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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!

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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}

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For the first part the joint distribution is $e^{-x}$ for the range $x\geq 0$ and $1\leq y\leq 2$, and 0 otherwise.

For the second part

$$\text{Pr}[X>Y]=\int_{y=1}^2\text{Pr}[X>y]\,dy=\int_{1}^2e^{-y}\,dy$$

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