Finding the pdf and cdf of this function

  • #1
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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 im on the right track. Thanks.
 

Answers and Replies

  • #2
HallsofIvy
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You need to change from "x" to "z"! Yes, you are given that the pdf for x is [tex]\frac{1}{\sqrt{2\pi\sigma^2}} e^{\frac{-(x- \mu)^2}{2\sigma^2}}=[/tex][tex] \frac{1}{\sqrt{2\pi\sigma^2}} e^{\frac{\left(\frac{x- \mu}{\sigma}\right)^2}{2}}[/tex][tex]= \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{z^2}{2}}[/tex]
 
  • #3
LCKurtz
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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 [tex]\frac{1}{\sqrt{2\pi\sigma^2}} e^{\frac{-(x- \mu)^2}{2\sigma^2}}=[/tex][tex] \frac{1}{\sqrt{2\pi\sigma^2}} e^{\frac{\left(\frac{x- \mu}{\sigma}\right)^2}{2}}[/tex][tex]= \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{z^2}{2}}[/tex]
And there should be no ##\sigma## here.
 

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