Joint PDF transformation

Question

I think I know how to do this but I just want to make sure that it's 100% correct.

$X_1, X_2$ are i.i.d. random variables with distribution EXP(1)

f($x_1, x_2$) = $e^{(-x_1-x_2)}$I(x >= 0)

Question: Find the joint PDF of $T_1$ = min($X_1, X_2)$ and $T_2$ = $X_1$ + $X_2$

Here is what I did:

Partition into two: $B_1$ is for $X_1 <= X_2$, $B_2$ is for $X_1 > X_2$

On $B_1$: $X_1 = T_1$, $X_2 = T_2 - T_1$

Jacobian is {1,0}{-1,1}, Det Jacobian = 1

On $B_2$: $X_2 = T_1$, $X_1 = T_2 - T_1$

Jacobian is {-1, 1}{1,0}, Det Jacobian = -1

So g($t_1, t_2$) = $e^{(-t_2)}$*1*I($x_1 < x_2$) + $e^{(-t_2)}$*|-1|I($x_1 > x_2$)

Answer 1

You can write the joint probability distribution of $T_1$ and $T_2$ as follow:

$P_{T_1,T_2}(t_1,t_2)=P_{T_1,T_2}(t_1,t_2|x_1<x_2)p_{X_1,X_2}(x_1<x_2)+P_{T_1,T_2}(t_1,t_2|x_2<x_1)p_{X_1,X_2}(x_2<x_1)=\frac{1}{2}(P_{T_1,T_2}(t_1,t_2|x_1<x_2)+P_{T_1,T_2}(t_1,t_2|x_2<x_1))$

For the first case we have:

$X_1=T_1$ and $X_2=T_2-T_1$, hence in this case the determinant of the Jacobian matrix is equal to $1$.

For the second case we have:

$X_1=T_2-T_1$ and $X_2=T_1$, hence in this case the determinant of the Jacobian matrix is equal to $-1$ and its absolute value is $1$.

Hence we can say:

$P_{T_1,T_2}(t_1,t_2)=\frac{1}{2}(P_{T_1,T_2}(t_1,t_2|x_1<x_2)+P_{T_1,T_2}(t_1,t_2|x_2<x_1))=\frac{1}{2}(P_{X_1}(x_1)P_{X_2}(x_2)\times 1+P_{X_1}(x_1)P_{X_2}(x_2)\times 1)$

but notice that the first part in the above relation is due to the first assumption that $X_1<X_2$, and the second part is due to the assumption that $X_2<X_1$, hence:

$P_{T_1,T_2}(t_1,t_2)=\frac{1}{2}(P_{X_1}(T_1)P_{X_2}(T_2-T_1)+P_{X_1}(T_2-T_1)P_{X_2}(T_1))=\frac{1}{2}(e^{-t_1}e^{t_1-t_2}+e^{t_1-t_2}e^{-t_1})=e^{-t_2}$

Am I right?