Let $X\sim \exp(\lambda=1)$ and $Y \sim U(1,2) $ two independent random variables; what is the joint distribution function? Compute $P(X>Y)$
I started by calculating:
$$ f(x) = \begin{cases}
e^{-x} & x\geq 0\\
0 &x < 0
\end{cases} $$
$$ f(y) = \begin{cases}
1 & 1\leq y\leq 2\\
0 & \text{otherwise}
\end{cases} $$
Now I am not sure about the joint density. I do know that the fact that X,Y are independent means that I should some how multiply their densities, though I am not so sure what would be the product range.
In addition, I am not sure how I can from that derive $P(X>Y)$. Again, I have some clue about a 2-d integral but I can't figure out what's the logic behind it, or how to even define it.
Thanks!
Best Answer
Consider that
$$P(X>Y)=E\left[\mathbf1_{X>Y}\right]$$
, where $\mathbf1_{X>Y}$ equals $1$ if $X>Y$, and equals $0$ otherwise.
And for any (measurable) function $g$ of $(X,Y)$, courtesy of this theorem, we have
$$E\left[g(X,Y)\right]=\iint g(x,y)f_{X,Y}(x,y)\,\mathrm{d}x\,\mathrm{d}y$$
, where $f_{X,Y}$ is the joint density of $(X,Y)$.
You are right that independence of $X$ and $Y$ implies that $f_{X,Y}$ is just the product of the marginal densities of $X$ and $Y$. So you have to evaluate
\begin{align} P(X>Y)&=\iint \mathbf1_{x>y}\,e^{-x}\mathbf1_{x>0}\mathbf1_{1<y<2}\,\mathrm{d}x\,\mathrm{d}y \\&=\iint \mathbf1_{x>y,\,x>0,1<y<2}\,e^{-x}\,\mathrm{d}x\,\mathrm{d}y \end{align}