4.2: The Calculus of Vector-Valued Functions
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To study the calculus of vector-valued functions, we follow a similar path to the one we took in studying real-valued functions. First, we define the derivative, then we examine applications of the derivative, then we move on to defining integrals. However, we will find some interesting new ideas along the way as a result of the vector nature of these functions and the properties of space curves.
Derivatives of Vector-Valued Functions
Now that we have seen what a vector-valued function is and how to take its limit, the next step is to learn how to differentiate a vector-valued function. The definition of the derivative of a vector-valued function is nearly identical to the definition of a real-valued function of one variable. However, because the range of a vector-valued function consists of vectors, the same is true for the range of the derivative of a vector-valued function.
Definition: Derivative of Vector-Valued Functions
The derivative of a vector-valued function \(\vecs{r}(t)\) is
\[\vecs{r}′(t) = \lim \limits_{\Delta t \to 0} \dfrac{\vecs{r}(t+\Delta t)−\vecs{r}(t)}{ \Delta t} \label{eq1} \]
provided the limit exists. If \(\vecs{r}'(t)\) exists, then \(\vecs{r}(t)\) is differentiable at \(t\). If \(\vecs{r}′(t)\) exists for all \(t\) in an open interval \((a,b)\) then \(\vecs{r}(t)\) is differentiable over the interval \((a,b)\). For the function to be differentiable over the closed interval \([a,b]\), the following two limits must exist as well:
\[\vecs{r}′(a) = \lim \limits_{\Delta t \to 0^+} \dfrac{\vecs{r}(a+\Delta t)−\vecs{r}(a)}{ \Delta t} \]
and
\[\vecs{r}′(b) = \lim \limits_{\Delta t \to 0^-} \dfrac{\vecs{r}(b+\Delta t)−\vecs{r}(b)}{ \Delta t}\]
Graphical Interpretation of the Derivative: Recall that the derivative of a real-valued function can be interpreted as the slope of a tangent line or the instantaneous rate of change of the function. The derivative of a vector-valued function can be understood to be an instantaneous rate of change as well; for example, when the function represents the position of an object at a given point in time, the derivative represents its velocity at that same point in time.
To understand what this derivative represents geometrically, it helps to consider the diagram shown in Figure \(\PageIndex{1}\). For the purposes of this illustration, consider \(\vecs r(t)\) to the represent position of an object.
We see vectors \(\vecs r(t)\) and \(\vecs r(t + \Delta t)\), where \(\Delta t\) is a small change in time producing a second vector that is farther along in the direction of motion.
The difference of these two vectors gives us the displacement vector, \(\vecs r(t + \Delta t) - \vecs r(t)\), shown here in red.
As \(\Delta t \rightarrow 0\), \(\vecs r(t + \Delta t)\) will approach \(\vecs r(t)\), and the displacement vector \(\vecs r(t + \Delta t) - \vecs r(t)\) will become closer and closer to being tangent to the curve at the terminal point of \(\vecs r(t)\).
Although the length of the displacement vector is approaching \(0\), we are actually taking the limit not of the displacement vector itself, but of the average velocity over the time \(\Delta t\), found by dividing the displacement vector by the scalar \(\Delta t\).
\[\text{Average Velocity}\quad=\quad\frac{\vecs r(t + \Delta t) - \vecs r(t)}{\Delta t}\]
Since \(\Delta t\) is also approaching \(0\), dividing by it will tend to lengthen the average velocity vector (as long as \(\Delta t \lt 1\)).
The limit gives us the vector \(\vecs r′(t)\), the velocity of the motion along the curve, which is tangent to the curve at this point.
We now demonstrate taking the derivative of a vector-valued function.
Example \(\PageIndex{1}\): Finding the Derivative of a Vector-Valued Function
Use the definition to calculate the derivative of the function
\[\vecs{r}(t)=(3t+4) \,\hat{\mathbf{i}}+(t^2−4t+3) \,\hat{\mathbf{j}} .\nonumber\]
Solution
Let’s use Equation \ref{eq1}:
\[\begin{align*} \vecs{r}′(t) &= \lim \limits_{\Delta t \to 0} \dfrac{\vecs{r}(t+Δt)− \vecs{r} (t)}{Δt} \\[5pt]
&= \lim \limits_{\Delta t \to 0} \dfrac{[(3(t+Δt)+4)\,\hat{\mathbf{i}}+((t+Δt)^2−4(t+Δt)+3)\,\hat{\mathbf{j}}]−[(3t+4) \,\hat{\mathbf{i}}+(t^2−4t+3) \,\hat{\mathbf{j}}]}{Δt} \\[5pt]
&= \lim \limits_{\Delta t \to 0} \dfrac{(3t+3Δt+4)\,\hat{\mathbf{i}}−(3t+4) \,\hat{\mathbf{i}}+(t^2+2tΔt+(Δt)^2−4t−4Δt+3) \,\hat{\mathbf{j}}−(t^2−4t+3)\,\hat{\mathbf{j}}}{Δt} \\[5pt]
&= \lim \limits_{\Delta t \to 0} \dfrac{(3Δt)\,\hat{\mathbf{i}}+(2tΔt+(Δt)^2−4Δt)\,\hat{\mathbf{j}}}{Δt} \\[5pt]
&= \lim \limits_{\Delta t \to 0} (3 \,\hat{\mathbf{i}}+(2t+Δt−4)\,\hat{\mathbf{j}}) \\[5pt]
&=3 \,\hat{\mathbf{i}}+(2t−4) \,\hat{\mathbf{j}} \end{align*} \]
Exercise \(\PageIndex{1}\)
Use the definition to calculate the derivative of the function \(\vecs{r}(t)=(2t^2+3) \,\hat{\mathbf{i}}+(5t−6) \,\hat{\mathbf{j}}\).
- Hint
-
Use Equation \ref{eq1}.
- Answer
-
\[\vecs{r}′(t)=4t \,\hat{\mathbf{i}}+5 \,\hat{\mathbf{j}} \nonumber\]
Notice that in the calculations in Example \(\PageIndex{1}\), we could also obtain the answer by first calculating the derivative of each component function, then putting these derivatives back into the vector-valued function. This is always true for calculating the derivative of a vector-valued function, whether it is in two or three dimensions. We state this in the following theorem. The proof of this theorem follows directly from the definitions of the limit of a vector-valued function and the derivative of a vector-valued function.
Theorem \(\PageIndex{1}\): Differentiation of Vector-Valued Functions
Let \(f\), \(g\), and \(h\) be differentiable functions of \(t\).
- If \(\vecs{r}(t)=f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}}\) then \[\vecs{r}′(t)=f′(t) \,\hat{\mathbf{i}}+g′(t) \,\hat{\mathbf{j}}.\]
- If \(\vecs{r}(t)=f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}} + h(t) \,\hat{\mathbf{k}}\) then \[\vecs{r}′(t)=f′(t) \,\hat{\mathbf{i}}+g′(t) \,\hat{\mathbf{j}} + h′(t) \,\hat{\mathbf{k}}.\]
Example \(\PageIndex{2}\): Calculating the Derivative of Vector-Valued Functions
Use Theorem \(\PageIndex{1}\) to calculate the derivative of each of the following functions.
- \(\vecs{r}(t)=(6t+8) \,\hat{\mathbf{i}}+(4t^2+2t−3) \,\hat{\mathbf{j}}\)
- \(\vecs{r}(t)=3 \cos t \,\hat{\mathbf{i}}+4 \sin t \,\hat{\mathbf{j}}\)
- \(\vecs{r}(t)=e^t \sin t \,\hat{\mathbf{i}}+e^t \cos t \,\hat{\mathbf{j}}−e^{2t} \,\hat{\mathbf{k}}\)
Solution
We use Theorem \(\PageIndex{1}\) and what we know about differentiating functions of one variable.
- The first component of \[\vecs r(t)=(6t+8) \,\hat{\mathbf{i}}+(4t^2+2t−3) \,\hat{\mathbf{j}} \nonumber\] is \(f(t)=6t+8\). The second component is \(g(t)=4t^2+2t−3\). We have \(f′(t)=6\) and \(g′(t)=8t+2\), so Theorem \(\PageIndex{1}\) gives \[\vecs r′(t)=6 \,\hat{\mathbf{i}}+(8t+2)\,\hat{\mathbf{j}} \nonumber\]
- The first component is \(f(t)=3 \cos t\) and the second component is \(g(t)=4 \sin t\). We have \(f′(t)=−3 \sin t\) and \(g′(t)=4 \cos t\), so we obtain \[\vecs r′(t)=−3 \sin t \,\hat{\mathbf{i}}+4 \cos t \,\hat{\mathbf{j}} \nonumber\]
- The first component of \(\vecs r(t)=e^t \sin t \,\hat{\mathbf{i}}+e^t \cos t \,\hat{\mathbf{j}}−e^{2t} \,\hat{\mathbf{k}}\) is \(f(t)=e^t \sin t\), the second component is \(g(t)=e^t \cos t\), and the third component is \(h(t)=−e^{2t}\). We have \(f′(t)=e^t(\sin t+\cos t)\), \(g′(t)=e^t (\cos t−\sin t)\), and \(h′(t)=−2e^{2t}\), so the theorem gives \[\vecs r′(t)=e^t(\sin t+\cos t)\,\hat{\mathbf{i}}+e^t(\cos t−\sin t)\,\hat{\mathbf{j}}−2e^{2t} \,\hat{\mathbf{k}} \nonumber\]
Exercise \(\PageIndex{2}\)
Calculate the derivative of the function
\[\vecs{r}(t)=(t \ln t)\,\hat{\mathbf{i}}+(5e^t) \,\hat{\mathbf{j}}+(\cos t−\sin t) \,\hat{\mathbf{k}}. \nonumber\]
- Hint
-
Identify the component functions and use Theorem \(\PageIndex{1}\).
- Answer
-
\[\vecs{r}′(t)=(1+ \ln t) \,\hat{\mathbf{i}}+5e^t \,\hat{\mathbf{j}}−(\sin t+\cos t)\,\hat{\mathbf{k}} \nonumber\]
Describing Motion with Vector-Valued Functions
As has been mentioned already in this section, vector-valued functions are often used to describe motion in the plane or in space. Now that we understand the derivative of a vector-valued function, let's make the relationships clear.
First, since the components of a vector-valued function represent the coordinates of a point on the corresponding curve, it may help to write
\[\vecs r(t) = x(t) \,\,\hat{\mathbf{i}} + y(t) \,\,\hat{\mathbf{j}} + z(t) \, \,\hat{\mathbf{k}}\]
This approach has already been used in Section 12.1.
Now we can define how to calculate the velocity, speed, acceleration, and jerk of the object's motion.
Definition: Velocity, speed, Acceleration, and Jerk
Let \(\vecs r(t)\) be a twice-differentiable vector-valued function of the parameter \(t\) that represents the position of an object as a function of time.
The velocity vector \(\vecs v(t)\) of the object is given by
\[\text{Velocity}\,=\vecs v(t)=\vecs r′(t). \label{Eq1}\]
The speed is defined to be
\[\text{Speed}\,=‖\vecs v(t)‖=‖\vecs r′(t)‖=\dfrac{ds}{\,dt}. \label{Eq2}\]
The acceleration vector \(\vecs a(t)\) is defined to be
\[\text{Acceleration}\,=\vecs a(t)=\vecs v′(t)=\vecs r″(t). \label{Eq3}\]
If \(\vecs r(t)\) has a third derivative, we can also define the jerk vector \(\vecd{\text{jerk}}(t)\) to be
\[\text{Jerk}\,=\vecd{\text{jerk}}(t) = \vecs a'(t)=\vecs r\,'''(t). \label{Eq4}\]
Now let's see how this looks in the context of a particular position function.
 Determine the velocity, speed, acceleration, and jerk as functions of time for this position function.
b. Evaluate each of these functions, including the position function, at times \(t = 0\) sec and \(t = 2\) sec.
c. Sketch a graph of this position function and draw in each of the non-zero vectors from part b on the graph of \(\vecs r\), starting at the location of the object on the curve at the corresponding time.
Solution
a. Since \(\vecs r(t) = (1 - t^2) \, \hat{\mathbf{i}} + t \, \hat{\mathbf{j}}\), we have
\[\begin{align*} \vecs v(t) &= \vecs r\,'(t) = -2t \, \hat{\mathbf{i}} + 1 \, \hat{\mathbf{j}},\\[5pt] \text{speed}(t) &= \|\vecs v(t) \| = \sqrt{(-2t)^2 +(1)^2} = \sqrt{4t^2 + 1}, \\[5pt] \vecs a(t) &= \vecs v'(t) = -2 \, \hat{\mathbf{i}}, \\[5pt] \vecd{\text{jerk}}(t) &= \vecs 0 \end{align*}\]
b. Evaluating these functions at time \(t = 0\), we get:
\[\begin{align*} \vecs r(0) &= (1 - 0^2) \, \hat{\mathbf{i}} + 0 \, \hat{\mathbf{j}} = \hat{\mathbf{i}} \\
\vecs v(0) &= -2(0) \, \hat{\mathbf{i}} + 1 \, \hat{\mathbf{j}} = \hat{\mathbf{j}} \\
\text{speed}(0) &= \sqrt{4(0)^2 + 1} = \sqrt{1} = 1 \, \text{unit/sec} \\
\vecs a(0) &= -2 \, \hat{\mathbf{i}}, \\
\vecd{\text{jerk}}(0) &= \vecs 0 \end{align*}\]
And for time \(t = 2\), we have
\[\begin{align*} \vecs r(2) &= (1 - 2^2) \, \hat{\mathbf{i}} + 2 \, \hat{\mathbf{j}} = -3 \, \hat{\mathbf{i}} + 2 \, \hat{\mathbf{j}} \\
\vecs v(2) &= -2(2) \, \hat{\mathbf{i}} + 1 \, \hat{\mathbf{j}} = -4 \, \hat{\mathbf{i}} + \hat{\mathbf{j}} \\
\text{speed}(2) &= \sqrt{4(2)^2 + 1} = \sqrt{17} \, \text{units/sec} \\
\vecs a(0) &= -2 \, \hat{\mathbf{i}}, \\
\vecd{\text{jerk}}(0) &= \vecs 0 \end{align*}\]
c. The graph of this curve is shown below in Figure \(\PageIndex{2}\). We then drew in and labeled the vectors \(\vecs v(0)\), \(\vecs a(0)\), \(\vecs v(2)\), and \(\vecs a(2)\).
