrite each in an equivalent way that makes the universal quantifier explicit while possibly suppressing the implication.
Problem 2.70. Consider the proposition “If ε > 0, then there exists N ∈ N such that 1/N < ε.” Assume the universe of discourse is the set R .
(a) Express the statement in logical and mathematical symbols. Is the statement true?
(b) Reverse the order of the quantifiers to get a new statement. Does the meaning change? If so, how? Is the new statement true?
The symbolic expression (∀x)(∀y) can be abbreviated as \(∀x,y\) as long as x and y are elements of the same universe.
Problem 2.71.
Problem 2.72. Rewrite each of the following propositions in words and determine whether the proposition is true or false.
 Consider the proposition (∀x)(∃y)(xy = 1).
(a) Provide an example of a universe of discourse where this proposition is true.
(b) Provide an example of a universe of discourse where this proposition is false.
To whet your appetite for the next section, consider how you might prove a true proposition of the form “For all x. . . .” If a proposition is false, then its negation is true. How would you go about negating a statement involving quantifiers?


2.21: Language as an Obstacle
- Discuss strategies to avoid language that can confuse or exclude readers
Each of us has a variety of ways of speaking and writing depending on circumstances. You write a thank-you note to your aunt for the socks she sent much differently than you write a thank-you note after a job interview with the vice president of the division. In business communication, the key is to choose language that is direct and easy for your audience to understand.
When you’re writing, it’s important to consider your audience’s understanding compared to your own. For example, if you’re writing a newsletter for customers, you would use much different language than you would if you were writing a product status update to the engineers who initially created the product.
As you write, keep your eye out for these common language challenges:
- Clichés
- Jargon
- Slang
- Euphemisms
- Doublespeak
Cliché
Clichés that we use in everyday conversation ( green with envy , face the music , add insult to injury , etc.) can make your writing sound boring. Merriam-Webster’s dictionary defines a cliché as “a trite phrase or expression; a hackneyed theme, characterization, or situation; something (such as a menu item) that has become overly familiar or commonplace.” [1]
Consider the paragraph and revision below.
Original Paragraph
When I started thinking about getting a new job, I was completely clueless. I knew I wanted to do something really cool, but I was lost about what might fit the bill.
Revised Paragraph
When I started thinking about getting a new job, I was overwhelmed by my options and unsure of what to choose. While I knew I wanted to do something interesting, I was uncertain of what that might be.
| Avoid | Try | Why |
|---|---|---|
| Please resolve this ASAP. | Please resolve this by March 30. | ASAP is not a time we each understand the same way. You might be thinking you will get a call back tomorrow. I might be thinking this research and forms needed will take a week, so seven days is sufficient. |
| Thanks for your help. Now just kick it through the goal posts. | Thanks for your help. Now you just need to finish submitting the paperwork to process the deal. | “Kick it through the goals posts” is common enough, and overused enough, but it is a sport reference that many may not understand. It is not clear what step must be taken next. |
Jargon
The dictionary defines jargon as “the technical terminology or characteristic idiom of a special activity or group.” [2] Since these terms are used within an activity, group, or profession, they’re typically not well understood outside that context. Within the context of a specific group, jargon may help members of the group refer to very specialized concepts, but those outside the group may find the jargon incomprehensible or may misunderstand the intended meaning.
For example, plumbers might use terms like elbow , ABS , sweating the pipes , reducer , flapper , snake , and rough-in . Other plumbers will understand those terms in the context of plumbing, but to non-plumbers, those terms may have different meanings or be entirely nonsensical.
Always keep your audience in mind. If you are addressing only members of a specialized field, using field-specific jargon will signal to your audience that you are also a member of the field and may allow you to talk about higher or specific concepts. However, if you are speaking to a broader group with no specialized knowledge, using jargon will only alienate or confuse your audience. If some technical terms are absolutely necessary when speaking to an audience, be sure to explain each term and its context.
For example, if you are writing a department-wide memo, it would be acceptable to use terms specific to your company or department. If you are writing a newsletter for customers, however, it would be better to avoid jargon and use broader language.
| Avoid | Try | Why |
|---|---|---|
| Your help is appreciated.10-4. | Your help is appreciated. I know exactly what do do with the claim now. | 10-4 started in police and military circles as a radio shorthand. While many may know that it means “ok” or “I understand”, it is not clear to all outside those circles. |
| ATM will work well in this situation. | Asynchronous Transfer Mode will work well in this situation. | If the receiver is not from the technical side of communications, someone might have been thinking about a bank machine. |
Slang
Avoid slang or idiomatic expressions in formal business writing or in academic writing. Slang and idiomatic expressions make your writing sound informal and less credible. They can also make it harder for non-native English speakers to understand you.
| Avoid | Try |
|---|---|
| That cart did not work any more. It was all jacked up. | That cart did not work any more. It had a broken wheel and bent frame. |
| She was dead after all that overtime last weekend. | She was exhausted after all that overtime last week. |
However, there are occasions when slang and idiomatic expressions may be appropriate, depending on who your audience is. If you are writing informally or humorously, slang and idiomatic language may help you better express yourself.
