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

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    Yes. $exp(a + b)$ means $e^{a + b}$.2012-11-26

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$\mu$ 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 $w=\sigma$.

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    Them typos, edited the question2012-11-25