Smooth Vector-Valued Functions
Definition: smoothness
Let \(\vecs{r}(t)=f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}} + h(t) \,\hat{\mathbf{k}}\) be the parameterization of a curve that is differentiable on an open interval \(I\).
Then \(\vecs r(t)\) is smooth on the open interval \(I\), if
\(\vecs r\,'(t) \neq \vecs 0\), for any value of \(t\) in the interval \(I\).
To put this another way, \(\vecs r(t)\) is smooth on the open interval \(I\) if:
- \(\vecs r(t)\) is defined and continuous on \(I\), and
- \(f', \, g'\) and \(h'\) are all continuous on \(I\), that is, \(\vecs r\,'(t)\) is defined and continuous on \(I\), and
- \(\vecs r\,'(t) \neq \vecs 0\) for all values of \(t\) in the interval \(I\).
The concept of smoothness is very closely connected to the application of vector-valued functions to motion. If you have ever ridden rides at an amusement park, you can probably identify some where the ride felt smooth, and others where there are moments during the ride where it was not smooth. These moments often correspond to points in the motion where the relevant parameterization is not smooth by the definition above.
To identify open intervals in the domain of \(\vecs r\) on which \(\vecs r\) is smooth, we usually identify values of \(t\) for which \(\vecs r\) is not smooth .
In addition to using the definition above to identify these points, it is often apparent from the graph of the vector-valued function \(\vecs r(t)\). A cusp , i.e., a sharp turning point, is a common occurrence at points where a curve is not smooth. In Figure \(\PageIndex{3}\) below, you can quickly identify points at which the parameterization fails to be smooth by locating cusps in the graph.
When we consider the motion described by \(\vecs r(t)\), there are at least two additional events that can occur at points where the parameterization fails to be smooth: a reversal of motion or a momentary pause in the motion. To summarize these events:
If a vector-valued function \(\vecs r(t)\) is not smooth at time \(t\) , we will observe that:
- There is a cusp at the associated point on the graph of \(\vecs r(t)\), or
- The motion reverses itself at the associated point, causing the motion to travel back along the same path in the opposite direction, or
- The motion actually stops and starts up again, with no visual cue, that is, where the curve appears smooth.
Let's consider a couple examples to see how this can look.
Example \(\PageIndex{4}\): Determining open intervals on which a vector-valued function is smooth
Determine the open intervals on which each of the following vector-valued functions is smooth:
- \(\vecs r(t) = \left( 2t^3 - 3t^2\right) \, \,\hat{\mathbf{i}} + \left( t^2 - 2t\right) \, \,\hat{\mathbf{j}} + 2\,\,\hat{\mathbf{k}}\)
- \(\vecs r(t) = t^3 \, \,\hat{\mathbf{i}} + \sqrt{t^2 - 1} \,\hat{\mathbf{j}}\)
- \(\vecs r(t) = t^4 \, \,\hat{\mathbf{i}} + \ln(t^2 + 1) \,\hat{\mathbf{j}}\)
- \(\vecs r(t) = \left( 6\cos t - \cos(6t)\right) \,\hat{\mathbf{i}} + \left( 6\sin t - \sin(6t) \right) \,\hat{\mathbf{j}}\), for \(0 \le t \le 2\pi\)
Solution
a. First we need to consider the domain of \(\vecs r\). In this case, \(\text{D}_{\vecs r} : (-\infty, \infty)\).
Then we need to find \(\vecs r\,'(t)\). Since \(\vecs r(t) = \left( 2t^3 - 3t^2\right) \, \,\hat{\mathbf{i}} + \left( t^2 - 2t\right) \, \,\hat{\mathbf{j}} + 2\,\,\hat{\mathbf{k}}\),
\[\vecs r\,'(t) = \left( 6t^2 - 6t\right) \, \,\hat{\mathbf{i}} + \left( 2t - 2\right) \, \,\hat{\mathbf{j}}\nonumber\]
Next we need to consider the domain of \(\vecs r\,'\). Here, \(\text{D}_{\vecs r\,'} : (-\infty, \infty)\).
Finally we need to determine if there are any values of \(t\) in \(\text{D}_{\vecs r}\) where \(\vecs r\,'(t) = \vecs 0\).
To do this, we set each component of \(\vecs r\,'(t)\) equal to \(0\) and solve for \(t\).
\[\begin{align*} 6t^2 - 6t = 0 & \quad \text{and} \quad & 2t - 2 = 0 \\ 6t(t -1) = 0 && 2(t - 1) = 0 \\ t = 0 \quad \text{or} \quad t = 1 & & t = 1 \end{align*} \]
Since \(t = 1\) makes both components of \(\vecs r\,'\) equal 0, we have \(\vecs r\,'(1) = \vecs 0\), so we know that \(\vecs r\) is not smooth at \(t = 1\).
Note however that \(\vecs r\) is smooth at \(t = 0\), because although one of the components of \(\vecs r\,'\) is \(0\), the other one is not.
Therefore we can conclude that \(\vecs r\) is smooth on the open intervals:
\[ (-\infty, 1) \quad \text{and} \quad (1, \infty) \nonumber\]
Note in the figure that there is a cusp in the curve when \(t = 1\).
b. Next let \(\vecs r(t) = t^3 \, \,\hat{\mathbf{i}} + \sqrt{t^2 - 1} \,\hat{\mathbf{j}}.\)
The domain of \(\vecs r\) is \(\text{D}_{\vecs r} : (-\infty, -1] \cup [1, \infty) \), since \(t^2 - 1\) is less than \(0\) between \(-1\) and \(1\).
Now we calculate \(\vecs r\,'(t)\). Since \(\vecs r(t) = t^3 \, \,\hat{\mathbf{i}} + \sqrt{t^2 - 1} \,\hat{\mathbf{j}}\), we have
\[\vecs r\,'(t) = 3t^2 \, \,\hat{\mathbf{i}} + \frac{t}{\sqrt{t^2-1}} \, \,\hat{\mathbf{j}}\nonumber\]
Then, since we cannot have zero in the denominator, \(t = -1\) and \(t = 1\) are not in the domain of \(\vecs r\,'\). Thus, \(\text{D}_{\vecs r\,'} : (-\infty, -1) \cup (1, \infty) \).
Lastly we determine if there are any values of \(t\) in \(\text{D}_{\vecs r}\) where \(\vecs r\,'(t) = \vecs 0\).
Setting each component of \(\vecs r\,'(t)\) equal to \(0\) and solving for \(t\) we get:
\[\begin{align*} 3t^2 = 0 & \quad \text{and} \quad & \frac{t}{\sqrt{t^2-1}} = 0 \\ t = 0 && t = 0 \end{align*}\nonumber\]
If \(t = 0\) were in the domain of \(\vecs r\,'\), then \(\vecs r\) would not be smooth there, but since it is not in the domain, we have nothing to remove, and we can conclude that \(\vecs r\) is smooth on the entire domain of its derivative \(\vecs r\,'\), that is,
\[\vecs r \, \text{is smooth on the open intervals:} \quad (-\infty, -1) \quad \text{and} \quad (1, \infty).\nonumber\]
Observe the two separate smooth pieces of this function in the graph below.
c. Now let's consider, \(\vecs r(t) = t^4 \, \,\hat{\mathbf{i}} + \ln(t^2 + 1) \,\hat{\mathbf{j}}.\)
The domain of \(\vecs r\) is \(\text{D}_{\vecs r} : (-\infty, \infty)\), since the argument of the natural log function, \(t^2 + 1,\) is always greater than \(0\) (In fact, it's always at least \(1\)).
Now we calculate \(\vecs r\,'(t)\). Here
\[\vecs r\,'(t) = 4t^3 \, \,\hat{\mathbf{i}} + \frac{2t}{t^2+1} \, \,\hat{\mathbf{j}}\nonumber\]
Then, since \(t^2+1\) cannot be \(0,\) the domain of \(\vecs r\,'\) is also all real numbers, \(\text{D}_{\vecs r\,'} : (-\infty, \infty) \).
Lastly we determine if there are any values of \(t\) in \(\text{D}_{\vecs r}\) where \(\vecs r\,'(t) = \vecs 0\).
Setting each component of \(\vecs r\,'(t)\) equal to \(0\) and solving for \(t\) we get:
\[\begin{align*} 4t^3 = 0 & \quad \text{and} \quad & \frac{2t}{t^2+1} = 0 \\ t = 0 && t = 0 \nonumber \end{align*}\]
Note that in this example, we have \(\vecs r\,'(0) = \vecs 0,\) so \(\vecs r\) is not smooth when \(t = 0.\)
Therefore we conclude that \(\vecs r\) is smooth on the open intervals: \[ (-\infty, 0) \quad \text{and} \quad (0, \infty) \nonumber\]
In this example, note that the motion reverses itself when \(t = 0.\) We can't see this very well just by looking at its graph (see Figure \(\PageIndex{6}\)). But if you evaluate \(\vecs r(t)\) for various values of \(t\) from \(-2\) to \(2\), you'll see that the motion comes in towards the origin when \(t\) is negative, arrives at the origin when \(t=0,\) and then moves away from the origin as \(t\) becomes more positive.
d. Finally, let's look at a really interesting example: \[\vecs r(t) = \left( 6\cos t - \cos(6t)\right) \,\hat{\mathbf{i}} + \left( 6\sin t - \sin(6t) \right) \,\hat{\mathbf{j}}.\nonumber\]
Its graph is the epicycloid shown above in Figure \(\PageIndex{3}\).
Again, \(\vecs r\) has no issues with its domain. We have \(\text{D}_{\vecs r} : (-\infty, \infty).\)
Calculating \(\vecs r\,'(t),\) we get:
\[\vecs r\,'(t) = \left( -6\sin t + 6\sin(6t)\right) \,\hat{\mathbf{i}} + \left( 6\cos t - 6\cos(6t)\right) \,\hat{\mathbf{j}}\nonumber\]
There are still no issues, so \(\text{D}_{\vecs r\,'} : (-\infty, \infty).\)
Now for the most interesting part! For what values of \(t\) is \(\vecs r\,'(t) = \vecs 0\)?
Setting the two components equal to \(0,\) we get:
\[ -6\sin t + 6\sin(6t)= 0 \quad \text{and} \quad 6\cos t - 6\cos(6t) = 0 \nonumber\]
This time it's not as easy to quickly solve each of these equations for all values of \(t\) that make these component functions equal to \(0\). But if you read carefully, you'll see that it's really not that difficult to work out.
Rewriting the first equation gives us
\[\sin(6t) = \sin t\label{eqSin6t}\]
To solve this equation, note that the periodic nature of the sine function means that its value is the same every \(2\pi\) radians.
That is, for any integer \(n\), we know
\[\sin(t+2n\pi) = \sin t \nonumber\]
This means that one way for Equation \ref{eqSin6t} to be true would be for us to have
\[ t + 2n\pi = 6t\nonumber\]
Solving this equation for \(t\), we have:
\( \qquad t = \dfrac{2n\pi}{5} \text{ for any integer }n\)
On the interval \(0 \le t \le 2\pi,\) this would give us the following solutions of Equation \ref{eqSin6t}:
\[t = 0, \,\frac{2\pi}{5}, \,\frac{4\pi}{5}, \,\frac{6\pi}{5}, \,\frac{8\pi}{5}, \,\frac{10\pi}{5}=2\pi\nonumber\]
Verifying that \(t = \frac{2\pi}{5}\) is a solution, we have:
\[ \sin\left(6\cdot \frac{2\pi}{5}\right) = \sin\left(2\pi +\frac{2\pi}{5}\right) = \sin\left(\frac{2\pi}{5}\right). \quad\checkmark \nonumber\]
Note that the sine value (as the \(y\)-coordinate of a point on the unit circle) is the same for reflected angles about the \(y)-axis (e.g., in the first and second quadrants). That is,
\[\sin(\pi - t) = \sin t \nonumber\]
Incorporating the periodic nature of the sine function again, we obtain
\[\sin\Big(\big(2n+1\big)\pi - t \Big) = \sin t \nonumber\]
Applying this identity to Equation \ref{eqSin6t}, yields
\[\sin\Big(\big(2n+1\big)\pi - t \Big) = \sin (6t) \nonumber\]
So, another way for Equation \ref{eqSin6t} to be true would be for us to have
\[(2n+1)\pi - t = 6t \nonumber\]
Solving this equation for \(t\), we have:
\( \qquad t = \dfrac{(2n+1)\pi}{7} \text{ for any integer }n\)
On the interval \(0 \le t \le 2\pi,\) this would give us the following solutions of Equation \ref{eqSin6t}:
\[t = \frac{\pi}{7}, \,\frac{3\pi}{7}, \,\frac{5\pi}{7}, \,\frac{7\pi}{7}=\pi, \,\frac{9\pi}{7}, \,\frac{11\pi}{7}, \,\frac{13\pi}{7} \nonumber\]
Verifying that \(t = \frac{\pi}{7}\) is a solution, we have:
\[ \sin\left(6\cdot \frac{\pi}{7}\right) = \sin\left(\pi -\frac{\pi}{7}\right) = \sin\left(\frac{\pi}{7}\right). \quad\checkmark \nonumber\]
Next we need to solve the equation from the setting the other component of \(\vecs r\,'(t)\) to \(0.\) That is
\[6\cos t - 6\cos(6t) = 0 \nonumber\]
Remember that not all values of \(t\) that make one of these components equal to \(0\) will cause \(\vecs r\) not to be smooth, but only those values of \(t\) that make both components equal to \(0\) at the same time . Once we solve this equation, we'll need to look for any values of \(t\) that are solutions to both of these equations.
