I came across this very interesting question in a forum:

If both X and Y are independent and exponentially distributed with parameter $\lambda$, find $E[X^2|X+Y]$

Someone gave the solution and stated that $X|X+Y$ ~ Uniform$[0,X+Y]$. Why does the distribution change to uniform distribution in this case?


Let $Z=X+Y$. $$f_{X|X+Y=z}(x)=\frac{f_{Z,X}(z,x)}{f_Z(z)}=\frac{f_{Y,X}(z-x,x)}{\lambda^2z e^{-\lambda z}}=\frac{\lambda e^{-\lambda (z-x)}\lambda e^{-\lambda x}}{\lambda^2z e^{-\lambda z}}=\frac{1}{z}$$

Assuming $z-x\geq 0$ and $x\geq 0$, which means $0\leq x\leq z$ and the PDF is $1/z$. This is $U[0,z]$, i.e. $U[0,X+Y]$.

Note: the variable change in joint PDFs requires a Jacobian multiplier, but it is $1$ in this case.

  • $\begingroup$ Thank you. Can you give more details on how to calculate E[X^2|X+Y]? I would like to see your way of solving it if you don't mind sharing. $\endgroup$ – Dennis Sep 9 '20 at 0:14
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    $\begingroup$ It's the second moment of uniform distribution where $a=0,b=z$: $z^2/3$ en.wikipedia.org/wiki/Uniform_distribution_(continuous)#/… $\endgroup$ – gunes Sep 9 '20 at 8:17
  • $\begingroup$ @Dennis is the answer clear for you? $\endgroup$ – gunes Sep 15 '20 at 21:20
  • $\begingroup$ Thanks for the explanation. Now I understand the second moment part. But I am stuck at the transformation from f_{Z,X} (z,x) to f_{Y,X} (z-x, x). Z = X+Y, so shouldn't it be transformed to f_{Y,X} (x+y, x)? $\endgroup$ – Dennis Sep 18 '20 at 0:40
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    $\begingroup$ @dennis we want to write the joint pdf of $Z,X$ in terms of joint pdf of $Y,X$. So, here $f_{Z,X}$ is unknown and $f_{Y,X}$ is known, which means we transform $Y,X$ to $Z,X$ and try to find the joint dist. of transformed pair. $\endgroup$ – gunes Sep 18 '20 at 20:53

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