Two statistical random variables combine to give Gaussian?

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June 9th 2008, 02:58 AM
Isomorphism

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} \frac1{\pi \sqrt{1 - \alpha^2}} &\mbox{ if $-1 \leq \alpha \leq 1$} \\ 0 &\mbox{ otherwise} \end{array} \right.$

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

Show that $z = xy$ has a Gaussian density function.
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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.(Doh)

Thanks,
Iso

P.S: I wonder whether I should have posted the integral in the Calculus forum (Thinking)
June 9th 2008, 04:07 AM
mr fantastic

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Quote:

Originally Posted by Isomorphism

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} \frac1{\pi \sqrt{1 - \alpha^2}} &\mbox{ if $-1 \leq \alpha \leq 1$} \\ 0 &\mbox{ otherwise} \end{array} \right.$

$p_y(\beta) =\left\{ \begin{array}{rl} \beta e^{- \beta^2} &\mbox{ if $\beta \geq 0$} \\ 0 &\mbox{ otherwise} \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.(Doh)

Thanks,
Iso

P.S: I wonder whether I should have posted the integral in the Calculus forum (Thinking)

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 (Rofl)
June 10th 2008, 01:00 AM
Isomorphism

Quote:

Originally Posted by mr fantastic

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 :)
June 10th 2008, 02:37 AM
mr fantastic

Quote:

Originally Posted by Isomorphism

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 :)

(Yes) (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.
June 10th 2008, 11:13 PM
mr fantastic

Quote:

Originally Posted by mr fantastic

(Yes) (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.
June 11th 2008, 04:01 AM
Isomorphism

Quote:

Originally Posted by mr fantastic

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 :)