# Compute $E(X_1|X_1+X_2)$ $X_1, X_2$ both iid $Exponential(1)$

I recently stumbled across this question on CV:

Conditional expectation conditional on exponential random variable

And really liked the answer provided by @Rush, but I wanted to try to compute this a different way to brush up on some old forgotten statistics skills. However, when I try to go about computing $$E(X_1|X_1+X_2)$$ (assuming $$X_1, X_2$$ are both iid $$Exponential(1)$$ random variables) another way, when I try to compute the marginal PDF of $$Y_2=X_1+X_2$$, so I can compute the conditional PDF of $$X_1|X_1+X_2$$, I get infinity. Maybe someone can help me understand what I'm doing wrong? Here is my approach:

Since $$X_1, X_2$$ are independent random variables then:

$$\begin{eqnarray*} f_{X_{1},X_{2}}(x_{1},x_{2}) & = & e^{-x_{1}-x_{2}},\,x_{1,\,}x_{2}\ge0 \end{eqnarray*}$$

Then I take the transformation $$Y_{1}=X_{1}$$ and~$$Y_{2}=X_{1}+X_{2}$$. Since this is a one-to-one transformation, the inverse transformation is: $$X_{1}=Y_{1}$$ and $$X_{2}=Y_{2}-Y_{1}$$ and the support of the transformed random variables is on $${\mathcal{B}=(0,\infty)\,\times(0,\infty)}$$. Next, I calculate the Jacobian, $$\left|J\right|:$$

$$\begin{eqnarray*} \left|J\right| & = & \begin{vmatrix}1 & 0\\ -1 & 1 \end{vmatrix}=1 \end{eqnarray*}$$

and then I find the joint pdf of $$Y_{1}$$ and $$Y_{2}$$:

$$\begin{eqnarray*} f_{Y_{1},Y_{2}}(y_{1},y_{2}) & = & f_{X_{1},X_{2}}(y_{1},y_{2}-y_{1})\left|J\right|\\ & = & e^{-y_{1}-(y_{2}-y_{1})}\\ & = & e^{-y_{2}},\,y_{1},y_{2}\ge0 \end{eqnarray*}$$

So, here is where I think I get tripped up as I try to compute the marginal PDF of $$Y_{2}$$:

$$\begin{eqnarray*} f_{,Y_{2}}(y_{2}) & = & \int_{0}^{\infty}f_{Y_{1},Y_{2}}(y_{1},y_{2})dy_{1}\\ & & \int_{0}^{\infty}e^{-y_{2}}dy_{1}\\ & = & \left.y_{1}e^{-y_{2}}\right|_{y_{1=0}}^{y_{1}=\infty}\\ & = & \infty \end{eqnarray*}$$

So, I realize at this step, I've obviously made a mistake somewhere, but I'm not sure where. At this point, I am unable to obtain the marginal PDF so I can use that to compute the conditional PDF to find the conditional expectation. Can you help me understand where I've gone off track?

• Hint: Your mistake occurs right after "...the support of the transformed random variables...." since it should be obvious that $X_1 = Y_1 \leq X_1+X_2 = Y_2$ and so your assertion about the support of the joint density of $Y_1$ and $Y_2$ is insupportable. – Dilip Sarwate May 1 '19 at 3:56
• Have you noticed that $$E(X_1)=E(X_2)$$ and that $$E(X_1+X_2\mid X_1+X_2)=X_1+X_2$$ enable you to check your answer with simple algebra? – whuber May 1 '19 at 13:45
• Yes, @whuber, thanks. That method was included as a very clever solution in the original post I referenced. I wanted to try to prove this using another method though -- I'm trying to reinforce my learning by proving results with multiple approaches. Thank you! – StatCurious May 1 '19 at 13:56
• I'm sorry I overlooked that reference. To atone, let me suggest another approach you might find instructive: obtain the distribution of $X_1/(X_1+X_2)$ directly and use that to find $E(X_1\mid X_1+X_2).$ This result is well-known and useful in its own right. – whuber May 1 '19 at 14:00

Ahhh, yes. Thanks. you @DilipSarwate. It seems I didn't compute the bounds of the transformed support correctly. Since $$x_1\ge 0$$ and $$x_2\ge0$$, then this implies that $$y_1\ge 0$$ and $$y_2-y_1 \ge 0\implies y_2\ge y_1\ge 0.$$ So bounds of integration for $$y_1$$ should obviously be between $$0\le y_1\le y_2$$. Once I replace the bounds of integration, the rest of the problem works out as I'd expect and is identical to the other problem referenced.