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Math Help - Two statistical random variables combine to give Gaussian?

  1. #1
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    Two statistical random variables combine to give Gaussian?

    Hi all,

    Sorry for such a strange title. The question I am stuck at is...

    Let x and y be statistically independent random variables with p.d.fs:

    p_x(\alpha) =\left\{ \begin{array}{rl}<br />
                     \frac1{\pi \sqrt{1 - \alpha^2}}  &\mbox{ if $-1 \leq \alpha \leq 1$} \\<br />
                      0 &\mbox{ otherwise}<br />
 \end{array} \right.


    p_y(\beta) =\left\{ \begin{array}{rl}<br />
\beta e^{- \beta^2}  &\mbox{ if $\beta \geq 0$} \\<br />
                      0 &\mbox{ otherwise}<br />
 \end{array} \right.

    Show that z = xy has a Gaussian density function.
    ------------------------------------------------------------------------------------------------------
    I tried solving it by manipulating it like this:

    Let y = \beta where \beta is fixed. Then the random variable x is transformed to z=\beta x. We then, know that

    p_z(\alpha) = \frac1{|\beta|} p_x \left(\frac{\alpha}{\beta}\right)

    Now p_z(\gamma) = \int_{-\infty}^{\infty} \frac1{|\beta|} p_{x,y} \left(\frac{\gamma}{\beta}, \beta \right) \, d\beta

    Following this I used Independence Property and wrote the joint pdf as the product of pdfs. However the resulting integral was too strange. I could not solve it(I am intimidated by it). So I think there must be some other ways to do it.

    So if anyone has any idea on the problem, please enlighten me.

    Thanks,
    Iso

    P.S: I wonder whether I should have posted the integral in the Calculus forum
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  2. #2
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    Quote Originally Posted by Isomorphism View Post
    Hi all,

    Sorry for such a strange title. The question I am stuck at is...

    Let x and y be statistically independent random variables with p.d.fs:

    p_x(\alpha) =\left\{ \begin{array}{rl}<br />
\frac1{\pi \sqrt{1 - \alpha^2}} &\mbox{ if $-1 \leq \alpha \leq 1$} \\<br />
0 &\mbox{ otherwise}<br />
\end{array} \right.


    p_y(\beta) =\left\{ \begin{array}{rl}<br />
\beta e^{- \beta^2} &\mbox{ if $\beta \geq 0$} \\<br />
0 &\mbox{ otherwise}<br />
\end{array} \right.

    Show that z = xy has a Gaussian density function.
    ------------------------------------------------------------------------------------------------------
    I tried solving it by manipulating it like this:

    Let y = \beta where \beta is fixed. Then the random variable x is transformed to z=\beta x. We then, know that

    p_z(\alpha) = \frac1{|\beta|} p_x \left(\frac{\alpha}{\beta}\right)

    Now p_z(\gamma) = \int_{-\infty}^{\infty} \frac1{|\beta|} p_{x,y} \left(\frac{\gamma}{\beta}, \beta \right) \, d\beta

    Following this I used Independence Property and wrote the joint pdf as the product of pdfs. However the resulting integral was too strange. I could not solve it(I am intimidated by it). So I think there must be some other ways to do it.

    So if anyone has any idea on the problem, please enlighten me.

    Thanks,
    Iso

    P.S: I wonder whether I should have posted the integral in the Calculus forum
    According to Rohatgi (1976) (Amazon.com: An Introduction to Probability Theory and Mathematical Statistics (Wiley Series in Probability & Mathematical Statistics): Vijay K. Rohatgi, V. K. Rohatgi: Books) you have the correct result. So you have:

    p_z(\gamma) = \int_{\gamma}^{+\infty} \frac{1}{\beta} \, \frac{1}{\pi \sqrt{1 - \left( \frac{\gamma}{\beta} \right)^2}} \cdot \beta \, e^{-\beta^2} \, d \beta  = \frac{1}{\pi} \int_{\gamma}^{+\infty} \frac{\beta}{\sqrt{\beta^2 - \gamma^2}} \, e^{-\beta^2} \, d \beta.

