# Thread: [SOLVED] joint pdfs

1. ## [SOLVED] joint pdfs

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

Let $\displaystyle X_1,\,X_2$ denote a random sample from a distribution $\displaystyle \chi^2\left(2\right)$. Find the joint pdf of $\displaystyle Y_1=X_1$ and $\displaystyle Y_2=X_1+X_2$. Note that the support of $\displaystyle Y_1,\,Y_2$ is $\displaystyle 0<y_1<y_2<\infty$. Also find the marginal pdf of each $\displaystyle Y_1$ and $\displaystyle Y_2$. Are $\displaystyle Y_1$ and $\displaystyle Y_2$ 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).

3. Care to enlighten us? Save the thread, post your solution!

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

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

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

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

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

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

5. Here's my solution:

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

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