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Math Help - [SOLVED] joint pdfs

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    [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<y_1<y_2<\infty. 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!!!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    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).
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  3. #3
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    Care to enlighten us? Save the thread, post your solution!
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    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}<br />
\frac{\partial X_1}{\partial Y_1} & \frac{\partial X_1}{\partial Y_2} \\<br />
 & \\<br />
\frac{\partial X_2}{\partial Y_1} & \frac{\partial X_2}{\partial Y_2}  \end{array}\right| = \left| \begin{array}{ccc}<br />
1 & 0 \\<br />
 & \\<br />
-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}.
    Last edited by mr fantastic; December 8th 2008 at 12:40 AM. Reason: Fixed some typos and latex
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    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.
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