$f(x,y) = \begin{cases} \frac{1}{18}e^{\frac{-(x+y)}{6}} &\mbox 0<x<y \\ 0, & \mbox otherwise \end{cases} $
Calculate $Var(Y|X=2)$
I first calculated the conditional probability function $f_{y|x=2}(y|x=2)=\frac{1}{6}e^{\frac{-(y-2)}{6}}$ and used integral to get the mean and the variance. The variance I got is 36.
Then I observed that $f_{y|x=2}(y|x=2)$ is an exponential distribution with variable $y-2$ And I'm inclined to think that instead of doing the tedious integral, I could just use the definition of exponential distribution to get the mean and variance, which are $6$ and $36$ respectively. However I'm not so sure about the $y-2$ part, does a constant value shift of the variable effects the value of mean and variance in anyway?
Probably not:
$$\int \frac{1}{6} e^{\frac{-(y-2)}{6}} dy$$
$$u = y-2, du = dy$$
$$ = \int \frac{1}{6} e^{\frac{-(u)}{6}} du$$