    The indefinite integral is given below, so all you need to do is evaluate the definite integral. As for getting the indefinite integral, I could probably get it in under an hour, whereas an inte-killer could probably get it under a minute. But if an inte-killer is indisposed and you have no luck yourself, I could probably manage something ..... (Let's see what the next 24 hours brings).

    And I don't think it'd be double posting to post the integral itself (and result) in the calculus forum. That's where you'll must likely find an inte-killer
    Attached Thumbnails Attached Thumbnails Two statistical random variables combine to give Gaussian?-msp22426207252956100785_1335.gif  
    Last edited by mr fantastic; June 14th 2008 at 09:42 PM. Reason: Footnote: I did get the solution in post #5 in under 1 hour as it happens. Under 5 minutes in fact, but not under 1 minute :)
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    According to Rohatgi (1976) (Amazon.com: An Introduction to Probability Theory and Mathematical Statistics (Wiley Series in Probability & Mathematical Statistics): Vijay K. Rohatgi, V. K. Rohatgi: Books) you have the correct result. So you have:

    p_z(\gamma) = \int_{\gamma}^{+\infty} \frac{1}{\beta} \, \frac{1}{\pi \sqrt{1 - \left( \frac{\gamma}{\beta} \right)^2}} \cdot \beta \, e^{-\beta^2} \, d \beta  = \frac{1}{\pi} \int_{\gamma}^{+\infty} \frac{\beta}{\sqrt{\beta^2 - \gamma^2}} \, e^{-\beta^2} \, d \beta.
    I actually didnt get the same integral. For me, the integral terminals are from -\infty to +\infty. But since x and y are defined only in (-1,1) and \mathbb{R}^{+}, I got the integral posted here. Can you explain how you got those terminals?

    Thank you,
    Srikanth

    EDIT: Oh I got it, you took x and y the other way round. So the integrator answer is encouraging me to work on it. Since the answer looks like

     \frac{1}{\pi} \int_{\gamma}^{+\infty} \frac{\beta}{\sqrt{\beta^2 - \gamma^2}} \, e^{-\beta^2} \, d \beta = \frac{e^{- \gamma^2}}{2\sqrt{\pi}}

    Thanks again

    P.S: I see Two different ways of looking at pdfs can be nice way to solve integrals
    Last edited by Isomorphism; June 10th 2008 at 01:40 AM.
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    Quote Originally Posted by Isomorphism View Post
    I actually didnt get the same integral. For me, the integral terminals are from -\infty to +\infty. But since x and y are defined only in (-1,1) and \mathbb{R}^{+}, I got the integral posted here. Can you explain how you got those terminals?

    Thank you,
    Srikanth

    EDIT: Oh I got it, you took x and y the other way round. So the integrator answer is encouraging me to work on it. Since the answer looks like

     \frac{1}{\pi} \int_{\gamma}^{+\infty} \frac{\beta}{\sqrt{\beta^2 - \gamma^2}} \, e^{-\beta^2} \, d \beta = \frac{e^{- \gamma^2}}{2\sqrt{\pi}}

    Thanks again

    P.S: I see Two different ways of looking at pdfs can be nice way to solve integrals
    (The resulting integral seemed easier to deal with than the other way around).

    Yes, an encouraging result! Let me know if you get stuck proving the integrator's answer.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    (The resulting integral seemed easier to deal with than the other way around).

    Yes, an encouraging result! Let me know if you get stuck proving the integrator's answer.
    Hi Isomorphism. You probably did this already, but if you make the substitution u = \sqrt{\beta^2 - \gamma^2} the desired result p_z(\gamma) = \frac{e^{- \gamma^2}}{2\sqrt{\pi}} falls out easily.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Hi Isomorphism. You probably did this already, but if you make the substitution u = \sqrt{\beta^2 - \gamma^2} the desired result p_z(\gamma) = \frac{e^{- \gamma^2}}{2\sqrt{\pi}} falls out easily.
    Yes I did.Thanks
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