Okay, so here's two problems from my book;
Problem 1) Let $f(x,y) = 2e^{-(x+2y)}$
$x,y>0$
Calculate $V[Y|X>3 \cap Y>3]$
Solution
since the joint density can be factored out into terms of x and y, we see that x and y are independent so $V[Y|X>3 \cap Y>3]$ reduces to $V[Y|Y>3]$. Also $f(x) = e^{-x}$ and $f(y) = 2e^{-2x}$
By the memoryless property of the exponential distribution tells us that $V[Y|Y>3]$ has the same distribution as $V[Y-3|Y>3]$. so $V[Y|Y>3] = V[Y] = 1/4$
Problem 2)
let $f(x) = e^{-x}$
Find $E[(X-1)^3|X \ge E(x)]$
Solution:
We can see that $E[X] = 1$ and $P(X \ge 1)= e^{-1}$
therefore $E[(X-1)^3|X \ge E(x)]$ = $E[(X-1)^3|X \ge 1]$ = $\frac{1}{P[X \ge 1]} \int_1^\infty e^{-x}(x-1)^3$ = 6.
Why do we use the memoryless property in one but not the other?
In the second problem, it is easier to use memorylessness. The distribution of $X-1$, given that $X\gt 1$, is the same as the unconditional distribution of $X^3$. Thus our conditional expectation is $\int_0^\infty x^3e^{-x}\,dx$.
In the first problem, I would prefer to say the distribution of $Y$, given that $Y\gt 3$, is the same as the distribution of $3+Y^\ast$, where $Y^\ast$ has exponential distribution with parameter $2$. But variance is unaffected by a constant shift, so the variance of $3+Y^\ast$ is $\frac{1}{4}$.