PDF and CDF

1. Feb 6, 2009

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

2. Feb 6, 2009

NoMoreExams

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

3. Feb 6, 2009

Satwant

Sorry, it is Y = X^2

4. Feb 6, 2009

HallsofIvy

Staff Emeritus
Then just replace $X^2$ in the formulas by Y!
$$fx(Y) = e^{-Y/2}/(2\pi)^{1/2}$$
$$0\le Y< \infty$$

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

5. Feb 6, 2009

Satwant

but if Y = modX

6. Feb 6, 2009

Satwant

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

7. Feb 6, 2009

xx yyy

8. Feb 6, 2009

HallsofIvy

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

9. Feb 6, 2009

Satwant

I mean
Y = |X|

10. Feb 6, 2009

Satwant

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

11. Feb 6, 2009

statdad

Try this. For the distribution function of $$X$$ write

$$F(x) = \Pr(X \le x) = \Phi(x)$$

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.

\begin{align*} G(y) & = \Pr(Y \le y) = P(X^2 \le y) \\ & =\Pr(-\sqrt{y} \le X \le \sqrt{y}) \\ & = \Phi(\sqrt{y}) - \Phi(-\sqrt{y}) \end{align*}

Since the standard Gaussian is symmetric around 0,

$$\Phi(-a) = 1 - \Phi(a)$$

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

\begin{align*} G(y) &= \Phi(\sqrt{y}) - \Phi(-\sqrt{y}) = \Phi(\sqrt{y}) - (1- \Phi(\sqrt{y}))\\ & = 2\Phi(\sqrt{y}) - 1 \end{align*}

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:
$$\Pr(|X| \le y) = \Pr(-y \le X \le y)$$

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

12. Feb 6, 2009

Satwant

thanks!!!!!!!!!!

13. Feb 8, 2009

mathclass

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

Thank you.

14. Feb 8, 2009

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.

15. Feb 9, 2009

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

16. Feb 9, 2009

statdad

$$g(y) = \frac{dG}{dy}$$

17. Feb 9, 2009

mathclass

Thanks

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