Find $\mathbb{P}(X<|Y|/2)$

Question

The joint density of $X$ and $Y$ are given by $$f(x,y)=C(y-x)e^{-y},\quad-y<x<y,\quad 0<y<\infty$$

1. Find C
2. Find the pdf for $X$.
3. Find the pdf for $Y$.
4. Find $\mathbb{P}(X<|Y|/2)$

My Attempt

1. I just need to find C such that $$\int^\infty_0\int^y_{-y}C(y-x)e^{-y}dxdy=1$$ Evaluating the integral gives $4C$, so $C=1/4$.
2. The pdf of $X$ is $$\int^\infty_0\frac{1}{4}(y-x)e^{-y}dy=-\frac{1}{4}(x-1)$$ I am concerned about this one because the pdf does not integrate to $1$ as it should, but I don't see any other way to integrate wrt y.
3. The pdf of $Y$ is $$\int^y_{-y}\frac{1}{4}(y-x)e^{-y}dx=\frac{y^2e^{-y}}{2}$$
4. I know that to find $\mathbb{P}(X<Y)$, you compute the following: $$\mathbb{P}(X<Y)=\int\mathbb{P}(X<y)f_Y(y)dy=\int F_X(y)f_Y(y)dy$$ How would this equation change if we were to compute $\mathbb{P}(X<|Y|/2)$? And what would the bounds of integration be?

Answer 1

When computing the PDF of $X$, the integral should be taken over the interval $(|x|, \infty)$ since those are the valid values of $y$ for a fixed value of $x$. Doing so leaves you with $$\frac{1}{4} e^{-|x|}(|x|-x + 1) = \begin{cases}\frac{1}{4} e^{-x} & x \ge 0 \\ \frac{1}{4}(-2x+1) e^x & x < 0\end{cases}$$

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In general, the probability that the point $(X,Y)$ lies in some region $A$ is $P((X,Y) \in A) = \iint_A f(x,y) \, dx \, dy$. Sketch a picture of the region consisting of points $(x,y)$ satisfying $x < |y|/2$. (It will look like an infinitely long pizza slice.) Then integrate the joint density over this region.