I am attempting the following question:
Suppose $X\sim\text{Expo}(1)$, $Y\sim\text{Expo}(1)$. Let $U = 6X + 8Y$ and $V = 2X + 3Y$. Find the joint density function of $U$ and $V$. Justify whether $U$ and $V$ are independent.
My attempt so far is: if $X\sim\text{Expo}(1)$, $Y\sim\text{Expo}(1)$, then $f_{XY}(x,y)=f_X(x)f_Y(y)=e^{-(x+y)}$. Then, for U=$6X+8Y$, $V=2X+3Y$, the Jacobian is $J=\begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix}=\begin{vmatrix} 6 & 8 \\ 2 & 3 \end{vmatrix}=18-16=2$, so $J^{-1}=\frac12$ and $f_{UV}(u,v)=J^{-1}f_{XY}(g_1(u,v),g_2(u,v))=\frac12 e^{-\frac{u}2+v}$ (since $x=\frac{3u}2-4v$ and $y=3v-u$).
The issue now is figuring out what the support is. My understanding is that since $0<x,y<\infty$ for the exponential variables, then $0<3v-u<\infty$ and $0<\frac32 u-4v<\infty$. All I can get from this is $\frac83 v<u<3v$, and integrating $f_{UV}(u,v)$ from $\frac83 v$ to $3v$ will give $2e^{-\frac5{12}v}\cosh\frac{v}{12}$, which is assume is $f_V(v)$. However, all I can get for $u$ is $0<u<\infty$ (which seems quite fishy) and I am not sure if this is the correct approach. Could someone please clear my doubts? Thank you!
Your conclusion is right. You can verify it here
I would add that you don't need to calculate the marginals. The fact that the support is not a rectangle is enough to prove the dependence.