Rewriting the equation above gives us
\[\cos(6t) = \cos t\label{eqCos6t}\]
We can solve this equation using a similar process to what was shown for Equation \ref{eqSin6t} above.
As the cosine function is periodic in the same way that sine is, we have
\(\cos(t+2n\pi) = \cos t\)
which tells us Equation \ref{eqCos6t} will be true when
\[ t + 2n\pi = 6t\nonumber\]
Solving this equation for \(t\), we have:
\( \qquad t = \dfrac{2n\pi}{5} \text{ for any integer }n\)
Note that these values for \(t\) also made the first component of \(\vecs r\,'(t)\) equal to \(0!\) Try verifying that one of these solutions satisfies Equation \ref{eqCos6t}, as we showed for Equation \ref{eqSin6t} above.
Since the cosine values (as the \(x\)-coordinates of points on the unit circle) are the same for reflected angles about the \(x\)-axis (e.g. in the first and fourth quadrants). This is represented in the identity:
\[ \cos(-t) = \cos(t) \nonumber\]
Incorporating the periodic nature of the cosine function (as we did before), we obtain
\[ \cos(2n\pi - t) = \cos(t) \nonumber\]
Applying this identity to Equation \ref{eqCos6t}, yields
\[ \cos(2n\pi - t) = \cos(6t) \nonumber\]
This is true when
\[ 2n\pi - t = 6t\nonumber\]
Solving this equation for \(t\), we have:
\( \qquad t = \dfrac{2n\pi}{7} \text{ for any integer }n\)
Although these may at first look like the same values as we found solved Equation \ref{eqSin6t} above, they are not.
On the interval \(0 \le t \le 2\pi,\) this would give us the following solutions of Equation \ref{eqCos6t}:
\[t = 0, \frac{2\pi}{7}, \,\frac{4\pi}{7}, \,\frac{6\pi}{7}, \,\frac{8\pi}{7}, \,\frac{10\pi}{7}, \,\frac{12\pi}{7}, \,\frac{14\pi}{7} = 2\pi \nonumber\]
Except for \(t = 0\) and \(t = 2\pi\) that are already contained in the first set of solutions \(t = \dfrac{2n\pi}{5}\) above, this second set of solutions does not add any additional values of \(t\) that would make both components of \(\vecs r\,'(t)\) equal to \(0\) at the same time.
We have now determined that \(\vecs r\,'(t) = \vecs 0\) and therefore \(\vecs r\) is not smooth when \(t = \dfrac{2n\pi}{5} \text{ for any integer }n.\)
Removing these values from the interval \(0 \le t \le 2\pi,\) we can say that \(\vecs r\) is smooth on the open intervals:
\[\left(0, \,\frac{2\pi}{5}\right), \; \left(\frac{2\pi}{5} , \, \frac{4\pi}{5} \right), \; \left(\frac{4\pi}{5} , \, \frac{6\pi}{5} \right), \; \left(\frac{6\pi}{5} , \, \frac{8\pi}{5} \right), \text{ and } \left(\frac{8\pi}{5} ,\, 2\pi \right). \nonumber\]
More generally, we can state that \(r\) is smooth on the open intervals:
\[ \Big( \dfrac{2n\pi}{5}, \, \dfrac{\big(2n+2\big)\pi}{5} \Big) \quad \text{for any integer }n\nonumber\]
Note in Figure \(\PageIndex{7},\) how there are cusps on the plot of this vector-valued function for each value of \(t\) at which \(\vecs r\) is not smooth.
Properties of the Derivative
We can extend to vector-valued functions the properties of the derivative that we presented previously for real-valued functions of a single variable. In particular, the constant multiple rule, the sum and difference rules, the product rule, and the chain rule all extend to vector-valued functions. However, in the case of the product rule, there are actually three extensions:
- for a real-valued function multiplied by a vector-valued function,
- for the dot product of two vector-valued functions, and
- for the cross product of two vector-valued functions.
Theorem: Properties of the Derivative of Vector-Valued Functions
Let \(\vecs{r}\) and \(\vecs{u}\) be differentiable vector-valued functions of \(t\), let \(f\) be a differentiable real-valued function of \(t\), and let \(c\) be a scalar.
\[\begin{array}{lrcll} \mathrm{i.} & \dfrac{d}{\,dt}[c\vecs r(t)] & = & c\vecs r′(t) & \text{Scalar multiple} \nonumber\\ \mathrm{ii.} & \dfrac{d}{\,dt}[\vecs r(t)±\vecs u(t)] & = & \vecs r′(t)±\vecs u′(t) & \text{Sum and difference} \nonumber\\ \mathrm{iii.} & \dfrac{d}{\,dt}[f(t)\vecs u(t)] & = & f′(t)\vecs u(t)+f(t)\vecs u′(t) & \text{Scalar product} \nonumber\\ \mathrm{iv.} & \dfrac{d}{\,dt}[\vecs r(t)⋅\vecs u(t)] & = & \vecs r′(t)⋅\vecs u(t)+\vecs r(t)⋅\vecs u′(t) & \text{Dot product} \nonumber\\ \mathrm{v.} & \dfrac{d}{\,dt}[\vecs r(t)×\vecs u(t)] & = & \vecs r′(t)×\vecs u(t)+\vecs r(t)×\vecs u′(t) & \text{Cross product} \nonumber\\ \mathrm{vi.} & \dfrac{d}{\,dt}[\vecs r(f(t))] & = & \vecs r′(f(t))⋅f′(t) & \text{Chain rule} \nonumber\\ \mathrm{vii.} & \text{If} \; \vecs r(t)·\vecs r(t) & = & c, \text{then} \; \vecs r(t)⋅\vecs r′(t) \; =0 \; . & \mathrm{} \nonumber \end{array} \nonumber \]
Proof
The proofs of the first two properties follow directly from the definition of the derivative of a vector-valued function. The third property can be derived from the first two properties, along with the product rule . Let \(\vecs u(t)=g(t)\,\hat{\mathbf{i}}+h(t)\,\hat{\mathbf{j}}\). Then
\[\begin{align*} \dfrac{d}{\,dt}[f(t)\vecs u(t)] \; & =\dfrac{d}{\,dt}[f(t)(g(t) \,\hat{\mathbf{i}}+h(t) \,\hat{\mathbf{j}})] \\[5pt] & =\dfrac{d}{\,dt}[f(t)g(t) \,\hat{\mathbf{i}}+f(t)h(t) \,\hat{\mathbf{j}}] \\[5pt] & =\dfrac{d}{\,dt}[f(t)g(t)] \,\hat{\mathbf{i}}+\dfrac{d}{\,dt}[f(t)h(t)] \,\hat{\mathbf{j}} \\[5pt] & =(f′(t)g(t)+f(t)g′(t)) \,\hat{\mathbf{i}}+(f′(t)h(t)+f(t)h′(t)) \,\hat{\mathbf{j}} \\[5pt] & =f′(t)\vecs u(t)+f(t)\vecs u′(t). \end{align*} \]
To prove property iv. let \(\vecs r(t)=f_1(t) \,\hat{\mathbf{i}}+g_1(t) \,\hat{\mathbf{j}}\) and \(\vecs u(t)=f_2(t) \,\hat{\mathbf{i}}+g_2(t) \,\hat{\mathbf{j}}\). Then
\[\begin{align*} \dfrac{d}{\,dt}[\vecs r(t)⋅\vecs u(t)] & =\dfrac{d}{\,dt}[f_1(t)f_2(t)+g_1(t)g_2(t)] \\[5pt] & =f_1′(t)f_2(t)+f_1(t)f_2′(t)+g_1′(t)g_2(t)+g_1(t)g_2′(t)=f_1′(t)f_2(t)+g_1′(t)g_2(t)+f_1(t)f_2′(t)+g_1(t)g_2′(t) \\[5pt] & =(f_1′ \,\hat{\mathbf{i}}+g_1′ \,\hat{\mathbf{j}})⋅(f_2 \,\hat{\mathbf{i}}+g_2 \,\hat{\mathbf{j}})+(f_1 \,\hat{\mathbf{i}}+g_1 \,\hat{\mathbf{j}})⋅(f_2′ \,\hat{\mathbf{i}}+g_2′ \,\hat{\mathbf{j}}) \\[5pt] & =\vecs r′(t)⋅\vecs u(t)+\vecs r(t)⋅\vecs u′(t). \end{align*} \]
The proof of property v. is similar to that of property iv. Property vi. can be proved using the chain rule. Last, property vii. follows from property iv:
\[\begin{align*} \dfrac{d}{\,dt}[\vecs r(t)·\vecs r(t)] \; & =\dfrac{d}{\,dt}[c] \\ \vecs r′(t)·\vecs r(t)+\vecs r(t)·\vecs r′(t) \; & = 0 \\ 2\vecs r(t)·\vecs r′(t) \; & = 0 \\ \vecs r(t)·\vecs r′(t) \; & = 0 \end{align*} \]
Now for some examples using these properties.
Example \(\PageIndex{5}\): Using the Properties of Derivatives of Vector-Valued Functions
Given the vector-valued functions
\[\vecs{r}(t)=(6t+8)\,\hat{\mathbf{i}}+(4t^2+2t−3)\,\hat{\mathbf{j}}+5t \,\hat{\mathbf{k}} \nonumber\]
and
\[\vecs{u}(t)=(t^2−3)\,\hat{\mathbf{i}}+(2t+4)\,\hat{\mathbf{j}}+(t^3−3t)\,\hat{\mathbf{k}}, \nonumber\]
calculate each of the following derivatives using the properties of the derivative of vector-valued functions.
- \(\dfrac{d}{\,dt}[\vecs{r}(t)⋅ \vecs{u}(t)]\)
- \(\dfrac{d}{\,dt}[ \vecs{u} (t) \times \vecs{u}′(t)]\)
Solution
We have \(\vecs{r}′(t)=6 \,\hat{\mathbf{i}}+(8t+2) \,\hat{\mathbf{j}}+5 \,\hat{\mathbf{k}}\) and \(\vecs{u}′(t)=2t \,\hat{\mathbf{i}}+2 \,\hat{\mathbf{j}}+(3t^2−3) \,\hat{\mathbf{k}}\). Therefore, according to property iv :
-
\[\begin{align*} \dfrac{d}{\,dt}[\vecs r(t)⋅\vecs u(t)] &= \vecs r′(t)⋅\vecs u(t)+\vecs r(t)⋅\vecs u′(t) \\[5pt]
&= (6 \,\hat{\mathbf{i}}+(8t+2) \,\hat{\mathbf{j}}+5 \,\hat{\mathbf{k}})⋅((t^2−3) \,\hat{\mathbf{i}}+(2t+4) \,\hat{\mathbf{j}}+(t^3−3t) \,\hat{\mathbf{k}}) \\[5pt]
& \, +((6t+8) \,\hat{\mathbf{i}}+(4t^2+2t−3) \,\hat{\mathbf{j}}+5t \,\hat{\mathbf{k}})⋅(2t \,\hat{\mathbf{i}}+2 \,\hat{\mathbf{j}}+(3t^2−3)\,\hat{\mathbf{k}}) \\[5pt]
&= 6(t^2−3)+(8t+2)(2t+4)+5(t^3−3t) \\[5pt]
& \;+2t(6t+8)+2(4t^2+2t−3)+5t(3t^2−3) \\[5pt]
&= 20t^3+42t^2+26t−16. \end{align*}\] -
First, we need to adapt
property v
for this problem:
\[\dfrac{d}{\,dt}[ \vecs{u}(t) \times \vecs{u}′(t)]=\vecs{u}′(t)\times \vecs{u}′(t)+ \vecs{u}(t) \times \vecs{u}′′(t). \nonumber\]
Recall that the cross product of any vector with itself is zero. Furthermore,\(\vecs u′′(t)\) represents the second derivative of \(\vecs u(t):\)
\[\vecs u′′(t)=\dfrac{d}{\,dt}[\vecs u′(t)]=\dfrac{d}{\,dt}[2t \,\hat{\mathbf{i}}+2 \,\hat{\mathbf{j}}+(3t^2−3) \,\hat{\mathbf{k}}]=2 \,\hat{\mathbf{i}}+6t \,\hat{\mathbf{k}}. \nonumber\]
Therefore,
\[\begin{align*} \dfrac{d}{\,dt}[\vecs u(t) \times \vecs u′(t)] &=0+((t^2−3)\,\hat{\mathbf{i}}+(2t+4)\,\hat{\mathbf{j}}+(t^3−3t)\,\hat{\mathbf{k}})\times (2 \,\hat{\mathbf{i}}+6t \,\hat{\mathbf{k}}) \\[5pt]
&= \begin{vmatrix} \,\hat{\mathbf{i}} & \,\hat{\mathbf{j}} & \,\hat{\mathbf{k}} \\ t^2-3 & 2t+4 & t^3 -3t \\ 2 & 0 & 6t \end{vmatrix} \\[5pt]
& =6t(2t+4) \,\hat{\mathbf{i}}−(6t(t^2−3)−2(t^3−3t)) \,\hat{\mathbf{j}}−2(2t+4) \,\hat{\mathbf{k}} \\[5pt]
& =(12t^2+24t) \,\hat{\mathbf{i}}+(12t−4t^3) \,\hat{\mathbf{j}}−(4t+8)\,\hat{\mathbf{k}}. \end{align*}\]
Exercise \(\PageIndex{3}\)
Calculate \(\dfrac{d}{\,dt}[\vecs{r}(t)⋅ \vecs{r}′(t)]\) and \( \dfrac{d}{\,dt}[\vecs{u}(t) \times \vecs{r}(t)]\) for the vector-valued functions:
- \(\vecs{r}(t)=\cos t \,\hat{\mathbf{i}}+ \sin t \,\hat{\mathbf{j}}−e^{2t} \,\hat{\mathbf{k}}\)
- \(\vecs{u}(t)=t \,\hat{\mathbf{i}}+ \sin t \,\hat{\mathbf{j}}+ \cos t \,\hat{\mathbf{k}}\),
- Hint
-
Follow the same steps as in Example \(\PageIndex{4}\).
