$f(x,y)=2e^{-(x+y)}$ for $0<x<y<\infty$
Determine the marginal density of $X$: $\int_x^\infty 2 \cdot e^{-(x+y)} dy= 2 e^{-2x}$ for $x>0$
Which is an exponential distribution with parameter $\lambda=2$
Determine the conditional density of $Y$ given $X=x$:
$f_{Y|X}(y|x)=(2e^{-(x+y)})/(2e^{-2x})=e^{-(y-x)}$ for $0<x<y<\infty$
$P(Y>2|X=1)=1/e=0.36788$
Given the support, we know that $X$ and $Y$ are independent.
Determine the conditional mean of $Y$ given $X=x$:
$E(Y|X)=\int_x^\infty y \cdot e^{-(y-x)}dy=x+1$
Determine the conditional variance of $Y$ given $X=x$.
Is what I have above correct? How do I find the conditional variance of $Y$ given $X=x$?
You can use the definition of the conditional variance or compute both the conditional expectation $E(Y|x)$ and $E(Y^2|x)$.
We have an analogous identity of $$ Var(X) = E(X^2) - E(X)^2. $$