# 30 Normal Distribution

## 30.1 Probability Distribution Function

A random variable $$X$$ is said to have a Normal Distribution with parameters $$\mu$$ and $$\sigma^2$$ if its probability density function is

$f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^\frac{-(x-\mu)^2}{2\sigma^2},\ \ x\in\Re,\ \mu\in\Re,\ 0<\sigma^2$

## 30.2 Cumulative Distribution Function

The cdf of the Normal Distribution cannot be written in closed form.

It is not uncommon in statistical practice to denote the cdf of the Standard Normal Distribution, that is, the Normal distribution with $$\mu=0$$ and $$\sigma^2=1$$ to be denoted by $$\Phi(\cdot)$$. Although it is rare, the $$\Phi(\cdot)$$ notation is sometimes used to denote any cumulative density function. Because such citations are rare they are usually accompanied by a statement about the distribution $$\Phi(\cdot)$$ represents. If no such statement is given, it is reasonably safe to assume that the function refers to the Standard Normal Distribution.

The importance of the Standard Normal Distribution is made evident by the fact that the area under any Normal curve is proportional to the distribution’s variance. A proof of this is provided in Theorem

## 30.3 Expected Values

\begin{aligned} E(X) &= \int\limits_{-\infty}^{\infty}x\frac{1}{\sqrt{2\pi}\sigma}\exp\Big\{-\frac{1}{2}\Big(\frac{x-\mu}{\sigma}\Big)^2\Big\} \\ ^{[1]} &= \frac{1}{\sqrt{2\pi}\sigma}\int\limits_{-\infty}^{\infty}x\exp\Big\{-\frac{1}{2}\Big(\frac{x-\mu}{\sigma}\Big)^2\Big\} \\ &= \frac{1}{\sqrt{2\pi}\sigma}\int\limits_{-\infty}^{\infty}(z\sigma+\mu)\exp\Big\{-\frac{1}{2}z^2\Big\}\sigma dz \\ &= \frac{1}{\sqrt{2\pi}\sigma}\int\limits_{-\infty}^{\infty}z\sigma^2\exp\Big\{-\frac{1}{2}z^2\Big\} +\mu\sigma\exp\Big\{-\frac{1}{2}z^2\Big\}dz \\ &= \frac{1}{\sqrt{2\pi}\sigma}\Bigg\lbrack\int\limits_{-\infty}^{\infty}z\sigma^2\exp\Big\{-\frac{1}{2}z^2\Big\}dz +\int\limits_{-\infty}^{\infty}\mu\sigma\exp\Big\{-\frac{1}{2}z^2\Big\}dz\Bigg\rbrack\\ &= \frac{1}{\sqrt{2\pi}\sigma}\Bigg\lbrack-\sigma^2\exp\Big\{-\frac{1}{2}z^2\Big\}\Bigg|_{-\infty}^{\infty} +\int\limits_{-\infty}^{\infty}\mu\sigma\exp\Big\{-\frac{1}{2}z^2\Big\}dz\Bigg\rbrack\\ &= \frac{1}{\sqrt{2\pi}\sigma}\Bigg\lbrack-0+0+\int\limits_{-\infty}^{\infty}\mu\sigma\exp\Big\{-\frac{1}{2}z^2\Big\}dz\Bigg\rbrack\\ &= \frac{1}{\sqrt{2\pi}\sigma}\Bigg\lbrack\mu\sigma\int\limits_{-\infty}^{\infty}\exp\Big\{-\frac{1}{2}z^2\Big\}dz\Bigg\rbrack \\ ^{[2]} &= \frac{1}{\sqrt{2\pi}\sigma}\Bigg\lbrack2\mu\sigma\int\limits_{0}^{\infty}\exp\Big\{-\frac{1}{2}z^2\Big\}dz\Bigg\rbrack \\ &= \frac{2\mu\sigma}{\sqrt{2\pi}\sigma}\Bigg\lbrack\int\limits_{0}^{\infty}\exp\Big\{-\frac{1}{2}z^2\Big\}dz\Bigg\rbrack \\ ^{[3]} &= \frac{2\mu}{\sqrt{2\pi}}\int\limits_{0}^{\infty}\frac{1}{2}u^{-\frac{1}{2}}e^{-\frac{u}{2}}du\\ &= \frac{2\mu}{2\sqrt{2\pi}}\int\limits_{0}^{\infty}u^{-\frac{1}{2}}e^{-\frac{u}{2}}du \\ &= \frac{\mu}{\sqrt{2\pi}}\int\limits_{0}^{\infty}u^{\frac{1}{2}-1}e^{-\frac{u}{2}}du\\ ^{[4]} &= \frac{\mu}{\sqrt{2\pi}}2^{\frac{1}{2}}\Gamma\Big(\frac{1}{2}\Big) \\ ^{[5]} &= \frac{\mu\sqrt{2\pi}}{\sqrt{2\pi}} \\ &= \mu \end{aligned}

