Conditional Distribution Question

Hi guys, I am stuck at some point in the exercise below and I have to ask for help.

Suppose the conditional pdf of y, given x is \(\displaystyle \displaystyle \frac {2(1+x^2)}{(1+x+y)^3}\) for \(\displaystyle \displaystyle 0<x<\infty,\ 0<y<\infty\)

Then compute E(1+x+y|x) for a fixed x, and use the result to compute E(y|x).

My main problem is that the resulting integral using these limits of integration is not convergent, unless of course I done something wrong.

\(\displaystyle \displaystyle \int_0^\infty \int_0^\infty \frac {2(1+x^2)}{(1+x+y)^2}\ \mathrm dy\mathrm dx\) after multiplying with 1+x+y.

The inner integral yields \(\displaystyle \ \displaystyle 2(1+x^2)\int_0^\infty \frac {1}{(1+x+y)^2}\mathrm dy = -2(1+x^2)[\frac{1}{(1+x+y)}]_0^\infty = \frac {2(1+x^2)}{(1+x)}\)

And now the outer integral: \(\displaystyle \displaystyle 2 \int_0^\infty \frac {1+x^2}{(1+x)} \mathrm dx\) which after integrating by parts twice gives \(\displaystyle \displaystyle ln(1+x)[(1+x^2)+1]-\frac{x}{1+x}\) which is not convergent if x tends to infinity. So the expectation does not exist. Is what I have done correct though?

Thanks in advance.

iocal said:

Hi guys, I am stuck at some point in the exercise below and I have to ask for help.

Suppose the conditional pdf of y, given x is \(\displaystyle \displaystyle \frac {2(1+x^2)}{(1+x+y)^3}\) for \(\displaystyle \displaystyle 0<x<\infty,\ 0<y<\infty\)

Then compute E(1+x+y|x) for a fixed x, and use the result to compute E(y|x).

My main problem is that the resulting integral using these limits of integration is not convergent, unless of course I done something wrong.

\(\displaystyle \displaystyle \int_0^\infty \int_0^\infty \frac {2(1+x^2)}{(1+x+y)^2}\ \mathrm dy\mathrm dx\) after multiplying with 1+x+y.

The inner integral yields \(\displaystyle \ \displaystyle 2(1+x^2)\int_0^\infty \frac {1}{(1+x+y)^2}\mathrm dy = -2(1+x^2)[\frac{1}{(1+x+y)}]_0^\infty = \frac {2(1+x^2)}{(1+x)}\)

Click to expand...

TOO MUCH WORK! I don't think there is any reason to integrate over \(\displaystyle x\) - at least not for the questions you have asked. \(\displaystyle x\) is a constant. To check the normalization of the given \(\displaystyle f(y | x)\)

Let.... \(\displaystyle u = (1 + x + y)\).... \(\displaystyle du = dy\)

\(\displaystyle \displaystyle \int_0^\infty \dfrac{2(1+x^2)}{(1+x+y)^3}\ dy= 2(1+x^2)\int_{1+x}^\infty \dfrac{du}{u^3} = -(1+x^2)\left[\dfrac{1}{u^2}\right]_{1+x}^\infty = \dfrac{(1 + x^2)}{(1+x)^2}\)

HMMM. This is supposed to be unity. Either we both made a mistake integrating, OR the ")" in the numerator is misplaced.

DrPhil said:

TOO MUCH WORK! I don't think there is any reason to integrate over \(\displaystyle x\) - at least not for the questions you have asked. \(\displaystyle x\) is a constant. To check the normalization of the given \(\displaystyle f(y | x)\)

Let.... \(\displaystyle u = (1 + x + y)\).... \(\displaystyle du = dy\)

\(\displaystyle \displaystyle \int_0^\infty \dfrac{2(1+x^2)}{(1+x+y)^3}\ dy= 2(1+x^2)\int_{1+x}^\infty \dfrac{du}{u^3} = -(1+x^2)\left[\dfrac{1}{u^2}\right]_{1+x}^\infty = \dfrac{(1 + x^2)}{(1+x)^2}\)

HMMM. This is supposed to be unity. Either we both made a mistake integrating, OR the ")" in the numerator is misplaced.

Click to expand...

You are right of course. Thanks I will do it again.
EDIT: The book had that typo in the Numerator, it was corrected in later editions.