# [SOLVED] joint pdfs

• Dec 7th 2008, 09:40 PM
Chris L T521
[SOLVED] joint pdfs
This problem has been irritating me for a couple days now. I can't figure it out.

Quote:

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!!! (Nod)
• Dec 8th 2008, 12:05 AM
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).
• Dec 8th 2008, 12:07 AM
Jameson
• Dec 8th 2008, 12:13 AM
mr fantastic
Quote:

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

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

$\displaystyle = \frac{\exp\left(-\frac{(y_1 + (y_2 - y_1))}{2}\right)}{4} = \frac{\exp\left(-\frac{y_2}{2}\right)}{4}$.
• Dec 8th 2008, 12:27 AM
Chris L T521
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.