- Answer
-
\(\dfrac{d}{\,dt}[\vecs{r}(t)⋅ \vecs{r}′(t)]=8e^{4t}\)
\( \dfrac{d}{\,dt}[ \vecs{u}(t) \times \vecs{r}(t)] =−(e^{2t}(\cos t+2 \sin t)+ \cos 2t) \,\hat{\mathbf{i}}+(e^{2t}(2t+1)− \sin 2t) \,\hat{\mathbf{j}}+(t \cos t+ \sin t− \cos 2t) \,\hat{\mathbf{k}}\)
Tangent Vectors and Unit Tangent Vectors
Recall that the derivative at a point can be interpreted as the slope of the tangent line to the graph at that point. In the case of a vector-valued function, the derivative provides a tangent vector to the curve represented by the function. Consider the vector-valued function
\[\vecs{r}(t)=\cos t \,\hat{\mathbf{i}} + \sin t \,\hat{\mathbf{j}} \label{eq10}\]
The derivative of this function is
\[\vecs{r}′(t)=−\sin t \,\hat{\mathbf{i}} + \cos t \,\hat{\mathbf{j}} \nonumber \]
If we substitute the value \(t=π/6\) into both functions we get
\[\vecs{r} \left(\dfrac{π}{6}\right)=\dfrac{\sqrt{3}}{2} \,\hat{\mathbf{i}}+\dfrac{1}{2}\,\hat{\mathbf{j}} \nonumber\]
and
\[ \vecs{r}′ \left(\dfrac{π}{6} \right)=−\dfrac{1}{2}\,\hat{\mathbf{i}}+\dfrac{\sqrt{3}}{2}\,\hat{\mathbf{j}}. \nonumber\]
The graph of this function appears in Figure \(\PageIndex{8}\), along with the vectors \(\vecs{r}\left(\dfrac{π}{6}\right)\) and \(\vecs{r}' \left(\dfrac{π}{6}\right)\).
Notice that the vector \(\vecs{r}′\left(\dfrac{π}{6}\right)\) is tangent to the circle at the point corresponding to \(t=\dfrac{π}{6}\). This is an example of a tangent vector to the plane curve defined by Equation \ref{eq10}.
Definition: principal unit tangent vector
Let \(C\) be a curve defined by a vector-valued function \(\vecs{r}\), and assume that \(\vecs{r}′(t)\) exists when \(t=t_0\) A tangent vector \(\vecs{r}\) at \(t=t_0\) is any vector such that, when the tail of the vector is placed at point \(\vecs r(t_0)\) on the graph, vector \(\vecs{r}\) is tangent to curve \(C\). Vector \(\vecs{r}′(t_0)\) is an example of a tangent vector at point \(t=t_0\). Furthermore, assume that \(\vecs{r}′(t)≠0\). The principal unit tangent vector at \(t\) is defined to be
\[\vecs{T}(t)=\dfrac{ \vecs{r}′(t)}{‖\vecs{r}′(t)‖},\]
provided \(‖\vecs{r}′(t)‖≠0\).
The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. To calculate a unit tangent vector, first find the derivative \(\vecs{r}′(t)\). Second, calculate the magnitude of the derivative. The third step is to divide the derivative by its magnitude.
Example \(\PageIndex{6}\): Finding a Unit Tangent Vector
Find the unit tangent vector for each of the following vector-valued functions:
- \(\vecs{r}(t)=\cos t \,\hat{\mathbf{i}}+\sin t \,\hat{\mathbf{j}}\)
- \(\vecs{u}(t)=(3t^2+2t) \,\hat{\mathbf{i}}+(2−4t^3)\,\hat{\mathbf{j}}+(6t+5)\,\hat{\mathbf{k}}\)
Solution
-
\(\begin{array}{lrcl} \text{First step:} & \vecs r′(t) & = & − \sin t \,\hat{\mathbf{i}}+ \cos t \,\hat{\mathbf{j}} \\ \text{Second step:} & ‖\vecs r′(t)‖ & = & \sqrt{(− \sin t)^2+( \cos t)^2} = 1 \\ \text{Third step:} & \vecs T(t) & = & \dfrac{\vecs r′(t)}{‖\vecs r′(t)‖}=\dfrac{− \sin t \,\hat{\mathbf{i}}+ \cos t \,\hat{\mathbf{j}}}{1}=− \sin t \,\hat{\mathbf{i}}+ \cos t \,\hat{\mathbf{j}} \end{array}\) -
\(\begin{array}{lrcl} \text{First step:} & \vecs r′(t) & = & (6t+2) \,\hat{\mathbf{i}}−12t^2 \,\hat{\mathbf{j}}+6 \,\hat{\mathbf{k}} \\ \text{Second step:} & ‖\vecs r′(t)‖ & = & \sqrt{(6t+2)^2+(−12t^2)^2+6^2} \\ \text{} & \text{} & = & \sqrt{144t^4+36t^2+24t+40} \\ \text{} & \text{} & = & 2 \sqrt{36t^4+9t^2+6t+10} \\ \text{Third step:} & \vecs T(t) & = & \dfrac{\vecs r′(t)}{‖\vecs r′(t)‖}=\dfrac{(6t+2) \,\hat{\mathbf{i}}−12t^2 \,\hat{\mathbf{j}}+6 \,\hat{\mathbf{k}}}{2 \sqrt{36t^4+9t^2+6t+10}} \\ \text{} & \text{} & = & \dfrac{3t+1}{\sqrt{36t^4+9t^2+6t+10}} \,\hat{\mathbf{i}} - \dfrac{6t^2}{\sqrt{36t^4+9t^2+6t+10}} \,\hat{\mathbf{j}} + \dfrac{3}{\sqrt{36t^4+9t^2+6t+10}} \,\hat{\mathbf{k}} \end{array}\)
Exercise \(\PageIndex{4}\)
Find the unit tangent vector for the vector-valued function
\[\vecs r(t)=(t^2−3)\,\hat{\mathbf{i}}+(2t+1) \,\hat{\mathbf{j}}+(t−2) \,\hat{\mathbf{k}}. \nonumber\]
- Hint
-
Follow the same steps as in Example \(\PageIndex{5}\).
- Answer
-
\[\vecs T(t)=\dfrac{2t}{\sqrt{4t^2+5}}\,\hat{\mathbf{i}}+\dfrac{2}{\sqrt{4t^2+5}}\,\hat{\mathbf{j}}+\dfrac{1}{\sqrt{4t^2+5}}\,\hat{\mathbf{k}} \nonumber\]
Integrals of Vector-Valued Functions
We introduced antiderivatives of real-valued functions in Antiderivatives and definite integrals of real-valued functions in The Definite Integral . Each of these concepts can be extended to vector-valued functions. Also, just as we can calculate the derivative of a vector-valued function by differentiating the component functions separately, we can calculate the antiderivative in the same manner. Furthermore, the Fundamental Theorem of Calculus applies to vector-valued functions as well.
The antiderivative of a vector-valued function appears in applications. For example, if a vector-valued function represents the velocity of an object at time t , then its antiderivative represents position. Or, if the function represents the acceleration of the object at a given time, then the antiderivative represents its velocity.
Definition: Definite and indefinite integrals of vector-valued functions
Let \(f\), \(g\), and \(h\) be integrable real-valued functions over the closed interval \([a,b].\)
-
The indefinite integral of a vector-valued function \(\vecs{r}(t)=f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}}\) is
\[\int [f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}}]\,dt= \left[ \int f(t)\,dt \right] \,\hat{\mathbf{i}}+ \left[ \int g(t)\,dt \right] \,\hat{\mathbf{j}}.\]
The definite integral of a vector-valued function is\[\int_a^b [f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}}]\,dt = \left[ \int_a^b f(t)\,dt \right] \,\hat{\mathbf{i}}+ \left[ \int_a^b g(t)\,dt \right] \,\hat{\mathbf{j}}.\]
-
The indefinite integral of a vector-valued function \(\vecs r(t)=f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}}+h(t) \,\hat{\mathbf{k}}\) is
\[\int [f(t) \,\hat{\mathbf{i}}+g(t)\,\hat{\mathbf{j}} + h(t) \,\hat{\mathbf{k}}]\,dt= \left[ \int f(t)\,dt \right] \,\hat{\mathbf{i}}+ \left[ \int g(t)\,dt \right] \,\hat{\mathbf{j}} + \left[ \int h(t)\,dt \right] \,\hat{\mathbf{k}}.\]
The definite integral of the vector-valued function is\[\int_a^b [f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}} + h(t) \,\hat{\mathbf{k}}]\,dt= \left[ \int_a^b f(t)\,dt \right] \,\hat{\mathbf{i}}+ \left[ \int_a^b g(t)\,dt \right] \,\hat{\mathbf{j}} + \left[ \int_a^b h(t)\,dt \right] \,\hat{\mathbf{k}}.\]
Since the indefinite integral of a vector-valued function involves indefinite integrals of the component functions, each of these component integrals contains an integration constant. They can all be different. For example, in the two-dimensional case, we can have
\[\int f(t)\,dt=F(t)+C_1 \; and \; \int g(t)\,dt=G(t)+C_2, \nonumber\]
where \(F\) and \(G\) are antiderivatives of \(f\) and \(g\), respectively. Then
\[\begin{align*} \int [f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}}]\,dt &= \left[ \int f(t)\,dt \right] \,\hat{\mathbf{i}}+ \left[ \int g(t)\,dt \right] \,\hat{\mathbf{j}} \\[5pt]
&= (F(t)+C_1) \,\hat{\mathbf{i}}+(G(t)+C_2) \,\hat{\mathbf{j}} \\[5pt]
&=F(t) \,\hat{\mathbf{i}}+G(t) \,\hat{\mathbf{j}}+C_1 \,\hat{\mathbf{i}}+C_2 \,\hat{\mathbf{j}} \\[5pt]
&= F(t) \,\hat{\mathbf{i}}+G(t) \,\hat{\mathbf{j}}+\vecs{C} \end{align*}\]
where \(\vecs{C}=C_1 \,\hat{\mathbf{i}}+C_2 \,\hat{\mathbf{j}}\). Therefore, the integration constants becomes a constant vector .
