# Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$

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 – 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

$\mu$ 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? – 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
The text passage contains a mathematical question about why a specific integral equals 1, and it provides an answer through a series of explanations and transformations.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Moment generating function: why is $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(z-t)^2}{2} dz = 1$

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

-
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

$\mu$ 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 $w=\sigma$.

-
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
user
The text passage contains a mathematical question about why a specific integral equals 1, and it provides an answer through a series of explanations and transformations.

TYPE: SINGLE_QUESTION_WITH_ANSWER