Given the joint-probability $$f_{XY}=x*e^{-x*(y+1)}1_{[0,\infty)}(x)1_{[0,\infty)}(y)$$
I am asked to compute the conditional expectation E[Y|X]. However, as the density of $f_Y$ is just $f_Y=\int f_{XY} dx$ and $$f_Y=\frac{1}{(y+1)^2}.$$ One can easily check, that $E[Y]$ is infinite and not integrable. As the formula $E[Y|X]=\int Y\frac{f_{XY}}{f_X}dy$ requires $Y$ to be integrable, in other words $E[Y]$ is bounded, how can I justify the use of the formula I mentioned.
On
you have a little mistake in your approach: it should be $$ \mathbb{E}[Y|X]= \int_{0}^\infty y \cdot \frac{f_{XY}(X,y)}{f_X(X)} \; dy. $$ so the denominator inside the integral should be $f_X$ not $f_Y$. Note that i have written upper case $X$ and lower case $y$ and i did this on purpose: $X$ is a random variable and the outcome of the integral is a random variable as well. On the other hand $y$ is a real number.
The integral on the right hand side can easily be determined since the conditional distribution of $Y$ given $X$ is exponential with parameter $X$. Therefore $$ \mathbb{E}[Y|X]= \frac{1}{X}. $$
Your formular still holds for non-negative, not necessarily integrable random variables.
To see this have a look at the proof of it… it's shown in three steps (usually)
1.) for simple random variables
2.) for non-negative ones
3.) for integrable ones
You use 1.) to proove 2.) and 2.) to proove third. But for 2.) you don't need integrability of Y…