Finding PDF of Y if X is a Gaussian PDF

  • Thread starter Thread starter Satwant
  • Start date Start date
  • Tags Tags
    Gaussian Pdf
AI Thread Summary
To find the PDF of Y when X is a Gaussian random variable, it is important to establish the relationship between X and Y, such as Y = X^2 or Y = |X|. For Y = X^2, the PDF can be derived using the CDF of X, resulting in fY(y) = e^{-Y/2}/(2π)^{1/2} for 0 ≤ Y < ∞. In the case of Y = |X|, the CDF G(y) is expressed as G(y) = 2Φ(√y) - 1, where Φ is the CDF of the standard Gaussian. To sketch the PDF and CDF of Y, the density is obtained by differentiating G(y) with respect to y, and technology may be needed for accurate graphing.
Satwant
Messages
8
Reaction score
0
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
 
Physics news on Phys.org
Satwant said:
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?
 
Sorry, it is Y = X^2
 
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
 
but if Y = modX
 
Also, to find PDF of Y of a Gaussian PDF, do I need to find mean and variance?
 
xx yyy
 
I don't know what you mean by "modX" or by "find PDF of Y of a Gaussian PDF".
 
I mean
Y = |X|
 
  • #10
you told me how to do it for Y = X^2 but in another part, how to do it for
Y = |X|
 
  • #11
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.)
 
  • #12
thanks!
 
  • #13
Any idea how to sketch the PDF, fY(y) and CDF, FY(y) for randon variable Y for this problem?

Thank you.
 
  • #14
once you have the expression for the density - graph it as any other function.
You'll need a technology aid to graph the CDF.
 
  • #15
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
 
  • #16
<br /> g(y) = \frac{dG}{dy}<br />
 
  • #17
Thanks
 
Back
Top