Example \(\PageIndex{7}\): Integrating Vector-Valued Functions
Calculate each of the following integrals:
- \( \displaystyle \int [(3t^2+2t) \,\hat{\mathbf{i}}+(3t−6) \,\hat{\mathbf{j}}+(6t^3+5t^2−4) \,\hat{\mathbf{k}}]\,dt\)
- \( \displaystyle \int [⟨t,t^2,t^3⟩ \times ⟨t^3,t^2,t⟩] \,dt\)
- \( \displaystyle \int_{0}^{\frac{\pi}{3}} [\sin 2t \,\hat{\mathbf{i}}+ \tan t \,\hat{\mathbf{j}}+e^{−2t} \,\hat{\mathbf{k}}]\,dt\)
Solution
-
We use the first part of the definition of the integral of a space curve:
\[\begin{align*} \int[(3t^2+2t)\,\hat{\mathbf{i}}+(3t−6) \,\hat{\mathbf{j}}+(6t^3+5t^2−4)\,\hat{\mathbf{k}}]\,dt &=\left[\int 3t^2+2t\,dt \right]\,\hat{\mathbf{i}}+ \left[\int 3t−6\,dt \right] \,\hat{\mathbf{j}}+ \left[\int 6t^3+5t^2−4\,dt \right] \,\hat{\mathbf{k}} \\[5pt]
&=(t^3+t^2) \,\hat{\mathbf{i}}+\left(\frac{3}{2}t^2−6t\right) \,\hat{\mathbf{j}}+\left(\frac{3}{2}t^4+\frac{5}{3}t^3−4t\right)\,\hat{\mathbf{k}}+\vecs C. \end{align*}\] -
First calculate \(⟨t,t^2,t^3⟩ \times ⟨t^3,t^2,t⟩:\)
\[\begin{align*} ⟨t,t^2,t^3⟩ \times ⟨t^3,t^2,t⟩ &= \begin{vmatrix} \hat{\mathbf{i}} & \,\hat{\mathbf{j}} & \,\hat{\mathbf{k}} \\ t & t^2 & t^3 \\ t^3 & t^2 & t \end{vmatrix} \\[5pt]
Next, substitute this back into the integral and integrate:
&=(t^2(t)−t^3(t^2)) \,\hat{\mathbf{i}}−(t^2−t^3(t^3))\,\hat{\mathbf{j}}+(t(t^2)−t^2(t^3))\,\hat{\mathbf{k}} \\[5pt]
&=(t^3−t^5)\,\hat{\mathbf{i}}+(t^6−t^2)\,\hat{\mathbf{j}}+(t^3−t^5)\,\hat{\mathbf{k}}. \end{align*} \]\[\begin{align*} \int [⟨t,t^2,t^3⟩ \times ⟨t^3,t^2,t⟩]\,dt &= \int (t^3−t^5) \,\hat{\mathbf{i}}+(t^6−t^2) \,\hat{\mathbf{j}}+(t^3−t^5)\,\hat{\mathbf{k}}\,dt \\[5pt]
&=\left(\frac{t^4}{4}−\frac{t^6}{6}\right)\,\hat{\mathbf{i}}+\left(\frac{t^7}{7}−\frac{t^3}{3}\right)\,\hat{\mathbf{j}}+\left(\frac{t^4}{4}−\frac{t^6}{6}\right)\,\hat{\mathbf{k}}+\vecs C. \end{align*}\] -
Use the second part of the definition of the integral of a space curve:
\[\begin{align*} \int_0^{\frac{\pi}{3}} [\sin 2t \,\hat{\mathbf{i}}+ \tan t \,\hat{\mathbf{j}}+e^{−2t} \,\hat{\mathbf{k}}]\,dt &=\left[\int_0^{\frac{π}{3}} \sin 2t \,dt \right] \,\hat{\mathbf{i}}+ \left[ \int_0^{\frac{π}{3}} \tan t \,dt \right] \,\hat{\mathbf{j}}+\left[\int_0^{\frac{π}{3}}e^{−2t}\,dt \right] \,\hat{\mathbf{k}} \\[5pt]
&= \left(−\tfrac{1}{2} \cos 2t \right) \Big\vert_{0}^{π/3} \,\hat{\mathbf{i}}−( \ln |\cos t|)\Big\vert_{0}^{π/3} \,\hat{\mathbf{j}}−\left(\tfrac{1}{2}e^{−2t}\right)\Big\vert_{0}^{π/3} \,\hat{\mathbf{k}} \\[5pt]
&=\left(−\tfrac{1}{2} \cos \tfrac{2π}{3}+\tfrac{1}{2} \cos 0\right) \,\hat{\mathbf{i}}−\left( \ln \left( \cos \tfrac{π}{3}\right)− \ln( \cos 0)\right) \,\hat{\mathbf{j}}−\left( \tfrac{1}{2}e^{−2π/3}−\tfrac{1}{2}e^{−2(0)}\right) \,\hat{\mathbf{k}} \\[5pt]
& =\left(\tfrac{1}{4}+\tfrac{1}{2}\right) \,\hat{\mathbf{i}}−(−\ln 2) \,\hat{\mathbf{j}}−\left(\tfrac{1}{2}e^{−2π/3}−\tfrac{1}{2}\right) \,\hat{\mathbf{k}} \\[5pt]
&=\tfrac{3}{4}\,\hat{\mathbf{i}}+(\ln 2) \,\hat{\mathbf{j}}+\left(\tfrac{1}{2}−\tfrac{1}{2}e^{−2π/3}\right)\,\hat{\mathbf{k}}. \end{align*}\]
Exercise \(\PageIndex{5}\)
Calculate the following integral:
\[\int_1^3 [(2t+4) \,\hat{\mathbf{i}}+(3t^2−4t) \,\hat{\mathbf{j}}]\,dt \nonumber\]
- Hint
-
Use the definition of the definite integral of a plane curve.
- Answer
-
\[\int_1^3 [(2t+4) \,\hat{\mathbf{i}}+(3t^2−4t) \,\hat{\mathbf{j}}]\,dt = 16 \,\hat{\mathbf{i}}+10 \,\hat{\mathbf{j}} \nonumber\]
Integration in the Context of Motion
Above, we defined velocity as the derivative of position and acceleration as the derivative of velocity. Integration allows us to go the other way! As you learned in first semester calculus, integration allows us to generate an object's velocity as a function of time given its acceleration and an initial velocity. We can also find the object's position function by integrating the velocity and using an initial position.
Let's consider an example.
Example \(\PageIndex{8}\): Finding an object's velocity from its acceleration
An object has an acceleration over time given by \(\vecs a(t) = \sin t \,\hat{\mathbf i} - 2t \,\hat{\mathbf j}\), and it's initial velocity was \(\vecs v(0) = 2 \,\hat{\mathbf i} + \,\hat{\mathbf j}\).
a. Find the object's velocity as a function of time.
b. Assuming the object was located at the point (2, 3, -1) when time \(t = 0\), determine the object's position function and find its location at time \(t = 3\) sec.
Solution
a. First, we find the velocity as the antiderivative of the acceleration.
\[\begin{align*} \vecs v(t) = \int \vecs a(t) \, \,dt &= \int \left( \sin t \,\hat{\mathbf i} - 2t \,\hat{\mathbf j}\right)\,dt \\
&= -\cos t \,\hat{\mathbf i} - t^2\,\hat{\mathbf j} + \vecs C_1 \end{align*}\]
Now we use the initial velocity to determine \(\vecs C_1\).
\[\begin{align*} \vecs v(0) = -\cos 0 \,\hat{\mathbf i} - (0)^2\,\hat{\mathbf j} + \vecs C_1 &= 2 \,\hat{\mathbf i} + \,\hat{\mathbf j} \\
-\hat{\mathbf i} + \vecs C_1 &= 2 \,\hat{\mathbf i} + \,\hat{\mathbf j} \\ \\
\text{And so} \quad \vecs C_1 &= 3 \,\hat{\mathbf i} + \,\hat{\mathbf j} \end{align*}\]
Incorporating this constant vector into our velocity function from above, we obtain the velocity describing this object's motion over time:
\[\vecs v(t) = \left( 3 -\cos t \right) \,\hat{\mathbf i} + \left(1 - t^2\right) \,\hat{\mathbf j}\]
b. Since the object's position is at the point (2, 3, -1) when time \(t = 0\), we know \(\vecs r(0) = 2 \,\hat{\mathbf i} + 3 \,\hat{\mathbf j} - \,\hat{\mathbf k}\).
Now, to determine the object's position function, we integrate it's velocity.
\[\begin{align*} \vecs r(t) = \int \vecs v(t) \, \,dt &= \int \bigg[\left( 3 -\cos t \right) \,\hat{\mathbf i} + \left(1 - t^2\right) \,\hat{\mathbf j}\bigg]\,dt \\
&= \left( 3t -\sin t \right) \,\hat{\mathbf i} + \left(t - \frac{t^3}{3}\right) \,\hat{\mathbf j} + \vecs C_2 \end{align*}\]
Now we use the initial position to determine \(\vecs C_2\).
\[\begin{align*} \vecs r(0) = \left( 3(0) -\sin 0 \right) \,\hat{\mathbf i} + \left( 0 - \frac{(0)^3}{3}\right) \,\hat{\mathbf j} + \vecs C_2 &= 2 \,\hat{\mathbf i} + 3 \,\hat{\mathbf j} - \,\hat{\mathbf k}\\ \\ \text{And so} \quad \vecs C_2 &= 2 \,\hat{\mathbf i} + 3 \,\hat{\mathbf j} - \,\hat{\mathbf k} \end{align*}\]
Incorporating this constant vector into our position function from above, we obtain:
\[\vecs r(t) = \left( 2 + 3t -\sin t \right) \,\hat{\mathbf i} + \left(3 + t - \frac{t^3}{3}\right) \,\hat{\mathbf j} - \,\hat{\mathbf k} \]
To find the object's position at time \(t = 3\) seconds, we just evaluate this position function at \(t = 3\).
\[\begin{align*} \vecs r(3) &= \left( 2 + 3(3) -\sin 3 \right) \,\hat{\mathbf i} + \left(3 + 3 - \frac{(3)^3}{3}\right) \,\hat{\mathbf j} - \,\hat{\mathbf k} \\
&= \left( 11 -\sin 3 \right) \,\hat{\mathbf i} - 3 \,\hat{\mathbf j} - \,\hat{\mathbf k} \end{align*}\]
This position vector indicates that the object will be located at the point \( (11 -\sin 3, -3, -1) \) at time \(t = 3\) seconds.
The Meaning of Definite Integrals of Vector-Valued Functions
Now let's turn our attention to the meaning of a definite integral of a vector-valued function. The context in which this will make the most sense is where the function we integrate is a velocity function. That is,
\[\int_a^b \vecs v(t) \, \,dt\]
We know that the antiderivative of velocity is position and that this definite integral gives us the change in position over the time interval, \(a \le t\le b\). In other words,
\[\int_a^b \vecs v(t) \, \,dt = \vecs r(b) - \vecs r(a)\]
Thus, the definite integral of velocity over a time interval \(a \le t\le b\) gives us the displacement vector that indicates the change in position over this time interval.
In general, the definite integral
\(\displaystyle \int_a^b \vecs r(t) \, \,dt = \vecs q(b) - \vecs q(a)\), where \(\vecs q(t)\) is the antiderivative of \(\vecs r(t)\),
gives us a change in the antiderivative of our vector-valued function over the given inteval \([a,b]\). This will always be a constant vector that would fit from tip to tip of the vectors given by the antiderivative function at \(t = a\) and \(t = b\), respectively (assuming the vectors were placed in standard position).
Summary
- To calculate the derivative of a vector-valued function, calculate the derivatives of the component functions, then put them back into a new vector-valued function.
- Many of the properties of differentiation of scalar functions also apply to vector-valued functions.
- The derivative of a vector-valued function \(\vecs r(t)\) is also a tangent vector to the curve. The unit tangent vector \(\vecs T(t)\) is calculated by dividing the derivative of a vector-valued function by its magnitude.
- The antiderivative of a vector-valued function is found by finding the antiderivatives of the component functions, then putting them back together in a vector-valued function.
- The definite integral of a vector-valued function is found by finding the definite integrals of the component functions, then putting them back together in a vector-valued function.
Key Equations
- Derivative of a vector-valued function \[\vecs r′(t) = \lim \limits_{\Delta t \to 0} \dfrac{\vecs r(t+\Delta t)−\vecs r(t)}{ \Delta t} \nonumber\]
- Principal unit tangent vector \[\vecs T(t)=\frac{\vecs r′(t)}{‖\vecs r′(t)‖}\]
- Indefinite integral of a vector-valued function \[\int [f(t) \,\hat{\mathbf{i}}+g(t)\,\hat{\mathbf{j}} + h(t) \,\hat{\mathbf{k}}]\,dt= \left[ \int f(t)\,dt \right] \,\hat{\mathbf{i}}+ \left[ \int g(t)\,dt \right] \,\hat{\mathbf{j}} + \left[ \int h(t)\,dt \right] \,\hat{\mathbf{k}}\nonumber\]
- Definite integral of a vector-valued function \[\int_a^b [f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}} + h(t) \,\hat{\mathbf{k}}]\,dt= \left[\int_a^b f(t)\,dt \right] \,\hat{\mathbf{i}}+ \left[ \int _a^b g(t)\,dt \right] \,\hat{\mathbf{j}} + \left[ \int _a^b h(t)\,dt \right] \,\hat{\mathbf{k}}\nonumber\]
Glossary
- definite integral of a vector-valued function
- the vector obtained by calculating the definite integral of each of the component functions of a given vector-valued function, then using the results as the components of the resulting function
- derivative of a vector-valued function
- the derivative of a vector-valued function \(\vecs{r}(t)\) is \(\vecs{r}′(t) = \lim \limits_{\Delta t \to 0} \frac{\vecs r(t+\Delta t)−\vecs r(t)}{ \Delta t}\), provided the limit exists
- indefinite integral of a vector-valued function
- a vector-valued function with a derivative that is equal to a given vector-valued function
- principal unit tangent vector
- a unit vector tangent to a curve C
- tangent vector
- to \(\vecs{r}(t)\) at \(t=t_0\) any vector \(\vecs v\) such that, when the tail of the vector is placed at point \(\vecs r(t_0)\) on the graph, vector \(\vecs{v}\) is tangent to curve C
Contributors
-
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .
- Paul Seeburger (Monroe Community College) edited this section and added content on visualizing the derivative of a vector-valued function and content on smooth vector-valued functions.


4.10: Conditional Expected Value Revisited
Conditional expected value is much more important than one might at first think. In fact, conditional expected value is at the core of modern probability theory because it provides the basic way of incorporating known information into a probability measure.
Basic Theory
Definition
As usual, our starting point is a random experiment modeled by a probability space \((\Omega, \mathscr F, \P)\), so that \( \Omega \) is the set of outcomes,, \( \mathscr F \) is the \( \sigma \)-algebra of events, and \( \P \) is the probability measure on the sample space \( (\Omega, \mathscr F) \). In our first elementary discussion, we studied the conditional expected value of a real-value random variable \(X\) given a general random variable \(Y\). The more general approach is to condition on a sub \(\sigma\)-algebra \( \mathscr G \) of \( \mathscr F \). The sections on \( \sigma \)-algebras and measure theory are essential prerequisites for this section.
Before we get to the definition, we need some preliminaries. First, all random variables mentioned are assumed to be real valued. next the notion of equivalence plays a fundamental role in this section. Next recall that random variables \( X_1 \) and \( X_2 \) are equivalent if \( \P(X_1 = X_2) = 1 \). Equivalence really does define an equivalence relation on the collection of random variables defined on the sample space. Moreover, we often regard equivalent random variables as being essentially the same object. More precisely from this point of view, the objects of our study are not individual random variables but rather equivalence classes of random variables under this equivalence relation. Finally, for \( A \in \mathscr F \), recall the notation for the expected value of \( X \) on the event \( A \) \[ \E(X; A) = \E(X \bs{1}_A)\] assuming of course that the expected value exists. For the remainder of this subsection, suppose that \( \mathscr G \) is a sub \( \sigma \)-algebra of \( \mathscr F \).
Suppose that \(X\) is a random variable with \( \E(|X|) \lt \infty \). The conditional expected value of \(X\) given \(\mathscr G\) is the random variable \(\E(X \mid \mathscr G)\) defined by the following properties:
- \(\E(X \mid \mathscr G)\) is measurable with repsect to \(\mathscr G\).
