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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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    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](http://en.wikipedia.org/wiki/Gaussian_integral)2012-11-25
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    @TenaliRaman does $exp(a+b)$ mean $e^{a+b}$2012-11-25
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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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    what does R.V. stand for?2012-11-25
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    Random variable2012-11-25
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    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 $\mu = t$...2012-11-25
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    @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$2012-11-25
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    this is easier to understand. thank you!2012-11-25
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    does this mean $\sigma$ can be, say, $w$ and we'll still get $1$ for the integral?2012-11-25
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    i thought the standard form is$\int_{-\infty}^{\infty}(\frac{1}{\sqrt{2\pi}}e^{\frac{-(x-\mu)^2}{2\sigma^2}})dx = 1$ which in our case is $\int_{-\infty}^{\infty}(\frac{1}{\sqrt{2\pi}}e^{\frac{-(x-t)^2}{2w^2}})dx = 1$ notice: no $\sqrt{w}$ at the denominator of 2$\pi$ and $2w^2$ at exponent's denominator2012-11-25
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    Them typos, edited the question2012-11-25