1. Let $$z=\frac{x-\mu}{\sigma}\\ \Rightarrow x=z\sigma+\mu \Rightarrow dx=\sigma dz$$
2. This change can be made because the function being integrated is an even function. See Theorem
3. Let $$u=z^2 \Rightarrow z=u^{\frac{1}{2}} \Rightarrow dz=\frac{1}{2}u^{-\frac{1}{2}}$$
4. $$\int\limits_{0}^{\infty}x^{\alpha-1}e^{-\frac{x}{\beta}}dx =\beta^\alpha\Gamma(\alpha)$$
5. $$\Gamma(\frac{1}{2})=\sqrt{\pi}$$

\begin{aligned} E(X^2) &= \int\limits_{-\infty}^{\infty}x^2\frac{1}{\sqrt{2\pi}\sigma} \exp\Big\{\frac{1}{2}\Big(\frac{x-\mu}{\sigma}\Big)^2\Big\}dx \\ ^{[1]} &= \int\limits_{-\infty}^{\infty}\frac{(z\sigma+\mu)^2}{\sqrt{2\pi}\sigma} \exp\Big\{\frac{z^2}{2}\Big\}\sigma dz \\ &= \int\limits_{-\infty}^{\infty}\frac{(z\sigma+\mu)^2}{\sqrt{2\pi}} \exp\Big\{\frac{z^2}{2}\Big\}dz \\ &= \int\limits_{-\infty}^{\infty}\frac{z^2\sigma^2+2z\sigma\mu+\mu^2}{\sqrt{2\pi}} \exp\Big\{\frac{z^2}{2}\Big\}dz \\ &= \int\limits_{-\infty}^{\infty}\Bigg\lbrack \frac{z^2\sigma^2}{\sqrt{2\pi}}\exp\Big\{\frac{z^2}{2}\Big\} +\frac{2z\sigma\mu}{\sqrt{2\pi}}\exp\Big\{\frac{z^2}{2}\Big\} +\frac{\mu^2}{\sqrt{2\pi}}\exp\Big\{\frac{z^2}{2}\Big\} \Bigg\rbrack dz \\ &= \int\limits_{-\infty}^{\infty}\frac{z^2\sigma^2}{\sqrt{2\pi}}\exp\Big\{\frac{z^2}{2}\Big\}dz +\int\limits_{-\infty}^{\infty}\frac{2z\sigma\mu}{\sqrt{2\pi}}\exp\Big\{\frac{z^2}{2}\Big\}dz +\int\limits_{-\infty}^{\infty}\frac{\mu^2}{\sqrt{2\pi}}\exp\Big\{\frac{z^2}{2}\Big\}dz\\ &= \sigma^2\int\limits_{-\infty}^{\infty}\frac{z^2}{\sqrt{2\pi}}\exp\Big\{\frac{z^2}{2}\Big\}dz +2\sigma\mu\int\limits_{-\infty}^{\infty}\frac{z}{\sqrt{2\pi}}\exp\Big\{\frac{z^2}{2}\Big\}dz +\mu^2\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\exp\Big\{\frac{z^2}{2}\Big\}dz \\ ^{[2]} &= \sigma^2\int\limits_{-\infty}^{\infty}\frac{z^2}{\sqrt{2\pi}}\exp\Big\{\frac{z^2}{2}\Big\}dz +2\sigma\mu\int\limits_{-\infty}^{\infty}\frac{z}{\sqrt{2\pi}}\exp\Big\{\frac{z^2}{2}\Big\}dz +\mu^2\cdot 1 \\ ^{[3]} &= \sigma^2\int\limits_{-\infty}^{\infty}\frac{z^2}{\sqrt{2\pi}}\exp\Big\{\frac{z^2}{2}\Big\}dz +2\sigma\mu E(Z)+\mu^2 \\ &= \sigma^2\int\limits_{-\infty}^{\infty}\frac{z^2}{\sqrt{2\pi}}\exp\Big\{\frac{z^2}{2}\Big\}dz +2\sigma\mu\cdot0+\mu^2 \\ ^{[4]} &= \sigma^2\Bigg(\frac{z}{\sqrt{2\pi}\cdot-\exp\Big\{-\frac{z^2}{2}\Big\}} -\int\limits_{-\infty}^{\infty}-\exp\Big\{-\frac{z^2}{2}\frac{1}{\sqrt{2\pi}}dz\Bigg)+\mu^2 \\ &= \sigma^2\Bigg(-0+0+\int\limits_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}\exp\Big\{-\frac{z^2}{2}\Big\}dz\Bigg)+\mu^2 \\ ^{[5]} &= \sigma^2(1)+\mu^2 \\ &=\sigma^2+\mu^2 \end{aligned}

