(adsbygoogle = window.adsbygoogle || []).push({}); 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

[tex]

\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*}

[/tex]

3. The attempt at a solution

[tex]

\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*}

[/tex]

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:

[tex]\[

E[X|Z]=\int_{0}^{2}xP(X=x|Z=z)[I(Z)]\][/tex]

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.

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# Homework Help: Calculating conditional expected value

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