Marginal Probability Density Functions

Question

Q) Let X and Y have the joint p.d.f.

\begin{equation} f(x,y)=\begin{cases} e^{-y}, \text{if } 0<x<y<\infty \\ 0, \text{otherwise } \end{cases} \end{equation}

Obtain the marginal p.d.f's for X and Y.

A)I know that $F_{x}(s)=\lim_{t \to \infty} F_{x,y}(s,t)$ So have found f(x) to be

\begin{equation} f(x)=\begin{cases} \lim_{y \to \infty} e^{-y} = 0, \text{if } 0<x<y<\infty \\ \lim_{y \to \infty} 0 = 0, \text{otherwise } \end{cases} \end{equation}

and f(y) to be

\begin{equation} f(y)=\begin{cases} \lim_{x \to \infty} e^{-y} = 0, \text{if } 0<x<y<\infty \\ \lim_{x \to \infty} 0 = 0, \text{otherwise } \end{cases} \end{equation}

(since if x is going to infinity then y is going to infinity.)

However, the next question is asking about calculating the co-variance matrix of X+Y and X-Y which makes me think that my first answer was wrong. I don't know how to fix it so any hints would be really helpful, thanks!

Answer 1

The following figure depicts the joint density which is zero outside the red region.

So the marginals are

$$f_X(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)\ dy=\int_x^{\infty}e^{-y}\ dy=e^{-x},$$ if $x\ge 0$

and

$$f_Y(y)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)\ dx=\int_0^{y}e^{-y}\ dx=ye^{-y},$$ if $y\ge 0$.

EDIT

$$E[XY]=\int_0^{\infty}\int_0^{\infty}xyf_{X,Y}(x,y)\ dxdy=$$ $$=\int_0^{\infty}x\int_x^{\infty}ye^{-y}\ dy\ dx.$$

For the internal integral, the antiderivative is $-e^{-y}(1+y)$. This antiderivative is gained by integrating by parts and using $v'=e^{-y}$ and $u=y$.

And $\left[-e^{-y}(1+y)\right]_x^{\infty}=e^{-x}(1+x).$ Then

$$E[XY]=\int_0^{\infty}xe^{-x}(1+x)\ dx=\int_0^{\infty}xe^{-x}\ dx+\int_0^{\infty}x^2e^{-x}\ dx$$

and you can use the antiderivative formula given above. But without that, you know that, the first integral is $1$ bacause we compute the mean of the exponential distribution of $\lambda=1$, also, the other integral is $2$ because we compute the second momentum of the same. So,

$$E[XY]=3.$$