If you are going to use slang, however, make sure that you’re using it correctly. Customers who are “native speakers” of slang (i.e., people who are a part of the demographic you may be targeting by using slang) can very easily spot a marketing professional who doesn’t understand the correct usage of a slang term.
Euphemisms and Doublespeak
Euphemisms are words or phrases used to talk about unpleasant or taboo topics in a more polite way. For example, instead of saying that you are leaving a meeting to urinate, you might say that you are “going to the restroom.” If you are talking about a person who just died, you might say that the person has “passed away.”
Doublespeak is a more deliberate form of euphemism that disguises the meaning of words so that the idea the words represent seems less unpleasant. For example, the act of terminating employment for many people at once may be referred to in doublespeak as downsizing or right-sizing, or a government dropping bombs might say it is servicing the target . Doublespeak language is usually used in bureaucracies and politics and should be avoided whenever possible. In business writing (and all writing), you should avoid using doublespeak.
In business writing, your purpose should not be to hide meaning but to communicate clearly.
| Avoid | Try |
|---|---|
| The department is being right-sized. There is a meeting for the entire staff next Friday. | There will be layoffs in that department. There is a meeting for the entire staff next Friday. |
| The month-end income statement showed a negative cash flow. | The month-end income statement showed a loss. |
- Merriam-Webster "Cliché." Merriam-Webster.com . n.d. Web. 12 June 2018.
- Merriam-Webster, "Jargon." Merriam-Webster.com . n.d. Web. 12 June 2018.
Contributors and Attributions
- Language as an Obstacle. Authored by : Susan Kendall. Provided by : Lumen Learning. License : CC BY: Attribution


4.8: Planes in Rⁿ
- Find the vector and scalar equations of a plane.
Much like the above discussion with lines, vectors can be used to determine planes in \(\mathbb{R}^n\). Given a vector \(\vec{n}\) in \(\mathbb{R}^n\) and a point \(P_0\), it is possible to find a unique plane which contains \(P_0\) and is perpendicular to the given vector.
Below is a video on graphing a plane using traces.
Let \(\vec{n}\) be a nonzero vector in \(\mathbb{R}^n\). Then \(\vec{n}\) is called a normal vector to a plane if and only if \[\vec{n} \bullet \vec{v} = 0\nonumber \] for every vector \(\vec{v}\) in the plane.
In other words, we say that \(\vec{n}\) is orthogonal (perpendicular) to every vector in the plane.
Consider now a plane with normal vector given by \(\vec{n}\), and containing a point \(P_0\). Notice that this plane is unique. If \(P\) is an arbitrary point on this plane, then by definition the normal vector is orthogonal to the vector between \(P_0\) and \(P\). Letting \(\overrightarrow{0P}\) and \(\overrightarrow{0P_0}\) be the position vectors of points \(P\) and \(P_0\) respectively, it follows that \[\vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) = 0\nonumber \] or \[\vec{n} \bullet \overrightarrow{P_0P} = 0\nonumber \]
The first of these equations gives the vector equation of the plane.
Let \(\vec{n}\) be the normal vector for a plane which contains a point \(P_0\). If \(P\) is an arbitrary point on this plane, then the vector equation of the plane is given by \[\vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) = 0\nonumber \]
Below is a video on determining the equation of a plane using a normal vector.
Notice that this equation can be used to determine if a point \(P\) is contained in a certain plane.
Let \(\vec{n} = \left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right]\) be the normal vector for a plane which contains the point \(P_0 = \left( 2, 1, 4 \right)\). Determine if the point \(P = \left( 5, 4, 1 \right)\) is contained in this plane.
Solution
By Definition \(\PageIndex{2}\) , \(P\) is a point in the plane if it satisfies the equation \[\vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) = 0\nonumber \]
Given the above \(\vec{n}\), \(P_0\), and \(P\), this equation becomes \[\begin{aligned} \left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right] \bullet \left( \left[ \begin{array}{r} 5 \\ 4 \\ 1 \end{array} \right] - \left[ \begin{array}{r} 2 \\ 1 \\ 4 \end{array} \right] \right) &= \left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right] \bullet \left( \left[ \begin{array}{r} 3 \\ 3 \\ -3 \end{array} \right] \right) \\ &= 3 + 6 - 9 = 0\end{aligned}\]
Therefore \(P = ( 5, 4, 1)\) is contained in the plane.
Suppose \(\vec{n} = \left[ \begin{array}{c} a \\ b \\ c \end{array} \right]\), \(P = \left( x,y,z\right)\) and \(P_0 = (x_0, y_0, z_0 )\).
