# Calculating conditional expected value

1. Apr 10, 2010

### thenava

1. The problem statement, all variables and given/known data
Hi All, I have a homework question that I would like some hints on. I am given X and Y with uniform densities on [0,2], and Z := X+Y. The end goal is to find E[X|Z]. Now I know I can use symmetry to argue that E[X|Z] = E[Y|Z], and that E[X|Z] + E[Y|Z] = E[Z|Z] = Z, so 2*E[X|Z] = Z, and E[X|Z] = Z/2. However, I want to also evaluate it the "long way". To do so, I first went and found the convolutions of the densities of X and Y

2. Relevant equations
$$\begin{eqnarray*} f_{x}(x),f_{y}(y) & = & \begin{cases} \frac{1}{2} & 0<x<2\\ 0 & otherwise\end{cases}\\\end{eqnarray*} \begin{eqnarray*} f_{z}(z) & = & (f_{x}*f_{y})(z)\\ & = & \int_{0}^{2}\left[f_{x}(z-y)\frac{1}{2}\right]dy\end{eqnarray*}$$

3. The attempt at a solution
$$\begin{eqnarray*} f_{z}(z) & = & \begin{cases} \frac{1}{4}\int_{0}^{z}dy & 0\leq z\leq2\\ \frac{1}{4}\int_{z-2}^{2}dy & 2<z\leq4\\ 0 & otherwise\end{cases}\end{eqnarray*} \begin{eqnarray*} f_{z}(z) & = & \begin{cases} \frac{z}{4} & 0\leq z\leq2\\ \frac{4-z}{4} & 2<z\leq4\\ 0 & otherwise\end{cases}\end{eqnarray*}$$

So I now have the density function of Z. Here is where I am getting confused; I can setup up E[X|Z] as follows:

$$$E[X|Z]=\int_{0}^{2}xP(X=x|Z=z)[I(Z)]$$$
where I is an indicator variable for Z. My problem is how do I fit the density I found above into P(X|Z=z), and how do I evaluate an integral with an indicator variable. I would greatly appreciate any hints that you guys could provide. Thanks.

P.S. Sorry if this question has already been posted.

2. Apr 11, 2010

### snipez90

Isn't the definition of P(X = x | Z = z) simply f_XZ_(x,z) / f_Z_(y), where f_XZ is the joint density and f_Z is the density of Z?

If you have an integral with an indicator variable, you typically just integrate over the set for which the image of indicator variable is 1 since elsewhere the indicator variable vanishes.