Joint probability distribution probability and law of total expectation

Question

So the question asks:

Let $X, Y$ be random variables, with the following joint probability density function:

$f_{X,Y}(x,y) = \left\{ \begin{array}{ 1 l } kye^{-y} & \mbox{if $0≤|x|≤y$}\\ 0 & \mbox{ otherwise} \end{array} \right.$

(a) Find the value of the constant $k$.

(b) Find the marginal densities $f_{X}(x)$ and $f_{Y}(y)$ ( including the range) . Identify the distribution of $Y$.

(c) Calculate $P[2X<Y]$.

(e) Explain why $X$ and $Y$ are dependent random variables.

So for (a), I got:

$ 1 = 2 \int_{x=0}^∞ \int_{ y=x }^{∞} kye^{-y} \,dy \, dx $ =1

$ k = 0.25$

b. $f_X(x) = \int_{ |x| }^{∞} 0.25ye^{-y} \,dy $ =$0.25e^{-|x|}(|x|+1)$

for $y>0$,

$f_Y(y) = \int_{ -y }^{y} 0.25ye^{-y} \,dx $ =$0.5e^{-y}y^2$

for $y≤0 , f_Y(y) =0$

Exponential distribution

c. I tried:

$P(2X<Y) = \int_{-y}^y\int_{ 2|x| }^{∞} 0.25ye^{-y} \,dy \, dx $

but the answer gets super comlicated and still involves with y, so am I using the wrong range?

d. $f_{X|Y =y} (x) = p(-y<x<y|Y=y) = \int_{ -y }^{y} 0.25ye^{-y} \,dx $ =$0.5e^{-y}y^2$

But how to apply the the law of total expectation to find $E[X^2]$?

e. I think the reason is going to be something like:

$E(Y|X) ≠ E(Y) $ and $E(Y|X) ≠ E(Y) $ I guess?

Answer 1

For c: the definition of pdf implies $-y<x<y$ and you are interested in $2x<y$ or $x<y/2$. Combining the two you get $-y<x<y/2$, hence $$\Pr(2X<Y)=\int_{y=0}^{\infty}\int_{x=-y}^{x=y/2}\frac14ye^{-y}dxdy=\frac34$$