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Apr 18

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

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

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

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

where $\bar{y}_j$ is a fixed value of $Y$. The drawback of this definition is its dependence on indexation of values $x_i,y_j$. 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 $1_{\{Y=\bar{y}_j\}}=\left\{\begin{array}{ll}1,&Y=\bar{y}_j;\\0,&\textrm{otherwise}.\end{array}\right.$

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

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

(the sum in the middle includes all points in the sample space). Using (1) and (3) we can rewrite (2) as $E(X|Y=\bar{y}_j)=\frac{E(X1_{\{Y=\bar{y}_j\}})}{P(Y=\bar{y}_j)}$.

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

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

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

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

where $p_z$ is the density.

Proof. From the expression of the density $\frac{dp_z(t)}{dt}=\frac{1}{\sqrt{2\pi}}e^{-t^2/2}(-t)=-tp_z(t)$.

Applying this equation and (4) we get $E(z|z\le a)=\frac{E(z1_{\{z\le a\}})}{P(z\le a)}=\frac{\int_{-\infty}^atp_z(t)dt}{\Phi(a)}=\frac{-\int_{-\infty}^a\frac{dp_z(t)}{dt}dt}{\Phi(a)}=\frac{p_z(-\infty)-p_z(a)}{\Phi(a)}=-\frac{p_z(a)}{\Phi(a)}$.