$X$ and $Y$ are Random Variables such that $f(x,y)=\frac{1}{2} e^{-x} \text{if} \ |y| \leq x$ , $\ x>0$

Question

$X$ and $Y$ are continuous Random Variables such that $f(x,y)=\frac{1}{2} e^{-x} \text{if} |y| \le x ,x>0$ Find $E(X|Y=-1)$

I tried to find marginal pdf of $Y$ by $f_{Y}(y)=\int_{y}^{\infty} \frac{1}{2}e^{-x} dx$ But something seems incorrect.

Answer 1

$$f_Y(y)=\int_{|y|}^{+\infty}\frac{1}{2} e^{-x}\,dx=\left[-\frac{1}{2}e^{-x}\right]_{|y|}^{+\infty}=0+\frac{1}{2}e^{-|y|}=\frac{1}{2}e^{-|y|}$$ Therefore $$f_{X|Y}(x,y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}=\frac{\frac{1}{2}e^{-x}}{\frac{1}{2}e^{-|y|}}=e^{-(x+|y|)}$$ Now we have $$\mathbf{E}(X\mid Y=y)=\int_0^\infty x\cdot e^{-(x+|y|)}\,dx=e^{-|y|}\int_0^\infty xe^{-x}\,dx=e^{-|y|}$$by recognising the second integral as the expectation of an $\text{Exp}(1)$-distributed random variable. Now substituting $y=-1$ gives $\mathbf{E}(X\mid Y=-1)=\boxed{ \dfrac{1}{e}}$.