1. ## [SOLVED] joint pdfs

This problem has been irritating me for a couple days now. I can't figure it out.

Let $X_1,\,X_2$ denote a random sample from a distribution $\chi^2\left(2\right)$. Find the joint pdf of $Y_1=X_1$ and $Y_2=X_1+X_2$. Note that the support of $Y_1,\,Y_2$ is $0. Also find the marginal pdf of each $Y_1$ and $Y_2$. Are $Y_1$ and $Y_2$ independent?
The only part I really need help with is determining the joint pdf. I can figure out the rest, given what the pdf is. I would appreciate any input!!!

2. I put some more thought into this and managed to figure it out on my own (it happened to be on change of variables...a topic my teacher never covered).

4. Originally Posted by Chris L T521
I put some more thought into this and managed to figure it out on my own (it happened to be on change of variables...a topic my teacher never covered).
I promised Chris in a pm that when I had a few moments I'd log on and reply. So here's my two cents (hopefully we both agree!):

1. I assume $X_1$ and $X_2$ are independent so that their joint pdf is given by $f(x_1, x_2) = \frac{\exp\left(-\frac{(x_1 + x_2)}{2}\right)}{4}$.

2. Now note that $X_1 = Y_1$ and $X_2 = Y_2 - Y_1$.

Therefore $J = \left| \begin{array}{ccc}
\frac{\partial X_1}{\partial Y_1} & \frac{\partial X_1}{\partial Y_2} \\
& \\
\frac{\partial X_2}{\partial Y_1} & \frac{\partial X_2}{\partial Y_2} \end{array}\right| = \left| \begin{array}{ccc}
1 & 0 \\
& \\
-1 & 1 \end{array}\right| = 1$
.

3. Therefore, by the change-of-variable formula, the joint pdf of $Y_1$ and $Y_2$ is given by

$g(y_1, y_2) = f(\, x_1(y_1, y_2), \, x_2(y_1, y_2) \, ) \, J$

$= \frac{\exp\left(-\frac{(y_1 + (y_2 - y_1))}{2}\right)}{4} = \frac{\exp\left(-\frac{y_2}{2}\right)}{4}$.

5. Here's my solution:

I figured out that the joint pdf of $X_1$ and $X_2$ is $f\left(x_1,x_2\right)=\frac{1}{\Gamma\left(1\right )2}x_1^0e^{-\frac{x_1}{2}}\cdot\frac{1}{\Gamma\left(1\right)2} x_2^0e^{-\frac{x_2}{2}}=\frac{1}{4}e^{-\frac{x_1+x_2}{2}}$. Applying the change in variables $x_1=y_1$ and $x_2=y_2-y_1$, we see that $J=\begin{vmatrix}\frac{\partial x_1}{\partial y_1}&\frac{\partial x_1}{\partial y_2}\\\frac{\partial x_2}{\partial y_1}&\frac{\partial x_2}{\partial y_2}\end{vmatrix}=\begin{vmatrix}1&0\\-1&1\end{vmatrix}=1$. Therefore, $f\left(y_1,y_2\right)=\left|1\right|\frac{1}{4}e^{-\frac{y_2}{2}}=\color{red}\boxed{\frac{1}{4}e^{-\frac{y_2}{2}}}$

Thus, the marginal pdf of $Y_1$ is $\int_{-\infty}^{\infty}f\left(y_1,y_2\right)\,dy_2=\frac{ 1}{4}\int_{y_1}^{\infty} e^{-\frac{y_2}{2}}\,dy_2=-\frac{1}{2}\left.\left[e^{-\frac{y_2}{2}}\right]\right|_{y_1}^{\infty}=\color{red}\boxed{\frac{1}{ 2}e^{-\frac{y_1}{2}}}$ and the marginal pdf of $Y_2$ is $\int_{-\infty}^{\infty}f\left(y_1,y_2\right)\,dy_1=\frac{ 1}{4}\int_{0}^{y_2} e^{-\frac{y_2}{2}}\,dy_1=\frac{1}{4}\left.\left[y_1e^{-\frac{y_2}{2}}\right]\right|_{0}^{y_2}=\color{red}\boxed{\frac{y_2}{4}e ^{-\frac{y_2}{2}}}$. With this, we see that $\frac{y_2}{4}e^{-\frac{y_2}{2}}\cdot\frac{1}{2}e^{-\frac{y_1}{2}}=\frac{y_2}{8}e^{-\left(y_1+y_2\right)}\neq\frac{1}{4}e^{-\frac{y_2}{2}}$. Thus, they are dependent.