Raising the bar

Conditional expectation generalized to continuous random variables

The conditional expectation definition needs to be generalized, to be applicable to continuous random variables. The generalization is accompanied with an example and later will be applied to expected shortfall.

Generalizing conditional expectation definition

Suppose $latex X$ can take values $latex x_1,...,x_I$ with probabilities $latex p_i^X=P(X=x_i)$ and $latex Y$ can take values $latex y_1,...,y_J$ with probabilities

(1) $latex p_j^Y=P(Y=y_j)$.

Denote the joint probabilities $latex p_{ij}=P(X=x_i,Y=y_j)$. The definition from this post gives

(2) $latex E(X|Y=\bar{y}_j)=\frac{\sum_{i=1}^Ip_{ij}x_i}{p_j^Y}$,

where $latex \bar{y}_j$ is a fixed value of $latex Y$. The drawback of this definition is its dependence on indexation of values $latex x_i,y_j$. Our purpose is to show how from this definition one can obtain a definition that does not use indexation and can therefore be applied to continuous random variables. Denote

$latex 1_{\{Y=\bar{y}_j\}}=\left\{\begin{array}{ll}1,&Y=\bar{y}_j;\\0,&\textrm{otherwise}.\end{array}
\right.$

Then the sum $latex \sum_{i=1}^Ip_{ij}x_i$ can be expanded by including zero terms:

(3) $latex \sum_{i=1}^Ip_{ij}x_i=\sum_{i=1}^I\sum_{j=1}^Jp_{ij}x_i1_{\{Y=\bar{y}_j\}}=E(X1_{\{Y=\bar{y}_j\}})$

(the sum in the middle includes all points in the sample space). Using (1) and (3) we can rewrite (2) as

$latex E(X|Y=\bar{y}_j)=\frac{E(X1_{\{Y=\bar{y}_j\}})}{P(Y=\bar{y}_j)}$.

Replacing here the conditioning on $latex Y=\bar{y}_j$ by conditioning on a general set $latex A$ whose probability is not zero we obtain the definition of conditional expectation:

(4) $latex E(X|A)=\frac{E(X1_A)}{P(A)}$.

Example. If $latex z$ is standard normal and $latex \Phi$ is its distribution function, then for any number $latex a$ one has

(5) $latex E(z|z\le a)=-p_z(a)/\Phi(a)$

where $latex p_z$ is the density.

Proof. From the expression of the density

$latex \frac{dp_z(t)}{dt}=\frac{1}{\sqrt{2\pi}}e^{-t^2/2}(-t)=-tp_z(t)$.

Applying this equation and (4) we get

$latex E(z|z\le a)=\frac{E(z1_{\{z\le a\}})}{P(z\le a)}=\frac{\int_{-\infty}^atp_z(t)dt}{\Phi(a)}=\frac{-\int_{-\infty}^a\frac{dp_z(t)}{dt}dt}{\Phi(a)}=\frac{p_z(-\infty)-p_z(a)}{\Phi(a)}=-\frac{p_z(a)}{\Phi(a)}$.