# bivariate transformation ends up with only one variable in the joint distribution?

I am given two random variables $X_1$ and $X_2$ drawn from identically distributed independent exponential distributions with $\beta = 1$. The question is whether $u=\frac{X_2-X_1}{2}$ and $v=\frac{X_2 + X_1}{2}$ are independent distributions.

Since the given two distributions are exponential and independent, I can write the joint distribution as:

$$f_{X_1, X_2 }(x_1, x_2) = e^{-(x_1+x_2)}$$

I took $h_1(U,V) = U - V$ and $h_2(U,V) = U + V$

Then the jacobian evaluates is calculated as:

\begin{vmatrix} \frac{\partial h_1}{\partial u} & \frac{\partial h_1}{\partial v} \\ \frac{\partial h_2}{\partial u} & \frac{\partial h_2}{\partial v} \end{vmatrix}

$= 2$

So finally my new joint function can be written as:

$$f_{U,V}(u,v) = f_{X_1,X_2}(h_1(u,v), h_2(u,v))\lvert J\rvert$$ $$f_{U,V}(u,v) = 2e^{-(u-v+u+v)} = 2e^{-2u}$$

The function I'm ended up with does not contain $v$ at all! Am I doing something wrong here? Or does the absence of the other variable imply independence? Any help would be appreciated.

• If the question is simply whether $u$ and $v$ are independent, then plotting a large random sample of values will show you that they aren't independent. – JimB Dec 15 '17 at 18:32
• @JimB sorry if my question is not clear. I'm supposed to show whether they're independent or not. – PPGoodMan Dec 15 '17 at 18:40
• @JimB I mean show it theoretically. Not with data. – PPGoodMan Dec 15 '17 at 18:49

## 1 Answer

I think you want $h_1 (U,V)=V-U$ as $V-U = x_1$. That makes the joint density function $2 e^{-2v}$ but only when $0\le v < \infty$ and $-v<=u<v$ or $-\infty<u<\infty$ and $v>|u|$. Otherwise it is zero.

So

$$\int _{-\infty }^{\infty }\int _{\left| \text{u}\right| }^{\infty }2 \exp (-2 \text{v})d\text{v}d\text{u}=\int _0^{\infty }\int _{-\text{v}}^{\text{v}}2 \exp (-2 \text{v})d\text{u}d\text{v}=1$$

From the above you can get the marginal distributions and show that their product is not $2 e^{-2v}$.

A random sample from the joint distribution of $U$ and $V$ looks like the following: 