compute marginal

Question

I have tried to solve this exercise

Let $X$ and $Y$ be random variables with joint probability density function given by: $f(x,y)=\frac{1}{8}(x^2-y^2)e^{-x}$ if $x>0$, $|y|<x$

Calculate $E(X\mid Y=1)$

so, the marginal $f_Y(y)$ is $\int_y^\infty \frac{1}{8}(x^2-y^2)e^{-x} dx +\int_{-y}^\infty \frac{1}{8}(x^2-y^2)e^{-x} dx\ $ ?

Is correct?

Answer 1

$$ \text{What you need for the marginal is } \begin{cases} \displaystyle \int_y^\infty & \text{if } y\ge 0, \\[10pt] \displaystyle \int_{-y}^\infty & \text{if } y<0. \end{cases} $$ Or you can just write it as $\displaystyle \int_{|y|}^\infty\!\!.~~$ At any rate in $f_{Y=1}(y)$ you'd have $\displaystyle\int_1^\infty$.

Answer 2

$$\begin{align}f(x,y)=&~\frac{1}{8}(x^2-y^2)\,\mathsf e^{-x}~\mathbf 1_{x>0, \lvert y\rvert<x}\\[2ex]f_{X\mid Y=1}(x) =&~ \dfrac{f(x,1)~\mathbf 1_{x>1}}{\int\limits_{1}^{\infty}f(s,1)\operatorname d s}\\[1ex] =&~ \dfrac{(x^2-1)\,\mathsf e^{-x}~\mathbf 1_{x>1}}{\int\limits_{1}^{\infty}(s^2-1)\,\mathsf e^{-s}\operatorname d s}\\[1ex] =&~ \tfrac{\mathsf e} 4(x^2-1)\mathsf e^{-x}\mathbf 1_{x>1}\\[3ex] \mathsf E(X\mid Y=1)~=&~\tfrac{\mathsf e} 4\int_1^\infty x(x^2-1)\mathsf e^{-x}\operatorname d x \\=&~\tfrac 7 2\end{align}$$