If $X,Y \sim \mathrm{Exp}(\lambda)$ independently, find $P(X\leq u,X+Y\leq V)$ $X,Y \sim Exp(\lambda)$ IID.
(a) Find $F_{X,X+Y}=P(X\leq u,X+Y\leq V)$
(b) Find $f_{X,X+Y}$
My solution :
(a) $$F_{X,X+Y}=P(X\leq u,X+Y\leq v)=P(X\leq u,Y\leq v-x)=P(X\leq u)\cdot P(Y\leq v-x)=(1-e^{-\lambda u})(1-e^{-\lambda(v-u)})=1-e^{-\lambda(v-u)}-e^{-\lambda u}+e^{-\lambda (u+v-u)}=1-e^{-\lambda(v-u)}-e^{-\lambda u}+e^{-\lambda v}$$
(b)$$ f_{X,X+Y}=\frac{d}{d(X)d(X+Y)}F_{X,X+Y}=\frac{d}{d(U)d(V)}F_{U,V}=\frac{\lambda e^{-\lambda(u-v)}+\lambda e^{-\lambda u}}{dV}  =-\lambda^2e^{-\lambda(u-v)}$$
Is my solution correct? I am not really sure about $(b)$.
 A: $f_{X,Y}(x,y)=\lambda^{2}e^{-\lambda(x+y)},\,x,y\geq 0$.
Let $U=X+Y$, $V=X$.
Then $|J|=1$.
So $f_{U,V}(u,v)=f_{X,Y}(x,y)|J|$.
So $f_{X+Y,X}(u,v)=\lambda^{2}e^{-\lambda u}\,,v\geq 0,u\geq v$.
From here you can calculate the CDF by integrating.
For the CDF you have to take cases.
if $u'\geq v'$.
Then $$F_{U,V}(u',v')=\int_{0}^{v'}\int_{0}^{u}f_{U,V}(u,v)dvdu+\int_{0}^{v'}\int_{v'}^{u'}f_{U,V}(u,v)\,dudv$$.
And if $u'<v'$ then:-
$$F_{U,V}(u'v')=\int_{0}^{u'}\int_{0}^{u}f_{U,V}(u,v)\,dvdu$$
A: Your mistake is in the following step.
$ \displaystyle P(X\leq u,Y\leq v-x) = P(X\leq u)\cdot P(Y\leq v-x) = \left(1-e^{-\lambda u}\right) \left(1-e^{-\lambda(v-u)}\right)$
Why do you think $P(Y \leq v - x) = \left(1-e^{-\lambda(v-u)}\right)$?
If you want to first find the CDF, note that
$U = X, V = X + Y \implies v \geq u \geq 0~$ as $~x, y \geq 0$
$ \displaystyle P(X\leq u,Y\leq v-x) = \int_0^u \int_0^{v-x} f(x, y) ~ dy ~ dx$
$$ \displaystyle = \lambda^2 \int_0^u e^{-\lambda x} \left[\int_0^{v-x} e^{-\lambda y} ~ dy \right] ~ dx$$
$$ \displaystyle = 1 - e^{- \lambda u} - \lambda u e^{- \lambda v}$$
Now compute pdf from it and you will get the correct answer. Please note the support of pdf is $v \geq u \geq 0$
