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Let $X$ and $Y$ be two independent random variables, exponentially distributed with parameter $\lambda=1$ and let $U=X$ and $V=X+Y$.

Determine the joint CDF of the random pair $(U,V)$ and determine the CDF of the random variable V.

I know that $f_{(U,V)}(U=u,V=v)=f_{(X,Y)}(X=u,Y=v-u)*1$, $1$ being the Jacobian. So given that $f_{(X,Y)}(X=x,Y=y)=e^{-(x+y)}$,
$f_{(U,V)}(U=u,V=v)=e^{-v}$ for $u>0, v>u$.

Now to get the joint CDF of the random pair $(U,V)$ I was thinking: $F_{(U,V)}(U \le a, V \le b)= \int_0^a\int_u^be^{-v}dudv$ (hopefully I am integrating over the correct region in the $UV$ plane)

My questions for this part is:

Is there another way to do it? We proved in class that $f_{(U,V)}(U=u,V=v)=f_{(X,Y)}(X=u,Y=v-u)*|J(u,v-u)|$ but don't have anything like that when it comes to the CDF of a random pair.

Now to determine the CDF of the random variable $V,$ $V$ follows a gamma distribution of parameters $\alpha=2$ and $\lambda=1$ since X and Y are independent. So from there we can just find the CDF. There is no need to go through the joint CDF of $U$ and $V$ to find the marginal density of $V$?

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