Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Problem: Z is normally distributed with mean $0$ standard deviation 1. Goal: obtain the moment generating function of Z.

So I started with $$E[e^{tz}] = M_z(t)= \int_{-\infty}^{\infty}e^{tz}(\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}})dz$$

$$\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-(\frac{z^2}{2}-tz)} dz = \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-(\frac{z^2-2tz}{2})} dz$$ completing the squares in the exponent we get

$$\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-(\frac{z^2-2tz+t^2}{2})}*e^{\frac{t^2}{2}} dz = \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{(\frac{-(z-t)^2}{2})}*e^{\frac{t^2}{2}} dz$$

then my notes says $e^{\frac{t^2}{2}} \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{(\frac{-(z-t)^2}{2})} dz $ where $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{(\frac{-(z-t)^2}{2})} =1$ so the mgf is $e^{\frac{t^2}{2}}$.

Why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{(\frac{-(z-t)^2}{2})} =1$?

I know that $\int_{-\infty}^{\infty}(\frac{1}{\sqrt{2\pi}}e^{\frac{-(x-\mu)^2}{2\sigma^2}})dx = 1$(prob. density function of a normal distribution) but no where in the problem did it mention $\mu = t$ in fact, $\mu=0$ how could $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{(\frac{-(z-t)^2}{2})} =1$??

share|improve this question
    
Put $\frac{z - t}{\sqrt{2}}$ as $x$, then your integral looks like $\int_{-\infty}^{\infty} \frac{1}{\sqrt{\pi}} exp(-x^2) dx$, this is just your old Gaussian Integral –  TenaliRaman Nov 25 '12 at 18:17
    
@TenaliRaman does $exp(a+b)$ mean $e^{a+b}$ –  user133466 Nov 25 '12 at 18:29
    
Yes. $exp(a + b)$ means $e^{a + b}$. –  TenaliRaman Nov 26 '12 at 0:00
add comment

1 Answer 1

up vote 2 down vote accepted

$\mu$ is just a dummy variable, that can be replaced by anything, such has $t,z,s,\triangle\ldots$ What you in fact have, namely $$ \frac{1}{\sqrt{2\pi}}e^{-\frac{(z-t)^2}{2}} $$ is just a normal random variable's density function with expected value $t$ and standard deviation $1$, which you integrate on the whole line.

Note that the density function of a normal random variable with expected value $\mu$ and standard deviation $\sigma$ is $$ \frac{1}{\sqrt{2\pi\sigma^2}}\text{e}^{-\frac{(x-\mu)^2}{2\sigma^2}}. $$ and that $\mu,\sigma$ are dummy variables and can be replaced by anything you want, namely $t=\mu, 1=\sigma, $ or $w=\sigma$.

share|improve this answer
    
what does R.V. stand for? –  user133466 Nov 25 '12 at 18:19
    
Random variable –  Jean-Sébastien Nov 25 '12 at 18:19
    
err... it's just no where does it say $mu = t$! the denominator fits because the problem stated that $\sigma = 1$ but its like asking me to take a leap when we replace $\mu = t$... –  user133466 Nov 25 '12 at 18:24
    
@user133466 I made a correction. $\mu$ isnt equal to $t$, but the integral you're left with is the integral of a normal distribution with mean $t$ and standard deviation $1$ –  Jean-Sébastien Nov 25 '12 at 18:28
    
this is easier to understand. thank you! –  user133466 Nov 25 '12 at 18:34
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.