How do I find a conditional expectation, $E(Y|X)$ for:

$f_{XY}(x,y)=2e^{-(x+y)}$ for $0<x<y$.

I researched that the formula for a conditional expectation is:

$E(Y|X) = \Sigma yf_{Y|X}(y|x)$

And I have already calculated the conditional density as:

$f_{Y|X}(y|x) = e^{x-y}$ for $0<x<y$


$E(Y|X) = \Sigma ye^{x-y}$ for $0<x<y$

  • $\begingroup$ You have nor written $f_{y|x}$ correctly. $\endgroup$ – Kavi Rama Murthy May 23 at 11:47
  • $\begingroup$ @KaviRamaMurthy sorry can you clarify what you mean that is is incorrect ? $\endgroup$ – i9-9980XE May 23 at 12:00
  • $\begingroup$ You have to specify that the conditional density is $e^{x-y}$ for $0<x<y$. Leaving out the condition $0<x<y$ will lead to wrong answers. $\endgroup$ – Kavi Rama Murthy May 23 at 12:02
  • $\begingroup$ I have re-edited the question for a more correct explanation. Thank you $\endgroup$ – i9-9980XE May 23 at 12:05

The conditional density you have found is for $0<x<y$. $E(Y|X)=\int_x^{\infty} ye^{x-y} dy$. Integrate by parts to find the exact value.

  • $\begingroup$ I have calculated the answer to be x + 1 but am unsure if I made a mistake with the integral $\endgroup$ – i9-9980XE May 23 at 11:50
  • $\begingroup$ The answer is $x+1$. $\endgroup$ – Kavi Rama Murthy May 23 at 11:51

First your density function $f_{Y\mid X=x}$ is wrong. Anyway, suppose it's true : what you computed (if you replace of cours $\sum_{y}$ by $\int_{\mathbb R}$) is $$\mathbb E[Y\mid X=x].$$ But $$\mathbb E[Y\mid X]=\int_{\mathbb R}ye^{X-y}\,\mathrm d y=e^X\int_{\mathbb R}ye^{-y}\,\mathrm d y.$$

  • $\begingroup$ Ahh.. so I evaluate the integral by way of integration by parts ? $\endgroup$ – i9-9980XE May 23 at 11:45
  • $\begingroup$ Yes indeed.$\ \ \ \ \ $ $\endgroup$ – Surb May 23 at 11:46
  • $\begingroup$ OP has written the conditional density wrongly. The integral is not over the whole line. $\endgroup$ – Kavi Rama Murthy May 23 at 11:47
  • $\begingroup$ @KaviRamaMurthy: Yes indeed. I'll let him adapt this point. $\endgroup$ – Surb May 23 at 11:48
  • $\begingroup$ @Surb sorry, isn’t the density function $\frac{f(x,y)}{f_X(x)} = \frac{2e^{-(x+y)}}{2e^{-2x}} = e^{x-y}$? $\endgroup$ – i9-9980XE May 23 at 11:57

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