- If \(A \in \mathscr G\) then \(\E[\E(X \mid \mathscr G); A] = \E(X; A)\)
The basic idea is that \( \E(X \mid \mathscr G) \) is the expected value of \( X \) given the information in the \( \sigma \)-algebra \( \mathscr G \). Hopefully this idea will become clearer during our study. The conditions above uniquely define \( \E(X \mid \mathscr G) \) up to equivalence. The proof of this fact is a simple application of the Radon-Nikodym theorem, named for Johann Radon and Otto Nikodym
Suppose again that \(X\) is a random variable with \( \E(|X|) \lt \infty \).
- There exists a random variable \( V \) satisfying the definition .
- If \( V_1 \) and \( V_2 \) satisfy the definition , then \( \P(V_1 = V_2) = 1 \) so that \( V_1 \) and \( V_2 \) are equivalent.
Proof
- Note that \( \nu(A) = \E(X; A) \) for \( A \in \mathscr G \) defines a (signed) measure on \( \mathscr G \). Moreover, if \( A \in \mathscr G \) and \( \P(A) = 0 \) then \( \nu(A) = 0 \). Hence \( \nu \) is absolutely continuous with respect to the restriction of \( \P \) to \( \mathscr G \). By the Radon-Nikodym theorem, there exists a random variable \( V \) that is measurable with respect to \( \mathscr G \) such that \( \nu(A) = \E(V; A) \) for \( A \in \mathscr G \). That is, \( V \) is the density or derivative of \( \nu \) with respect to \( \P \) on \( \mathscr G \).
- This follows from the uniqueness of the Radon-Nikodym derivative, up to equivalence.
The following characterization might seem stronger but in fact in equivalent to the definition.
Suppose again that \(X\) is a random variable with \( \E(|X|) \lt \infty \). Then \( \E(X \mid \mathscr G) \) is characterized by the following properties:
- \( \E(X \mid \mathscr G) \) is measurable with respect to \( \mathscr G \)
- If \( U \) is measurable with respect to \( \mathscr G \) and \( \E(|U X|) \lt \infty \) then \( \E[U \E(X \mid \mathscr G)] = \E(U X) \).
Proof
We have to show that part (b) in the definition is equivalent to part (b) here. First (b) here implies (b) in the definition since \( \bs{1}_A \) is \( \mathscr G \)-measurable if \( A \in \mathscr G \). Conversely suppose that (b) in the definition holds. We will show that (b) here holds by a classical bootstrapping argument. . First \( \E[U \E(X \mid \mathscr G)] = \E(U X) \) if \( U = \bs{1}_A \) for some \( A \in \mathscr G \). Next suppose that \( U \) is a simple random variable that is \( \mathscr G \)-measurable. That is, \( U = \sum_{i \in I} a_i \bs{1}_{A_i} \) where \( I \) is a finite index set, \( a_i \ge 0 \) for \( i \in I \), and \( A_i \in \mathscr G \) for \( i \in I \). then \[ \E[U \E(X \mid \mathscr G)] = \E\left[\sum_{i \in I} a_i \bs{1}_{A_i} \E(X \mid \mathscr G)\right] = \sum_{i \in I} a_i \E[\bs{1}_{A_i} \E(X \mid \mathscr G)] = \sum_{i \in I} a_i \E(\bs{1}_{A_i} X) = \E\left(\sum_{i \in I} a_i \bs{1}_{A_i} X\right) = \E(U X) \] Next suppose that \( U \) is nonnegative and \( \mathscr G \)-measurable. Then there exists a sequence of simple \( \mathscr G \)-measurable random variables \( (U_1, U_2, \ldots) \) with \( U_n \uparrow U \) as \( n \to \infty \). Then by the previous step, \( \E[U_n \E(X \mid \mathscr G)] = \E(U_n X) \) for each \( n \). Letting \( n \to \infty \) and using the monotone convergence theorem we have \( \E[U \E(X \mid \mathscr G)] = \E(U X) \). Finally, suppose that \( U \) is a general \( \mathscr G \)-measurable random variable. Then \( U = U^+ - U^- \) where \( U^+ \) and \( U^- \) are the usual positive and negative parts of \( U \). These parts are nonnegative and \( \mathscr G \)-measurable, so by the previous step, \( \E[U^+ \E(X \mid \mathscr G)] = \E(U^+ X) \) and \( \E[U^- \E(X \mid \mathscr G)] = \E(U^- X) \). hence \[ \E[U \E(X \mid \mathscr G)] = \E[(U^+ - U^-) \E(X \mid \mathscr G)] = \E[U^+ \E(X \mid \mathscr G)] - \E[U^- \E(X \mid \mathscr G)] = \E(U^+ X) - \E(U^- X) = \E(U X) \]
Properties
Our next discussion concerns some fundamental properties of conditional expected value. All equalities and inequalities are understood to hold modulo equivalence , that is, with probability 1. Note also that many of the proofs work by showing that the right hand side satisfies the properties in the definition for the conditional expected value on the left side. Once again we assume that \( \mathscr G \) is a sub \( sigma \)-algebra of \( \mathscr F \).
Our first property is a simple consequence of the definition: \( X \) and \( \E(X \mid \mathscr G) \) have the same mean.
Suppose that \(X\) is a random variable with \( \E(|X|) \lt \infty \). Then \( \E[\E(X \mid \mathscr G)] = \E(X) \).
Proof
This follows immediately by letting \( A = \Omega \) in the definition .
The result above can often be used to compute \( \E(X) \), by choosing the \( \sigma \)-algebra \( \mathscr G \) in a clever way. We say that we are computing \( \E(X) \) by conditioning on \( \mathscr G \). Our next properties are fundamental: every version of expected value must satisfy the linearity properties. The first part is the additive property and the second part is the scaling property .
Suppose that \( X \) and \( Y \) are random variables with \( \E(|X|) \lt \infty \) and \( \E(|Y|) \lt \infty \), and that \( c \in \R\). Then
- \( \E(X + Y \mid \mathscr G) = \E(X \mid \mathscr G) + \E(Y \mid \mathscr G) \)
- \( \E(c X \mid \mathscr G) = c \E(X \mid \mathscr G) \)
Proof
- Note that \( \E(|X + Y|) \le \E(|X|) + \E(|Y|) \lt \infty\) so \( \E(X + Y \mid \mathscr G) \) is defined. We show that \( \E(X \mid \mathscr G) + \E(Y \mid \mathscr G) \) satisfies the conditions in the definition for \( \E(X + Y \mid \mathscr G) \). Note first that \( \E(X \mid \mathscr G) + \E(Y \mid \mathscr G) \) is \( \mathscr G \)-measurable since both terms are. If \(A \in \mathscr G \) then \[ \E\{[\E(X \mid \mathscr G) + \E(Y \mid \mathscr G)]; A\} = \E[\E(X \mid \mathscr G); A] + \E[\E(Y \mid \mathscr G); A] = \E(X; A) + \E(Y; A) = \E[X + Y; A] \]
- Note that \( \E(|c X|) = |c| \E(|X|) \lt \infty \) so \( \E(c X \mid \mathscr G) \) is defined. We show that \( c \E(X \mid \mathscr G) \) satisfy the conditions in the definition for \( \E(c X \mid \mathscr G) \). Note first that \( c \E(X \mid \mathscr G) \) is \( \mathscr G \)-measurable since the second factor is. If \(A \in \mathscr G \) then \[ \E[c \E(X \mid \mathscr G); A] = c \E[\E(X \mid \mathscr G); A] = c \E(X; A) = \E(c X; A) \]
The next set of properties are also fundamental to every notion of expected value. The first part is the positive property and the second part is the increasing property .
Suppose again that \( X \) and \( Y \) are random variables with \( \E(|X|) \lt \infty \) and \( \E(|Y|) \lt \infty \).
- If \( X \ge 0 \) then \( \E(X \mid \mathscr G) \ge 0 \)
- If \( X \le Y \) then \( \E(X \mid \mathscr G) \le \E(Y \mid \mathscr G) \)
Proof
- Let \( A = \{\E(X \mid \mathscr G) \lt 0\} \). Note that \(A \in \mathscr G \) and hence \( \E(X; A) = \E[\E(X \mid \mathscr G); A] \). Since \( X \ge 0 \) with probability 1 we have \( E(X; A) \ge 0 \). On the other hand, if \( \P(A) \gt 0 \) then \( \E[\E(X \mid \mathscr G); A] \lt 0 \) which is a contradiction. Hence we must have \( \P(A) = 0 \).
- Note that if \( X \le Y \) then \( Y - X \ge 0 \). Hence by (a) and the additive property, \( \E(Y - X \mid \mathscr G) = \E(Y \mid \mathscr G) - \E(X \mid \mathscr G) \ge 0 \) so \( \E(Y \mid \mathscr G) \ge \E(X \mid \mathscr G) \).
The next few properties relate to the central idea that \( \E(X \mid \mathscr G) \) is the expected value of \( X \) given the information in the \( \sigma \)-algebra \( \mathscr G \).
Suppose that \( X \) and \( V \) are random variables with \( \E(|X|) \lt \infty \) and \( \E(|X V|) \lt \infty \) and that \( V \) is measurable with respect to \( \mathscr G \). Then \( \E(V X \mid \mathscr G) = V \E(X \mid \mathscr G) \).
Proof
We show that \( V \E(X \mid \mathscr G) \) satisfy the in properties that characterize \( \E(V X \mid \mathscr G) \). First, \( V \E(X \mid \mathscr G) \) is \( \mathscr G \)-measurable since both factors are. If \( U \) is \( \mathscr G \)-measurable with \( \E(|U V X|) \lt \infty \) then \( U V \) is also \( \mathscr G \)-measurable and hence \[ \E[U V \E(X \mid \mathscr G)] = \E(U V X) = \E[U (V X)] \]
Compare this result with the scaling property . If \( V \) is measurable with respect to \( \mathscr G \) then \( V \) is like a constant in terms of the conditional expected value given \( \mathscr G \). On the other hand, note that this result implies the scaling property , since a constant can be viewed as a random variable, and as such, is measurable with respect to any \( \sigma \)-algebra. As a corollary to this result, note that if \( X \) itself is measurable with respect to \( \mathscr G \) then \( \E(X \mid \mathscr G) = X \). The following result gives the other extreme.
Suppose that \( X \) is a random variable with \( \E(|X|) \lt \infty \). If \( X \) and \( \mathscr G \) are independent then \( \E(X \mid \mathscr G) = \E(X) \).
Proof
We show that \( \E(X) \) satisfy the properties in the definiton for \( \E(X \mid \mathscr G) \). First of course, \( \E(X) \) is \( \mathscr G \)-measurable as a constant random variable. If \( A \in \mathscr G \) then \( X \) and \( \bs{1}_A \) are independent and hence \[ \E(X; A) = \E(X) \P(A) = \E[\E(X); A] \]
Every random variable \( X \) is independent of the trivial \( \sigma \)-algebra \( \{\emptyset, \Omega\} \) so it follows that \( \E(X \mid \{\emptyset, \Omega\}) = \E(X) \).
The next properties are consistency conditions , also known as the tower properties . When conditioning twice, with respect to nested \( \sigma \)-algebras, the smaller one (representing the least amount of information) always prevails.
Suppose that \( X \) is a random variable with \( \E(|X|) \lt \infty \) and that \( \mathscr H \) is a sub \( \sigma \)-algebra of \( \mathscr G \). Then
- \( \E[\E(X \mid \mathscr H) \mid \mathscr G] = \E(X \mid \mathscr H) \)
- \( \E[\E(X \mid \mathscr G) \mid \mathscr H] = \E(X \mid \mathscr H) \)
Proof
- Note first that \( \E(X \mid \mathscr H) \) is \( \mathscr H \)-measurable and hence also \( \mathscr G \)-measurable. Thus by (7) , \( \E[\E(X \mid \mathscr H) \mid \mathscr G] = \E(X \mid \mathscr H) \).
- We show that \( \E(X \mid \mathscr H) \) satisfies the coonditions in the definition for \( \E[\E(X \mid \mathscr G) \mid \mathscr H]\). Note again that \( \E(X \mid \mathscr H) \) is \( \mathscr H \)-measurable. If \(A \in \mathscr H \) then \( A \in \mathscr G \) and hence \[ \E[\E(X \mid \mathscr G); A] = \E(X; A) = \E[\E(X \mid \mathscr H); A] \]
The next result gives Jensen's inequality for conditional expected value, named for Johan Jensen.
Suppose that \( X \) takes values in an interval \( S \subseteq \R \) and that \( g: S \to \R \) is convex. If \( \E(|X|) \lt \infty \) and \( \E(|g(X)| \lt \infty \) then \[ \E[g(X) \mid \mathscr G] \ge g[\E(X \mid \mathscr G)] \]
Proof
As with Jensen's inequality for ordinary expected value, the best proof uses the characterization of convex functions in terms of supporting lines : For each \( t \in S \) there exist numbers \( a \) and \( b \) (depending on \( t \)) such that
- \( a + b t = g(t) \)
- \( a + b x \le g(x) \) for \( x \in S \)
Random variables \( X \) and \( \E(X \mid \mathscr G) \) takes values in \( S \). We can construct a random supporting line at \( \E(X \mid \mathscr G) \). That is, there exist random variables \( A \) and \( B \), measurable with respect to \( \mathscr G \), such that
- \( A + B \E(X \mid \mathscr G) = g[\E(X \mid \mathscr G)] \)
- \( A + B X \le g(X) \)
We take conditional expected value through the inequality in (b) and then use properties of conditional expected value and property (a): \[ \E[g(X) \mid \mathscr G] \ge \E(A + B X \mid \mathscr G) = A + B \E(X \mid \mathscr G) = g[\E(X \mid \mathscr G] \] Note that the second step uses the fact that \( A \) and \( B \) are measurable with respect to \( \mathscr G \).
