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!
Best Answer
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,
$$E[XY]=3.$$