$\mathbb{E}(X)$ for joint pdf where $0<y<x< \infty$

Question

On this problem, X and Y have joint pdf

$$f(x,y) = \begin{cases} \frac{2y}{x^2}e^{-x} & ,0 < y < x<+ \infty \\ 0 & , \text{otherwise}\end{cases}$$

So I'm supposed to find $\mathbb{E}(X)$ in order to find the covariance

$\mathbb{E}(Y)$ kinda makes intuitive sense, as you integrate $y$ from $0$ to $x$ and $x$ from $0$ to $\infty$, but I'm confused about what the bounds for the double integral of $\mathbb{E}(X)$ would be.

Apparently the answer is $1$.

Thanks!

Answer 1

To find the $E(x)$, we just need to find the probability distribution of $x$. Now you know the joint probability of $x$ and $y$, so you can get the probability distribution of $x$:$f(x)=\int_{-\infty}^{\infty}\!f(x,y)\,\mathrm{d}x=\int_{0}^{x} \dfrac{{2y}}{x^2} e^{-x}dy=\dfrac{{e^{-x}}}{x^2}\int_{0}^{x}\!y^2\,\mathrm{d}y=e^{-x}$. Then $E(x)=\int_{0}^{\infty}xe^{-x}dx=\int_{0}^{\infty}xd{-e^{-x}}=\int_{0}^{\infty}e^{-x}dx=\int_{0}^{\infty}d{-e^{-x}}=1$!

Answer 2

Let

$$f(x,y)= \left\{ \begin{array}{ll} \frac{2 y }{x^2} e^{-x} & 0 < y < x < \infty \\ 0 & \mbox{otherwise} \end{array} \right. $$

We want to compute the marginal distribution of $X$. We get,

$$f(x) = \int_0^x \frac{2 y}{x^2}e^{-x} \, dy = e^{-x}.$$

Observe that

$$E[X] = \int_0^{\infty} x f(x) \, dx=\int_0^{\infty} x e^{-x} \, dx = 1$$

Or, similarly, using a double integral...

$$E[X]=\int _0^{\infty }\int _0^x x \frac{ 2y }{x^2} e^{-x}dydx$$

Answer 3

Just as you wrote the integral bounds, $$ \mathbb E[X] = \int_0^\infty \int_0^x x\cdot \frac{2y}{x^2}e^{-x} \,dy\, dx = \int_0^\infty \frac{1}{x}e^{-x} \cdot x^2 \,dx= \int_0^\infty xe^{-x} \,dx =1. $$ If you want to integrate by $x$ first, then by $y$, you get exponential integral which is not an elementary function: $$ \mathbb E[X] = \int_0^\infty \int_y^\infty x\cdot \frac{2y}{x^2}e^{-x} \,dx\, dy = \int_0^\infty 2y \underbrace{\int_y^\infty \frac{e^{-x}}{x} \,dx}_{-\text{Ei}(-y)}\, dy. $$ So, this way is improper.