The question is as follows:
Let $X_1, X_2$ be jointly distributed with joint density function given by $f(x_1, x_2) = 2e^{-x_1 – x_2}$ for $0<x_1<x_2<\infty$. Compute the joint density under the transformation $Y_1 = X_1, Y_2 = X_1 + X_2$.
I have computed the determinant of the Jacobian to be $1$, fairly certain this is correct. I have inverted the transformation to yield $x_1 = y_1, x_2 = y_2 – y_1$. However, if you substitute these into the original distribution $f(x_1, y_2)$ and multiply by the determinant of $J$, it simply reduces to $2e^{-y_1 – y_2 + y_1}$, or simply $2e^{- y_2}$. My concern is that, because one of the dependencies has now vanished, this is not a joint density function of $y_1, y_2$. Have I made some sort of error, or is this solution an acceptable one? The instructor for the course gave me very unhelpful response when I ask her this question, so I am not sure how to proceed. Any help would be much appreciated!
Best Answer
I approached it through the cdf
\begin{eqnarray*} P\left(X_1 \geq y_1, X_2 \geq y_2 - X_1\right) &=& \int^{\infty}_{y_1} \int^{\infty}_{\max\left(y_2-x_1,x_1\right)} 2e^{-x_1}e^{-x_2}dx_1dx_2,\\ &=& \int^{\infty}_{y_1} 2e^{-x_1}e^{-\max\left(y_2-x_1,x_1\right)}dx_1. \end{eqnarray*}
where I assume $y_2 \geq 2y_1$ due to the constraints on the support. The lower limit on the inner integral is due to the fact that you want $X_2 \geq y_2-X_1$ but also you must have $X_2 \geq X_1$. Note that $y_2 - x_1 \geq x_1$ if $x_1 \leq \frac{1}{2}y_2$, so we can split the integral into two parts,
$$\int^{\frac{1}{2}y_2}_{y_1} 2e^{-x_1}e^{-y_2}e^{x_1}dx_1 + \int^{\infty}_{\frac{1}{2}y_2}2e^{-2x_1}dx_1 = 2e^{-y_2}\left(\frac{1}{2}y_2-y_1\right)+e^{-y_2}.$$
Now you can differentiate to obtain the density. Indeed, the density works out to $2e^{-y_2}$, which does not explicitly depend on $y_1$ because the cdf is linear in that variable. However, $y_1$ may play a role in determining the support.