Conditional Probability
For our next discussion, suppose as usual that \( \mathscr G \) is a sub \( \sigma \)-algebra of \( \mathscr F \). The conditional probability of an event \(A\) given \( \mathscr G \) can be defined as a special case of conditional expected value. As usual, let \(\bs{1}_A\) denote the indicator random variable of \(A\).
For \( A \in \mathscr F \) we define \[ \P(A \mid \mathscr G) = \E(\bs 1_A \mid \mathscr G) \]
Thus, we have the following characterizations of conditional probability, which are special cases of the definition and the alternate version :
If \( A \in \mathscr F \) then \( \P(A \mid \mathscr G) \) is characterized (up to equivalence) by the following properties
- \( \P(A \mid \mathscr G) \) is measurable with respect to \( \mathscr G \).
- If \( B \in \mathscr G \) then \( \E[\P(A \mid \mathscr G); B] = \P(A \cap B) \)
Proof
For part (b), note that \[ \E[\bs{1}_B \P(A \mid \mathscr G)] = \E[\bs{1}_B \E(\bs{1}_A \mid \mathscr G)] = \E(\bs{1}_A \bs{1}_B) = \E(\bs{1}_{A \cap B}) = \P(A \cap B) \]
If \( A \in \mathscr F \) then \( \P(A \mid \mathscr G) \) is characterized (up to equivalence) by the following properties
- \( \P(A \mid \mathscr G) \) is measurable with respect to \( \mathscr G \).
- If \( U \) is measurable with respect to \( \mathscr G \) and \( \E(|U|) \lt \infty \) then \( \E[U \P(A \mid \mathscr G)] = \E(U; A) \)
The properties above for conditional expected value, of course, have special cases for conditional probability. In particular, we can compute the probability of an event by conditioning on a \( \sigma \)-algebra:
If \( A \in \mathscr F \) then \(\P(A) = \E[\P(A \mid \mathscr G)]\).
Proof
This is a direct result of the mean property since \( \E(\bs{1}_A) = \P(A) \).
Again, the last theorem is often a good way to compute \(\P(A)\) when we know the conditional probability of \(A\) given \(\mathscr G\). This is a very compact and elegant version of the law of total probability given first in the section on Conditional Probability in the chapter on Probability Spaces and later in the section on Discrete Distributions in the Chapter on Distributions. The following theorem gives the conditional version of the axioms of probability.
The following properties hold (as usual, modulo equivalence):
- \( \P(A \mid \mathscr G) \ge 0 \) for every \( A \in \mathscr F \)
- \( \P(\Omega \mid \mathscr G) = 1 \)
- If \( \{A_i: i \in I\} \) is a countable disjoint subset of \( \mathscr F \) then \( \P(\bigcup_{i \in I} A_i \bigm| \mathscr G) = \sum_{i \in I} \P(A_i \mid \mathscr G) \)
Proof
- This is a direct consequence of (6) .
- This is trivial since \( \bs{1}_\Omega = 1 \).
- We show that the right side satisfies the conditions in (11) that define the left side. Note that \( \sum_{i \in I} \P(A_i \mid \mathscr G) \) is \( \mathscr G \)-measurable since each term in the sum has this property. Let \( B \in \mathscr G \). then \[ \E\left[\sum_{i \in I} \P(A_i \mid \mathscr G); B\right] = \sum_{i \in I} \E[\P(A_i \mid \mathscr G); B] = \sum_{i \in I} \P(A_i \cap B) = \P\left(B \cap \bigcup_{i \in I} A_i\right) \]
From the last result, it follows that other standard probability rules hold for conditional probability given \( \mathscr G \) (as always, modulo equivalence). These results include
- the complement rule
- the increasing property
- Boole's inequality
- Bonferroni's inequality
- the inclusion-exclusion laws
However, it is not correct to state that \( A \mapsto \P(A \mid \mathscr G) \) is a probability measure, because the conditional probabilities are only defined up to equivalence, and so the mapping does not make sense. We would have to specify a particular
version
of \( \E(A \mid \mathscr G) \) for each \( A \in \mathscr F \) for the mapping to make sense. Even if we do this, the mapping may not define a probability measure. In part (c), the left and right sides are random variables and the equation is an event that has probability 1. However this event depends on the collection \( \{A_i: i \in I\} \). In general, there will be uncountably many such collections in \( \mathscr F \), and the intersection of all of the corresponding events may well have probability less than 1 (if it's measurable at all). It turns out that if the underlying probability space \( (\Omega, \mathscr F, \P) \) is sufficiently
nice
(and most probability spaces that arise in applications are nice), then there does in fact exist a
regular conditional probability
. That is, for each \( A \in \mathscr F \), there exists a random variable \( \P(A \mid \mathscr G) \) satisfying the conditions in
(12)
and such that with probability 1, \( A \mapsto \P(A \mid \mathscr G) \) is a probability measure.
The following theorem gives a version of Bayes' theorem , named for the inimitable Thomas Bayes.
Suppose that \( A \in \mathscr G \) and \( B \in \mathscr F \). then \[ \P(A \mid B) = \frac{\E[\P(B \mid \mathscr G); A]}{\E[\P(B \mid \mathscr G)]} \]
Proof
The proof is absolutely trivial. By definition of conditional probability given \( \mathscr G \), the numerator is \( \P(A \cap B) \) and the denominator is \( P(B) \). Nonetheless, Bayes' theorem is useful in settings where the expected values in the numerator and denominator can be computed directly
Basic Examples
The purpose of this discussion is to tie the general notions of conditional expected value that we are studying here to the more elementary concepts that you have seen before. Suppose that \( A \) is an event (that is, a member of \( \mathscr F \)) with \( \P(A) \gt 0 \). If \( B \) is another event, then of course, the conditional probability of \( B \) given \( A \) is \[ \P(B \mid A) = \frac{\P(A \cap B)}{\P(A)} \] If \( X \) is a random variable then the conditional distribution of \( X \) given \( A \) is the probability measure on \( \R \) given by \[ R \mapsto \P(X \in R \mid A) = \frac{\P(\{X \in R\} \cap A)}{\P(A)} \text{ for measurable } R \subseteq \R \] If \( \E(|X|) \lt \infty \) then the conditional expected value of \( X \) given \( A \), denoted \( \E(X \mid A) \), is simply the mean of this conditional distribution.
Suppose now that \( \mathscr{A} = \{A_i: i \in I\} \) is a countable partition of the sample space \( \Omega \) into events with positive probability. To review the jargon, \( \mathscr A \subseteq \mathscr F \); the index set \( I \) is countable; \( A_i \cap A_j = \emptyset \) for distinct \( i, \, j \in I \); \( \bigcup_{i \in I} A_i = \Omega \); and \( \P(A_i) \gt 0 \) for \( i \in I \). Let \( \mathscr G = \sigma(\mathscr{A}) \), the \( \sigma \)-algebra generated by \( \mathscr{A} \). The elements of \( \mathscr G \) are of the form \( \bigcup_{j \in J} A_j \) for \( J \subseteq I \). Moreover, the random variables that are measurable with respect to \( \mathscr G \) are precisely the variables that are constant on \( A_i \) for each \( i \in I \). The \( \sigma \)-algebra \( \mathscr G \) is said to be countably generated .
If \( B \in \mathscr F \) then \( \P(B \mid \mathscr G) \) is the random variable whose value on \( A_i \) is \( \P(B \mid A_i) \) for each \(i \in I \).
Proof
Let \( U \) denote the random variable that takes the value \( \P(B \mid A_i) \) on \( A_i \) for each \( i \in I \). First, \( U \) is measurable with respect to \( \scr G \) since \( U \) is constant on \( A_i \) for each \( i \in I \). So we just need to show that \( E(U ; A) = \P(A \cap B) \) for each \( A \in \mathscr G \). Thus, let \( A = \bigcup_{j \in J} A_j \) where \( J \subseteq I \). Then \[ \E(U; A) = \sum_{j \in J} \E(U ; A_j) = \sum_{j \in J} \P(B \mid A_j) \P(A_j) = \P(A \cap B)\]
In this setting, the version of Bayes' theorem in (15) reduces to the usual elementary formulation: For \( i \in I \), \( \E[\P(B \mid \mathscr G); A_i] = \P(A_i) \P(B \mid A_i) \) and \( \E[\P(B \mid \mathscr G)] = \sum_{j \in I} \P(A_j) \P(B \mid A_j) \). Hence \[ \P(A_i \mid B) = \frac{\P(A_i) \P(B \mid A_i)}{\sum_{j \in I} \P(A_j) \P(B \mid A_j)} \]
If \( X \) is a random variable with \( \E(|X|) \lt \infty \), then \( \E(X \mid \mathscr G) \) is the random variable whose value on \( A_i \) is \( \E(X \mid A_i) \) for each \( i \in I \).
Proof
Let \( U \) denote the random variable that takes the value \( \E(X \mid A_i) \) on \( A_i \) for each \( i \in I \). First, \( U \) is measurable with respect to \( \scr G \) since \( U \) is constant on \( A_i \) for each \( i \in I \). So we just need to show that \( E(U; A) = \E(X; A) \) for each \( A \in \mathscr G \). Thus, let \( A = \bigcup_{j \in J} A_j \) where \( J \subseteq I \). Then \[ \E(U; A) = \sum_{j \in J} \E(U; A_j) = \sum_{j \in J} \E(X \mid A_j) \P(A_j) = E(X; A) \]
The previous examples would apply to \( \mathscr G = \sigma(Y) \) if \( Y \) is a discrete random variable taking values in a countable set \( T \). In this case, the partition is simply \( \mathscr{A} = \{ \{Y = y\}: y \in T\} \). On the other hand, suppose that \( Y \) is a random variable taking values in a general set \( T \) with \( \sigma \)-algebra \( \mathscr{T} \). The real-valued random variables that are measurable with respect to \( \mathscr G = \sigma(Y) \) are (up to equivalence) the measurable, real-valued functions of \( Y \).
Specializing further, Suppose that \( X \) takes values in \( S \subseteq \R \), \( Y \) takes values in \( T \subseteq \R^n \) (where \( S \) and \( T \) are Lebesgue measurable) and that \( (X, Y) \) has a joint continuous distribution with probability density function \( f \). Then \( Y \) has probability density function \( h \) given by \[ h(y) = \int_S f(x, y) \, dx, \quad y \in T \] Assume that \( h(y) \gt 0 \) for \( y \in T \). Then for \( y \in T \), a conditional probability density function of \( X \) given \( Y = y \) is defined by \[ g(x \mid y) =\frac{f(x, y)}{h(y)}, \quad x \in S \] This is precisely the setting of our elementary discussion of conditional expected value. If \( \E(|X|) \lt \infty \) then we usually write \( \E(X \mid Y) \) instead of the clunkier \( \E[X \mid \sigma(Y)] \).
In this setting above suppose that \( \E(|X|) \lt \infty \). Then \[ \E(X \mid Y) = \int_S x g(x \mid Y) \, dx \]
Proof
Once again, we show that the integral on the right satisfies the properties in the definition for \( \E(X \mid Y) = \E[X \mid \sigma(Y)] \). First, \( y \mapsto \int_S x g(x \mid y) \, dx \) is measurable as a function from \( T \) into \( \R \) and hence the random variable \( \int_x g(x \mid Y) \, dx \) is a measurable function of \( Y \) and so is measurable with respect to \( \sigma (Y) \). Next suppose that \( B \in \sigma(Y) \). Then \( B = \{Y \in A\} \) for some \( A \in \mathscr F \). Then \begin{align*} \E\left[\int_S x g(x \mid Y) \, dx; B\right] & = \E\left[\int_S x g(x \mid Y) \, dx; Y \in A\right] \\ & = \E\left[\int_S x \frac{f(x, y)}{h(y)} \, dx; Y \in A\right] = \int_A \int_S x \frac{f(x, y)}{h(y)} h(y) \, dx \, dy \\ & = \int_{S \times A} x f(x, y) \, d(x, y) = \E(X; Y \in A) = \E(X; B) \end{align*}
Best Predictor
In our elementary treatment of conditional expected value, we showed that the conditional expected value of a real-valued random variable \( X \) given a general random variable \( Y \) is the best predictor of \( X \), in the least squares sense, among all real-valued functions of \( Y \). A more careful statement is that \( \E(X \mid Y) \) is the best predictor of \( X \) among all real-valued random variables that are measurable with respect to \( \sigma(Y) \). Thus, it should come as not surprise that if \( \mathscr G \) is a sub \( \sigma \)-algebra of \( \mathscr F \), then \( \E(X \mid \mathscr G) \) is the best predictor of \( X \), in the least squares sense, among all real-valued random variables that are measurable with respect to \( \mathscr G) \). We will show that this is indeed the case in this subsection. The proofs are very similar to the ones given in the elementary section. For the rest of this discussion, we assume that \( \mathscr G \) is a sub \( \sigma \)-algebra of \( \mathscr F \) and that all random variables mentioned are real valued.
Suppose that \( X \) and \( U \) are random variables with \( \E(|X|) \lt \infty \) and \( \E(|X U|) \lt \infty \) and that \( U \) is measurable with respect to \( \mathscr G \). Then \( X - \E(X \mid \mathscr G) \) and \( U \) are uncorrelated.
Proof
Note that \( X - \E(X \mid \mathscr G) \) has mean 0 by the mean property . Using the properties that characterize \( \E(X \mid \mathscr G) \) we have \[ \cov[X - \E(X \mid \mathscr G), U] = \E(U [X - \E(X \mid \mathscr G)]) = \E(U X) - \E[U \E(X \mid \mathscr G] = \E(U X) - \E(U X) = 0 \]
The next result is the main one: \( \E(X \mid \mathscr G) \) is closer to \( X \) in the mean square sense than any other random variable that is measurable with respect to \( \mathscr G \). Thus, if \( \mathscr G \) represents the information that we have, then \( \E(X \mid \mathscr G) \) is the best we can do in estimating \( X \).
