1
$\begingroup$

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 $0<x<y$

Thus,

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $$\mathbb E[Y\mid X]=\int_{\mathbb R}ye^{X-y}\,\mathrm d y=e^X\int_{\mathbb R}ye^{-y}\,\mathrm d y.$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.