The Analysis of Data, volume 1

Random Vectors: Conditional Expectations

4.7. Conditional Expectations

Definition 4.7.1. The conditional expectation of the RV $Y$ conditioned on $X=x$ is \begin{align*} \E(Y \c X=x)=\begin{cases} \int_{-\infty}^{\infty} y f_{Y \c X=x}(y)\, dy & Y \c X=x \text{ is a continuous RV}\\ \sum_{y} y p_{Y \c X=x}(y)& Y \c X=x \text{ is a discrete RV} \end{cases} \end{align*}

Intuitively, $\E(Y \c X=x)$ represents the average or expected value of $Y$ if we know that $X=x$. Definition 4.7.1 extends naturally to conditioning on multiple random variables, for example $\E(X_i \c \{X_j=x_j:j\neq i\})$ (replace in Definition 4.7.1 the appropriate conditional pdf or pmf).

For a given $x$, the conditional expectation $\E(Y \c X=x)$ is a real number. We can also consider the conditional expectation $\E(Y \c X=x)$ as a function of $x$: $g(x)=\E(Y \c X=x)$. The mapping $x\mapsto g(x)=\E(Y \c X=x)$ corresponds to the random variable $\E(Y \c X)$.

Definition 4.7.1. The conditional expectation $\E(Y \c X)$ is a random variable $\E(Y \c X):\Omega\to\R$ defined as follows: \[\E(Y \c X)(\omega) = \E(Y \c X=X(\omega)).\]

In other words, for every value $\omega\in\Omega$ we obtain a value $X(\omega)\in\R$, which we may denote as $x$, and this in turn leads to $\E(Y \c X=x)$. Note that $\E(Y \c X)$ is a RV that is a function of the random variable $X$, or in other words $\E(Y \c X)=g(X)$ for some function $g$.

Since $\E(Y \c X)$ is a random variable, we can compute its expectation. The following proposition discovers an interesting relationship between the expectation of $\E(Y \c X)$ and the expectation of $Y$: for any $X$, we have $\E(\E(Y \c X))=\E(Y)$.

Proposition 4.7.1. \[\E(\E(Y \c X))=\E(Y).\]
Proof. \begin{align*} \E(\E(Y \c X)) &= \int_{-\infty}^{\infty} \E(Y \c X=x)f_{X}(x)\, dx \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} y f_{Y \c X=x}(y) \, dy f_X(x)\, dx\\ &= \int_{-\infty}^{\infty} y \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dx dy \\ &=\int_{-\infty}^{\infty} y f_Y(y)\, dy \\ &= \E(Y) \end{align*} where the first equality holds since $\E(Y \c X)$ is a function of $X$ and $\E(g(X))=\int g(x)f_X(x)dx$. The proof in the discrete case is similar.

The proposition above is sometimes useful for simplifying the calculation of $\E X$. An example appears below.

Example 4.7.1. Recalling Example 4.5.3, where \begin{align*} X &\sim U([0,1])\\ \{Y \c X=x\} &\sim U([x,1]) \quad x\in(0,1),\end{align*} we have $\E(Y \c X=x)=(x+1)/2$ (as shown in Section 3.7 it is the middle point of the interval $[x,1]$). It follows that $\E(Y \c X)=(X+1)/2$ and by the linearity of the expectation, \[\E(Y)=\E(\E(Y \c X))=(\E(X)+1)/2=\left(\frac{1}{2}+1\right)/2=3/4.\]