A similar question to this was on an exam* and I'm still stuck on how to find an answer:
Suppose $X$ and $Y$ are jointly distributed random variables according to the distribution:
\begin{align*}
f(x,y) =
\begin{cases}
4e^{-(4x+y)} &\text{if } x\geq 0, y\geq 0 \\
0 & \text{else}
\end{cases}.
\end{align*}
What is the expected value $E[X| X \geq 2, Y \geq 2]$?
I would like to know two things:
first, what is the best approach to this problem from the point of view of handling the multiple conditional statements? And secondly, is an answer with an exponential ($e^{8}$) term reasonable?
If my reasoning is correct, it appears that the expectation is independent of $Y \geq 2$ and so boils down to finding $E[X| X \geq 2]$. I did this by "dividing" the distribution for $X$ by a normalizing factor which looks like:
\begin{align*}
\int_{2}^{\infty}{4e^{-4x} dx}
\end{align*}
and then finding the expectation using this new distribution.
This part is what I was having the hardest time justifying!
*Note: I have changed the problem enough so that it is not in the same form as it was on the exam.
Best Answer
Long route.
$$\begin{align}\mathsf E(X\mid X\geqslant 2, Y\geqslant 2] & = \dfrac{\mathsf E(X~\mathbf 1_{x\geqslant 2, y\geqslant 2})}{\mathsf P(X\geqslant 2, Y\geqslant 2)} \\[1ex] &= \dfrac{\int_2^\infty\int_2^\infty x (4 e^{-4x-y})\operatorname d x\operatorname d y}{\int_2^\infty\int_2^\infty (4 e^{-4x-y})\operatorname d x\operatorname d y}\\[1ex] &~\,~\vdots\end{align}$$
Short cut.
Notice that $f_X(x)=4e^{-4x}\mathbf 1_{x>0}, f_Y(y)=e^{-y}\mathbf 1_{y>0}$ and $X,Y$ are therefore independent exponential random variables, which have the memoryless property.
Justified by eyeballing the joint distribution and quickly assertaining that $f_X(x) = 4e^{-4x}\mathbf 1_{x>0}\cdot\int_0^\infty e^{-y}\operatorname d y$ and symmetrically $f_Y(y)=e^{-y}\mathbf 1_{y>0}\cdot\int_0^\infty 4e^{-4x}\operatorname d x$ so therefore $f_{X,Y}(x,y)=f_X(x)f_Y(y)$. This shall come with practice.