Published on *STAT 414 / 415* (https://onlinecourses.science.psu.edu/stat414)

Now that we've mastered the concept of a conditional probability mass function, we'll now turn our attention to finding conditional means and variances. We'll start by giving formal definitions of the conditional mean and conditional variance when *X* and *Y* are discrete random variables. And then we'll end by actually calculating a few!

\(\mu_{Y|X}=E[Y|x]=\sum\limits_y yh(y|x)\) And, the \(\mu_{X|Y}=E[X|y]=\sum\limits_x xg(x|y)\) The \(\sigma^2_{Y|x}=E\{[Y-\mu_{Y|x}]^2|x\}=\sum\limits_y [y-\mu_{Y|x}]^2 h(y|x)\) or, alternatively, using the usual shortcut: \(\sigma^2_{Y|x}=E[Y^2|x]-\mu^2_{Y|x}=\left[\sum\limits_y y^2 h(y|x)\right]-\mu^2_{Y|x}\) And, the \(\sigma^2_{X|y}=E\{[X-\mu_{X|y}]^2|y\}=\sum\limits_x [x-\mu_{X|y}]^2 g(x|y)\) or, alternatively, using the usual shortcut: \(\sigma^2_{X|y}=E[X^2|y]-\mu^2_{X|y}=\left[\sum\limits_x x^2 g(x|y)\right]-\mu^2_{X|y}\) |

As you can see by the formulas, a conditional mean is calculated much like a mean is, except you replace the probability mass function with a conditional probability mass function. And, a conditional variance is calculated much like a variance is, except you replace the probability mass function with a conditional probability mass function. Let's return to one of our examples to get practice calculating a few of these guys.

Let *X* be a discrete random variable with support *S*_{1} = {0, 1}, and let *Y* be a discrete random variable with support *S*_{2} = {0, 1, 2}. Suppose, in tabular form, that *X* and *Y* have the following joint probability distribution *f*(*x*,*y*):

What is the conditional mean of *Y *given *X* = *x*?

**Solution.** We previously determined that the conditional distribution of *Y* given *X* is:

Therefore, we can use it, that is, *h*(*y*|*x*), and the formula for the conditional mean of *Y* given *X* = *x* to calculate the conditional mean of *Y *given *X* = 0. It is:

\(\mu_{Y|0}=E[Y|0]=\sum\limits_y yh(y|0)=0\left(\dfrac{1}{4}\right)+1\left(\dfrac{2}{4}\right)+2\left(\dfrac{1}{4}\right)=1\)

And, we can use *h*(*y*|*x*) and the formula for the conditional mean of *Y* given *X* = *x* to calculate the conditional mean of *Y* given *X* = 1. It is:

\(\mu_{Y|1}=E[Y|1]=\sum\limits_y yh(y|1)=0\left(\dfrac{2}{4}\right)+1\left(\dfrac{1}{4}\right)+2\left(\dfrac{1}{4}\right)=\dfrac{3}{4}\)

Note that the conditional mean of *Y*|*X* = *x* depends on *x*, and depends on *x *alone. You might want to think about these conditional means in terms of sub-populations again. The mean of *Y* is likely to depend on the sub-population, as it does here. The mean of *Y* is 1 for the *X* = 0 sub-population, and the mean of *Y* is ¾ for the *X* = 1 sub-population. Intuitively, this dependence should make sense. Rather than calculating the average weight of an adult, for example, you would probably want to calculate the average weight for the sub-population of females and the average weight for the sub-population of males, because the average weight no doubt depends on the sub-population!

What is the conditional mean of *X *given *Y* = *y*?

**Solution.** We previously determined that the conditional distribution of *X* given *Y* is:

As the conditional distribution of *X* given *Y* suggests, there are three sub-populations here, namely the *Y* = 0 sub-population, the *Y* = 1 sub-population and the *Y* = 2 sub-population. Therefore, we have three conditional means to calculate, one for each sub-population. Now, we can use *g*(*x*|*y*) and the formula for the conditional mean of *X* given *Y* = *y* to calculate the conditional mean of *X *given *Y* = 0. It is:

\(\mu_{X|0}=E[X|0]=\sum\limits_x xg(x|0)=0\left(\dfrac{1}{3}\right)+1\left(\dfrac{2}{3}\right)=\dfrac{2}{3}\)

And, we can use *g*(*x*|*y*) and the formula for the conditional mean of *X* given *Y* = *y* to calculate the conditional mean of *X *given *Y* = 1. It is:

\(\mu_{X|1}=E[X|1]=\sum\limits_x xg(x|1)=0\left(\dfrac{2}{3}\right)+1\left(\dfrac{1}{3}\right)=\dfrac{1}{3}\)

And, we can use *g*(*x*|*y*) and the formula for the conditional mean of *X* given *Y* = *y* to calculate the conditional mean of *X *given *Y* = 2. It is:

\(\mu_{X|2}=E[X|2]=\sum\limits_x xg(x|2)=0\left(\dfrac{1}{2}\right)+1\left(\dfrac{1}{2}\right)=\dfrac{1}{2}\)

Note that the conditional mean of *X*|*Y* = *y* depends on *y*, and depends on *y *alone. The mean of *X* is 2/3 for the *Y* = 0 sub-population, the mean of *X* is 1/3 for the *Y* = 1 sub-population, and the mean of *X* is 1/2 for the *Y* = 2 sub-population.

What is the conditional variance of *Y* given *X* = 0?

**Solution.** We previously determined that the conditional distribution of *Y* given *X* is:

Therefore, we can use it, that is, *h*(*y*|*x*), and the formula for the conditional variance of *Y* given *X* = *x* to calculate the conditional variance of *Y *given *X* = 0. It is:

\begin{align}

\sigma^2_{Y|0} &= E\{[Y-\mu_{Y|0}]^2|x\}=E\{[Y-1]^2|0\}=\sum\limits_y (y-1)^2 h(y|0)\\

&= (0-1)^2 \left(\dfrac{1}{4}\right)+(1-1)^2 \left(\dfrac{2}{4}\right)+(2-1)^2 \left(\dfrac{1}{4}\right)=\dfrac{1}{4}+0+\dfrac{1}{4}=\dfrac{2}{4}

\end{align}

We could have alternatively used the shortcut formula. Doing so, we better get the same answer:

\begin{align}

\sigma^2_{Y|0} &= E[Y^2|0]-\mu_{Y|0}]^2=\left[\sum\limits_y y^2 h(y|0)\right]-1^2\\

&= \left[(0)^2\left(\dfrac{1}{4}\right)+(1)^2\left(\dfrac{2}{4}\right)+(2)^2\left(\dfrac{1}{4}\right)\right]-1\\

&= \left[0+\dfrac{2}{4}+\dfrac{4}{4}\right]-1=\dfrac{2}{4}

\end{align}

And we do! That is, no matter how we choose to calculate it, we get that the variance of *Y* is ½ for the *X* = 0 sub-population.