Finding the pdf and cdf of this function

[bonfire09]

Homework Statement

Let $X$ have the pdf $f_X(x)=\dfrac{1}{\sqrt{2\pi\sigma^2}}e^{\dfrac{-(x-\mu)^2}{2\sigma^2}}$ where $-\infty<x<\infty,-\infty<\mu<\infty,\sigma>0$. Let $Z=g(X)=\frac{X-\mu}{\sigma}$. Find the pdf and cdf of $Z$[/B]

Homework Equations

The Attempt at a Solution

Basically I noticed that $X$ has a normal distribution. I get for the cdf of $Z$ is $P((\dfrac{x-\mu}{\sigma})\leq t)$ $=P(x-\mu\leq t\sigma)=P(x\leq\mu+t\sigma)=F_Z(\mu+t\sigma)=\int_{-\infty}^{\mu+t\sigma} \dfrac{1}{\sqrt{2\pi\sigma^2}}e^{\dfrac{-(x-\mu)^2}{2\sigma^2}} dx$. The pdf of $Z$ I get that $f_z(x)=F'_Z(t)=\dfrac{1}{\sqrt{2\pi\sigma^2}}e^{\dfrac{-(x-\mu)^2}{2\sigma^2}} *\sigma$. Not sure if I am on the right track. Thanks.

 

[HallsofIvy]

You need to change from "x" to "z"! Yes, you are given that the pdf for x is $$\frac{1}{\sqrt{2\pi\sigma^2}} e^{\frac{-(x- \mu)^2}{2\sigma^2}}=$$$$\frac{1}{\sqrt{2\pi\sigma^2}} e^{\frac{\left(\frac{x- \mu}{\sigma}\right)^2}{2}}$$$$= \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{z^2}{2}}$$

 

[LCKurtz]

I get that $f_z(x)=F'_Z(t)=\dfrac{1}{\sqrt{2\pi\sigma^2}}e^{\dfrac{-(x-\mu)^2}{2\sigma^2}} *\sigma$. Not sure if I am on the right track. Thanks.

Note that the $\sigma$ cancels the $\sigma^2$ under the square root.

You need to change from "x" to "z"! Yes, you are given that the pdf for x is $$\frac{1}{\sqrt{2\pi\sigma^2}} e^{\frac{-(x- \mu)^2}{2\sigma^2}}=$$$$\frac{1}{\sqrt{2\pi\sigma^2}} e^{\frac{\left(\frac{x- \mu}{\sigma}\right)^2}{2}}$$$$= \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{z^2}{2}}$$

And there should be no $\sigma$ here.