# Finding the pdf and cdf of this function

## 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]

## 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 im on the right track. Thanks.

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HallsofIvy
Homework Helper
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
Homework Helper
Gold Member
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 im 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.