Can someone explain me how to reach to the solution of the conditional expectation, I am not able to understand how to approach the question.
The random variables $X$ and $Y$ are distributed according to the joint PDF
$f_{X,Y}(x,y) = ax^2$ if $1 ≤ x ≤ 2$ and $0 ≤ y ≤ x$
Here, I calculated $a = 4/15$, $f_Y(y)= 28/45$ for $0\leq y \leq 1$, and $f_Y(y)= (-4/45)*(y^3 - 8)$ for $1 < y \leq 2$.
I am supposed to find the conditional expectation of $1/(X^2 Y)$ , given that $Y=5/4$
My solution goes:
$$\frac{(5*x^2/14)}{(-4/45)*((125/64) - 8)} = \frac{64x^2}{129}$$
Applying integration with respect to $x$ from $1$ to $5/4$, on $1/(x^2 *y) * (64*x^2)/129$ the answer I get is $16/129y$, which is not correct.
Could anyone guide me on this?
Your $(5\cdot x^2/14)(−4/45)\cdot((125/64)−8)$ term should be $(4\cdot x^2/15)(−4/45)\cdot((125/64)−8)$, with integration limits from $5/4$ to $2$, also replace the "$y$" in your last expression with $5/4$.