# 19.3 - Conditional Means and Variances

19.3 - Conditional Means and Variances

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!

Definition. Suppose $$X$$ and $$Y$$ are discrete random variables. Then, the conditional mean of $$Y$$ given $$X=x$$ is defined as:

$$\mu_{Y|X}=E[Y|x]=\sum\limits_y yh(y|x)$$

And, the conditional mean of $$X$$ given $$Y=y$$ is defined as:

$$\mu_{X|Y}=E[X|y]=\sum\limits_x xg(x|y)$$

The conditional variance of $$Y$$ given $$X=x$$ is:

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

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 conditional variance of $$X$$ given $$Y=y$$ is:

$$\sigma^2_{X|y}=E\{[X-\mu_{X|y}]^2|y\}=\sum\limits_x [x-\mu_{X|y}]^2 g(x|y)$$

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.

## Example 19-3

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 $$\frac{3}{4}$$ 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 $$\frac{2}{3}$$ for the $$Y=0$$ sub-population, the mean of $$X$$ is $$\frac{1}{3}$$ for the $$Y=1$$ sub-population, and the mean of $$X$$ is $$\frac{1}{2}$$ for the $$Y=2$$ sub-population.

What is the conditional variance of $$X$$ 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 $$X$$ given $$X=x$$ to calculate the conditional variance of $$X$$ 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 $$\frac{1}{2}$$ for the $$X=0$$ sub-population.

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