Finding PDF of Y if X is a Gaussian PDF

[Satwant]

A question which I amnot able to do...please help:
Find the PDF of Y if X is a Gaussian PDF:
fx(x) = (1e-x^2/2)/(2pi)^1/2 ; -infnity<x<+infinity

Express your answer in terms of CDF of X gven by

Fi(x) = Integral -infnty to + infnity((1e-x^2/2)/(2pi)^1/2)

b) Sketch both the PDF, fY(y) and CDF, FY(y) for randon variable Y

 

[NoMoreExams]

A question which I amnot able to do...please help:
Find the PDF of Y if X is a Gaussian PDF:
fx(x) = (1e-x^2/2)/(2pi)^1/2 ; -infnity<x<+infinity

Express your answer in terms of CDF of X gven by

Fi(x) = Integral -infnty to + infnity((1e-x^2/2)/(2pi)^1/2)

b) Sketch both the PDF, fY(y) and CDF, FY(y) for randon variable Y

You haven't given a relationship between X and Y have you?

 

[Satwant]

Sorry, it is Y = X^2

 

[HallsofIvy]

Then just replace X^2 in the formulas by Y!
fx(Y) = e^{-Y/2}/(2\pi)^{1/2}
0\le Y&lt; \infty

Fi(x) = \int_{-\infty}^\infty (e^{-Y/2}/(2\pi)^{1/2})dy

 

[Satwant]

but if Y = modX

 

[Satwant]

Also, to find PDF of Y of a Gaussian PDF, do I need to find mean and variance?

 

[Satwant]

xx yyy

 

[HallsofIvy]

I don't know what you mean by "modX" or by "find PDF of Y of a Gaussian PDF".

 

[Satwant]

I mean
Y = |X|

 

[Satwant]

you told me how to do it for Y = X^2 but in another part, how to do it for
Y = |X|

 

[statdad]

Try this. For the distribution function of X write

<br /> F(x) = \Pr(X \le x) = \Phi(x)<br />

Here \Phi(x) is the usual notation for the CDF of the standard Gaussian.
I'll use G(y) as the CDF for your new random variable.

<br /> \begin{align*}<br /> G(y) &amp; = \Pr(Y \le y) = P(X^2 \le y) \\<br /> &amp; =\Pr(-\sqrt{y} \le X \le \sqrt{y}) \\<br /> &amp; = \Phi(\sqrt{y}) - \Phi(-\sqrt{y})<br /> \end{align*}<br />

Since the standard Gaussian is symmetric around 0,

<br /> \Phi(-a) = 1 - \Phi(a)<br />

for any number a. From the place where I left off:

<br /> \begin{align*}<br /> G(y) &amp;= \Phi(\sqrt{y}) - \Phi(-\sqrt{y}) = \Phi(\sqrt{y}) - (1- \Phi(\sqrt{y}))\\<br /> &amp; = 2\Phi(\sqrt{y}) - 1<br /> \end{align*}<br />

Now use these facts:

* The density of Y is the derivative of G(y)
* You need to use the chain rule when you take the derivative of \Phi(\sqrt{y})
* The derivative of \Phi(x) is the density of the standard Gaussian
* The random variable Y is defined on (0, \infty)
* The distribution of Y is one you should be able to recognize

Edited to add:
the method for your second question is similar:
<br /> \Pr(|X| \le y) = \Pr(-y \le X \le y)<br />

go from here. (I have a second \le between X and y above, but it isn't showing.)

 

[Satwant]

thanks!

 

[mathclass]

Any idea how to sketch the PDF, fY(y) and CDF, FY(y) for randon variable Y for this problem?

Thank you.

 

[statdad]

once you have the expression for the density - graph it as any other function.
You'll need a technology aid to graph the CDF.

 

[mathclass]

When finding the density of Y by taking the derivative of G(y) as described earlier, we want to take the derivative of G(y) in respect to y correct? Or is it in respect to phi?

Thank you for all the help

 

[statdad]

<br /> g(y) = \frac{dG}{dy}<br />

 

[mathclass]

Thanks