random process (Gaussian distribution)

**Rezast** · January 26th 2007, 01:52 PM

Hi,

Well as you see I’m a newbie..
This was one of the problems on my telecommunications exam but it’s pure math:

Assuming that X is a random Gaussian variable and ${Y}=X^2$ ,find $f_{Y}\left({y}\right)$

$f_{X}\left({x}\right)=\frac{1}{\sqrt{2\pi}\sigma}e ^{\frac{{-}x^2}{2\sigma^2}}$

any help appreciated

**CaptainBlack** · January 26th 2007, 02:28 PM

Originally Posted by Rezast

Hi,

Well as you see I’m a newbie..
This was one of the problems on my telecommunications exam but it’s pure math:

Assuming that X is a random Gaussian variable and ${Y}=X^2$ ,find $f_{Y}\left({y}\right)$

$f_{X}\left({x}\right)=\frac{1}{\sqrt{2\pi}\sigma}e ^{\frac{{-}x^2}{2\sigma^2}}$

any help appreciated

One rather clunky way of doing it is this:

if $y \ne 0$

$P(y_0-\delta/2 < y < y_0+\delta/2)=P(y_0-\delta/2 < x^2 < y_0+\delta/2)$

..... $= P(\sqrt{y_0-\delta/2}<x<\sqrt{y_0+\delta/2}) + P(-\sqrt{y_0+\delta/2}<x<-\sqrt{y_0-\delta/2})$

..... $=2 P(\sqrt{y_0-\delta/2}<x<\sqrt{y_0+\delta/2})$

(this last since the pdf of X is symmetric)

..... $=2 P(\sqrt{y_0}-(1/4)\delta/ \sqrt{y_0}<x<\sqrt{y_0}-(1/4)\delta/ \sqrt{y_0})$

..... $\approx 2 p_X(\sqrt{y_0}) \left( \frac{1}{2}\, \frac{\delta}{\sqrt{y_0}}\right)$

..... $= p_X(\sqrt{y_0}) \left( \frac{\delta}{\sqrt{y_0}}\right)$

..... $= p_Y(y_0) \delta$

So we have:

$p_Y(y_0)=\frac{p_X(\sqrt{y_0})}{\sqrt{y_0}}$

So the density of $Y$ is:

$p_Y(y)=\frac{p_X(\sqrt{y})}{\sqrt{y}}$ ,

where $p_X(x)$ is the density of $X$ .

(I'm pretty sure that this has gone wrong somewhere, but you should get the idea)

RonL

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