Then \[\begin{aligned} \vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) &= 0 \\ \left[ \begin{array}{c} a \\ b \\ c \end{array} \right] \bullet \left( \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] - \left[ \begin{array}{c} x_0 \\ y_0 \\ z_0 \end{array} \right] \right) &= 0 \\ \left[ \begin{array}{c} a \\ b \\ c \end{array} \right] \bullet \left[ \begin{array}{c} x - x_0 \\ y - y_0 \\ z - z_0 \end{array} \right] &= 0 \\ a(x - x_0) + b (y - y_0) + c (z-z_0) &= 0 \end{aligned}\]
We can also write this equation as \[ax + by + cz = ax_0 + by_0 + cz_0\nonumber \]
Notice that since \(P_0\) is given, \(ax_0+by_0+cz_0\) is a known scalar, which we can call \(d\). This equation becomes \[ax + by + cz = d\nonumber \]
Let \(\vec{n} = \left[ \begin{array}{c} a \\ b \\ c \end{array} \right]\) be the normal vector for a plane which contains the point \(P_0 = (x_0, y_0, z_0)\).Then if \(P=(x,y,z)\) is an arbitrary point on the plane, the scalar equation of the plane is given by \[ax + by + cz = d\nonumber \] where \(a,b,c,d \in \mathbb{R}\) and \(d = ax_0 + by_0 + cz_0\).
Consider the following equation.
Find an equation of the plane containing \(P_0 = (3, -2, 5)\) and orthogonal to \(\vec{n} = \left[ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right]\).
Solution
The above vector \(\vec{n}\) is the normal vector for this plane. Using Definition \(\PageIndex{2}\) , we can determine the vector equation for this plane. \[\begin{aligned} \vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) &= 0 \\ \left[ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right] \bullet \left(\left[ \begin{array}{c} x \\ y \\ z \end{array} \right] - \left[ \begin{array}{r} 3 \\ -2 \\ 5 \end{array} \right] \right) &= 0 \\ \left[ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right] \bullet \left[ \begin{array}{c} x - 3 \\ y + 2 \\ z - 5 \end{array} \right] &= 0 \end{aligned}\]
Using Definition \(\PageIndex{3}\) , we can determine the scalar equation of the plane. \[-2x + 4y + 1z = -2(3) + 4(-2) + 1(5) = -9\nonumber \]
Hence, the vector equation of the plane is \[\left[ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right] \bullet \left[ \begin{array}{c} x - 3 \\ y + 2 \\ z - 5 \end{array} \right] = 0\nonumber \] and the scalar equation is \[-2x + 4y + 1z = -9\nonumber \]
Below is a video on finding the equation of a plane given a point and a line.
Below is a video on finding the equation of a plane given three points on the plane.
Below is a video on finding the equation of the plane that is orthogonal to a given line.
Suppose a point \(P\) is not contained in a given plane. We are then interested in the shortest distance from that point \(P\) to the given plane. Consider the following example.
Find the shortest distance from the point \(P = (3,2,3)\) to the plane given by
\(2x + y + 2z = 2\), and find the point \(Q\) on the plane that is closest to \(P\).
Solution
Pick an arbitrary point \(P_0\) on the plane. Then, it follows that \[\overrightarrow{QP} = proj_{\vec{n}}\overrightarrow{P_0P}\nonumber \] and \(\| \overrightarrow{QP} \|\) is the shortest distance from \(P\) to the plane. Further, the vector \(\overrightarrow{0Q} = \overrightarrow{0P} - \overrightarrow{QP}\) gives the necessary point \(Q\).
From the above scalar equation, we have that \(\vec{n} = \left[ \begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right]\). Now, choose \(P_0 = (1, 0, 0)\) so that \(\vec{n} \bullet \overrightarrow{0P} = 2 = d\). Then, \(\overrightarrow{P_0P} = \left[ \begin{array}{c} 3 \\ 2 \\ 3 \end{array} \right] - \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right] = \left[ \begin{array}{c} 2 \\ 2 \\ 3 \end{array} \right]\).
Next, compute \(\overrightarrow{QP} = proj_{\vec{n}}\overrightarrow{P_0P}\). \[\begin{aligned} \overrightarrow{QP} &= proj_{\vec{n}}\overrightarrow{P_0P} \\ &= \left( \frac{ \overrightarrow{P_0P} \bullet \vec{n}}{\| \vec{n} \| ^2}\right)\vec{n} \\ &= \frac{12}{9} \left[ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right] \\ &= \frac{4}{3} \left[ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right] \end{aligned}\]
Then, \(\| \overrightarrow{QP} \| = 4\) so the shortest distance from \(P\) to the plane is \(4\).
Next, to find the point \(Q\) on the plane which is closest to \(P\) we have \[\begin{aligned} \overrightarrow{0Q} &= \overrightarrow{0P} - \overrightarrow{QP} \\ &= \left[ \begin{array}{r} 3 \\ 2 \\ 3 \end{array} \right] - \frac{4}{3} \left[ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right] \\ &= \frac{1}{3} \left[ \begin{array}{r} 1 \\ 2 \\ 1 \end{array} \right]\end{aligned}\]
Therefore, \(Q = (\frac{1}{3}, \frac{2}{3}, \frac{1}{3} )\).
Below is a video on determining the distance between a point and a plane.