Suppose that \( X \) and \( U \) are random variables with \( \E(X^2) \lt \infty \) and \( \E(U^2) \lt \infty\) and that \( U \) is measurable with respect to \( \mathscr G \). Then
- \( \E([X - \E(X \mid \mathscr G)]^2) \le \E[(X - U)^2] \).
- Equality holds if and only if \(\P[U = \E(X \mid \mathscr G)] = 1 \), so \( U \) and \( \E(X \mid \mathscr G) \) are equivalent.
Proof
- Note that \begin{align} \E[(X - U)^2] & = \E([X - \E(X \mid \mathscr G) + \E(X \mid \mathscr G) - U]^2) \\ & = \E([X - \E(X \mid \mathscr G)]^2) + 2 \E([X - \E(X \mid \mathscr G)][\E(X \mid \mathscr G) - U]) + \E([\E(X \mid \mathscr G) - U]^2 ) \end{align} By mean property , \( X - \E(X \mid \mathscr G) \) has mean 0, so the middle term in the displayed equation is \( 2 \cov[X - \E(X \mid \mathscr G), \E(X \mid \mathscr G) - U] \). But \( \E(X \mid \mathscr G) - U \) is \( \mathscr G \)-measurable and hence this covariance is 0 by uncorrelated proerty . Therefore \[ \E[(X - U)^2] = \E([X - \E(X \mid \mathscr G)]^2) + \E([\E(X \mid \mathscr G) - U]^2 ) \ge \E([X - \E(X \mid \mathscr G)]^2) \]
- Equality holds if and only if \( \E([\E(X \mid \mathscr G) - U]^2 ) = 0 \) if and only if \(\P[U = \E(X \mid \mathscr G)] = 1 \)
Conditional Variance
Once again, we assume that \( \mathscr G \) is a sub \( \sigma \)-algebra of \( \mathscr F \) and that all random variables mentioned are real valued, unless otherwise noted. It's natural to define the conditional variance of a random variable given \( \mathscr G \) in the same way as ordinary variance, but witl all expected values conditioned on \( \mathscr G \).
Suppose that \( X \) is a random variable with \( \E(X^2) \lt \infty \). The conditional variance of \( X \) given \(\mathscr G\) is \[ \var(X \mid \mathscr G) = \E\left([X - \E(X \mid \mathscr G)]^2 \biggm| \mathscr G\right) \]
Like all conditional expected values relative to \( \mathscr G \), \( \var(X \mid \mathscr G) \) is a random variable that is measurable with respect to \( \mathscr G \) and is unique up to equivalence. The first property is analogous to the computational formula for ordinary variance.
Suppose again that \( X \) is a random variable with \( \E(X^2) \lt \infty \). Then \[\var(X \mid \mathscr G) = \E(X^2 \mid \mathscr G) - [\E(X \mid \mathscr G)]^2\]
Proof
Expanding the square in the definition and using basic properties of conditional expectation, we have
\begin{align} \var(X \mid \mathscr G) & = \E(X^2 - 2 X \E(X \mid \mathscr G) + [\E(X \mid \mathscr G)]^2 \biggm| \mathscr G ) = \E(X^2 \mid \mathscr G) - 2 \E[X \E(X \mid \mathscr G) \mid \mathscr G] + \E([\E(X \mid \mathscr G)]^2 \mid \mathscr G) \\ & = \E(X^2 \mid \mathscr G) - 2 \E(X \mid \mathscr G) \E(X \mid \mathscr G) + [\E(X \mid \mathscr G)]^2 = \E(X^2 \mid \mathscr G) - [\E(X \mid \mathscr G)]^2 \end{align}Next is a formula for the ordinary variance in terms of conditional variance and expected value.
Suppose again that \( X \) is a random variable with \( \E(X^2) \lt \infty \). Then \[\var(X) = \E[\var(X \mid \mathscr G)] + \var[\E(X \mid \mathscr G)]\]
Proof
From the previous theorem and properties of conditional expected value we have \( \E[\var(X \mid \mathscr G)] = \E(X^2) - \E([\E(X \mid \mathscr G)]^2) \). But \( \E(X^2) = \var(X) + [\E(X)]^2 \) and similarly, \(\E([\E(X \mid \mathscr G)]^2) = \var[\E(X \mid \mathscr G)] + (\E[\E(X \mid \mathscr G)])^2 \). But also, \( \E[\E(X \mid \mathscr G)] = \E(X) \) so subsituting we get \( \E[\var(X \mid \mathscr G)] = \var(X) - \var[\E(X \mid \mathscr G)] \).
So the variance of \( X \) is the expected conditional variance plus the variance of the conditional expected value. This result is often a good way to compute \(\var(X)\) when we know the conditional distribution of \(X\) given \(\mathscr G\). In turn, this property leads to a formula for the mean square error when \( \E(X \mid \mathscr G) \) is thought of as a predictor of \( X \).
Suppose again that \( X \) is a random variable with \( \E(X^2) \lt \infty \). \[ \E([X - \E(X \mid \mathscr G)]^2) = \var(X) - \var[\E(X \mid \mathscr G)] \]
Proof
From the definition and from the mean property and variance formula , \[ \E([X - \E(X \mid \mathscr G)]^2) = \E[\var(X \mid \mathscr G)] = \var(X) - \var[\E(X \mid \mathscr G)] \]
Let us return to the study of predictors of the real-valued random variable \(X\), and compare them in terms of mean square error.
Suppose again that \( X \) is a random variable with \( \E(X^2) \lt \infty \).
- The best constant predictor of \(X\) is \(\E(X)\) with mean square error \(\var(X)\).
- If \(Y\) is another random variable with \( \E(Y^2) \lt \infty \), then the best predictor of \(X\) among linear functions of \(Y\) is \[ L(X \mid Y) = \E(X) + \frac{\cov(X,Y)}{\var(Y)}[Y - \E(Y)] \] with mean square error \( \var(X)[1 - \cor^2(X,Y)]\).
- If \(Y\) is a (general) random variable, then the best predictor of \(X\) among all real-valued functions of \(Y\) with finite variance is \(\E(X \mid Y)\) with mean square error \( \var(X) - \var[\E(X \mid Y)]\).
- If \(\mathscr G\) is a sub \( \sigma \)-algebra of \( \mathscr F \), then the best predictor of \(X\) among random variables with finite variance that are measurable with respect to \(\mathscr G\) is \(\E(X \mid \mathscr G)\) with mean square error \(\var(X) - \var[\E(X \mid \mathscr G)]\).
Of course, (a) is a special case of (d) with \( \mathscr G = \{\emptyset, \Omega\} \) and (c) is a special case of (d) with \( \mathscr G = \sigma(Y) \). Only (b), the linear case, cannot be interpreted in terms of conditioning with respect to a \( \sigma \)-algebra.
Conditional Covariance
Suppose again that \( \mathscr G \) is a sub \( \sigma \)-algebra of \( \mathscr F \). The conditional covariance of two random variables is defined like the ordinary covariance, but with all expected values conditioned on \( \mathscr G \).
Suppose that \( X \) and \( Y \) are random variables with \( \E(X^2) \lt \infty \) and \( \E(Y^2) \lt \infty \). The conditional covariance of \(X\) and \( Y \) given \(\mathscr G\) is defined as \[ \cov(X, Y \mid \mathscr G) = \E\left([X - \E(X \mid \mathscr G)] [Y - \E(Y \mid \mathscr G)] \biggm| \mathscr G \right) \]
So \( \cov(X, Y \mid \mathscr G) \) is a random variable that is measurable with respect to \( \mathscr G \) and is unique up to equivalence. As should be the case, conditional covariance generalizes conditional variance.
Suppose that \( X \) is a random variable with \( \E(X^2) \lt \infty \). Then \( \cov(X, X \mid \mathscr G) = \var(X \mid \mathscr G) \).
Proof
This follows immediately from the two definitions.
Our next result is a computational formula that is analogous to the one for standard covariance—the covariance is the mean of the product minus the product of the means, but now with all expected values conditioned on \( \mathscr G \):
Suppose again that \( X \) and \( Y \) are random variables with \( \E(X^2) \lt \infty \) and \( \E(Y^2) \lt \infty \). Then \[\cov(X, Y \mid \mathscr G) = \E(X Y \mid \mathscr G) - \E(X \mid \mathscr G) E(Y \mid \mathscr G)\]
Proof
Expanding the product in the definition and using basic properties of conditional expectation, we have
\begin{align} \cov(X, Y \mid \mathscr G) & = \E\left(X Y - X \E(Y \mid \mathscr G) - Y E(X \mid \mathscr G) + \E(X \mid \mathscr G) E(Y \mid \mathscr G) \biggm| \mathscr G \right) = \E(X Y \mid \mathscr G) - \E\left[X \E(Y \mid \mathscr G) \mid \mathscr G\right] - \E\left[Y \E(X \mid \mathscr G) \mid \mathscr G\right] + \E\left[\E(X \mid \mathscr G) \E(Y \mid \mathscr G) \mid \mathscr G\right] \\ & = \E\left(X Y \mid \mathscr G\right) - \E(X \mid \mathscr G) \E(Y \mid \mathscr G) - \E(X \mid \mathscr G) \E(Y \mid \mathscr G) + \E(X \mid \mathscr G) \E(Y \mid \mathscr G) = \E\left(X Y \mid \mathscr G\right) - \E(X \mid \mathscr G) E(Y \mid \mathscr G) \end{align}Our next result shows how to compute the ordinary covariance of \( X \) and \( Y \) by conditioning on \( X \) .
Suppose again that \( X \) and \( Y \) are random variables with \( \E(X^2) \lt \infty) \) and \( \E(Y^2 \lt \infty) \). Then \[\cov(X, Y) = \E\left[\cov(X, Y \mid \mathscr G)\right] + \cov\left[\E(X \mid \mathscr G), \E(Y \mid \mathscr G) \right]\]
Proof
From (29) and properties of conditional expected value we have \[ \E\left[\cov(X, Y \mid \mathscr G)\right] = \E(X Y) - \E\left[\E(X\mid \mathscr G) \E(Y \mid \mathscr G) \right] \] But \( \E(X Y) = \cov(X, Y) + \E(X) \E(Y)\) and similarly, \[\E\left[\E(X \mid \mathscr G) \E(Y \mid \mathscr G)\right] = \cov[\E(X \mid \mathscr G), \E(Y \mid \mathscr G) + \E[\E(X\mid \mathscr G)] \E[\E(Y \mid \mathscr G)]\] But also, \( \E\left[\E(X \mid \mathscr G)\right] = \E(X) \) and \( \E[\E(Y \mid \mathscr G)] = \E(Y) \) so subsituting we get \[ \E\left[\cov(X, Y \mid \mathscr G)\right] = \cov(X, Y) - \cov\left[\E(X \mid \mathscr G), \E(Y \mid \mathscr G)\right] \]
Thus, the covariance of \( X \) and \( Y \) is the expected conditional covariance plus the covariance of the conditional expected values. This result is often a good way to compute \(\cov(X, Y)\) when we know the conditional distribution of \((X, Y)\) given \(\mathscr G\).


15.1: Why It Matters- Human Resource Management
Why explain the role of human resource management in planning, recruiting, and managing a workforce?
It’s the early 1900s and you are twelve years old, living in an American city where industrialization has begun to boom. On Monday morning at 5 a.m., you line up with the rest of your family to begin your long workday at a textile factory. If it isn’t terribly busy you will end your day around 8:00 p.m. You visit with your parents and siblings during your fifteen-minute lunch break. The factory where you work is not heated, has no ventilation system, and the windows can’t be opened to let out the exhaust fumes from machinery. Since you are small, your responsibility is to climb in between the machinery and dislodge pieces of material that get caught in the gears and belts—while the machines are still running. You have to be especially careful because the only light in the factory is the sunlight that comes through the dirty windows. If you are not killed or maimed by the equipment, chances are good that your life will be cut short by the toxic fumes you inhale while in the factory. Every member of your family works there, including your four-year-old sister.  Fortunately you don’t have to worry about doing homework, because you don’t attend school, and you’ll learn to read and write only if your parents teach you on Sunday—the one day of the week when you don’t go to work.
Similar grim working conditions continued for decades in America until labor unions formed and activists began to lobby for worker protections. It’s hard, today, to imagine what those conditions really must have been like. We have always worked in conditions regulated for health and safety. We aren’t forced to work for pennies per day. Understanding how far we have come in terms of employee rights and protections is an important context for thinking about human resource management. As you work through this section, try to keep this historical perspective in mind.


7.2: Reactions in Fluorous Solvents
Fluorous solvents having suitable boiling and melting points can be used as solvent. Importantly, the fluorous solvents are different from the corresponding hydrocarbons and form two layers with conventional organic solvents. Thus, some catalysts can be immobilized in fluorous solvents in biphase system and can be recovered and recycled. In addition, in some combination, the fluorous and organic solvents on heating are miscible at elevated temperature leading to a homogeneous mixture, which, after the reaction, on cooling to room temperature lead to the formation of a biphase system. The products stay in organic phase and the catalysts move to fluorous phase that can be recovered and recycled.
7.2.1 Cyclopropanation
Fluorous complex tetrakis-dirhodium(II)-( S )-N-(n-perfluorooctylsulfonyl) pollinate L-6 exhibits good chemo- and diastereoselectivity in cyclopropanation of styrene (Scheme \(\PageIndex{1}\)). The advantage of the protocol is that the catalyst can be separated from the reaction mixture and recycled.