1. $$z=\frac{x-\mu}{\sigma} \\ \Rightarrow x=z\sigma+\mu\Rightarrow dx=\sigma dz$$
2. Theorems and with $$\mu=0$$ and $$\sigma^2=1$$
3. Using the previous result with $$\mu=0$$ and $$\sigma^2=1$$, $$E(Z)=0$$.
4. Integration by parts: $$\int\limits_{a}^{b}u\ dv =\lbrack u\cdot v\rbrack_a^b-\int\limits_{a}^{b}v\ du$$ with $$u=\frac{z}{\sqrt{2\pi}}\Rightarrow du=\frac{1}{\sqrt{2\pi}}dz$$\ and $$dv=z\exp\{-\frac{z^2}{2}\}\Rightarrow v=-\exp\{-\frac{z^2}{2}\}$$
5. See footnote 2.

\begin{aligned} \mu &= E(X) \\ &= \mu \\ \\ \\ \sigma^2 &= E(X^2)-E(X)^2 \\ &= \sigma^2+\mu^2-\mu^2 \\ &= \sigma^2 \end{aligned}

## 30.4 Moment Generating Function

\begin{aligned} E(e^{tX}) &= \int\limits_{-\infty}^{\infty}\exp\{tx\}\frac{1}{\sqrt{2\pi}\sigma} \exp\Big\{\frac{1}{2\sigma^2}-(x-\mu)^2\Big\}dx \\ &= \int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma} \exp\Big\{tx-\frac{1}{2}\Big(\frac{x-\mu}{\sigma}\Big)^2\Big\}dx \\ ^{[1]} &= \int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma} \exp\Big\{t(z\cdot\sigma+\mu-\frac{1}{2}z^2\Big\}\sigma dz \\ &= \int\limits_{-\infty}^{\infty}\frac{\sigma}{\sqrt{2\pi}\sigma} \exp\Big\{t(z\cdot\sigma+\mu)-\frac{1}{2}z^2\Big\}dz \\ &= \int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma} \exp\{\mu t\}\exp\Big\{z\sigma t-\frac{1}{2}z^2\Big\}dz \\ &= \exp\{\mu t\}\int\limits_{-\infty}^{\infty}\frac{\sigma}{\sqrt{2\pi}\sigma} \exp\Big\{z\sigma t-\frac{1}{2}z^2\Big\}dz \\ &= \exp\{\mu t\}\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} \exp\Big\{z\sigma t-\frac{1}{2}z^2-\frac{1}{2}t^2\sigma^2+\frac{1}{2}t^2\sigma^2\Big\}dz \\ &= \exp\{\mu t\}\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} \exp\Big\{-\frac{1}{2}z^2+z\sigma t-\frac{1}{2}t^2\sigma^2\Big\} \exp\Big\{\frac{1}{2}t^2\sigma^2\Big\}dz \\ &= \exp\{\mu t\}\exp\Big\{\frac{1}{2}t^2\sigma^2\Big\}\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} \exp\Big\{z^2-2z\sigma t+t^2\sigma^2\Big\}dz \\ &= \exp\Big\{\mu t+\frac{t^2\sigma^2}{2}\Big\}\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} \exp\Big\{\frac{1}{2}(z-\sigma t)^2\Big\}dz \\ ^{[2]} &= \exp\Big\{\mu t+\frac{t^2\sigma^2}{2}\Big\}\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} \exp\Big\{\frac{1}{2}(u)^2\Big\}du \\ &= \exp\Big\{\mu t+\frac{t^2\sigma^2}{2}\Big\}\cdot 1 \\ &= \exp\Big\{\mu t+\frac{t^2\sigma^2}{2}\Big\} \end{aligned}

1. $$z=\frac{x-\mu}{\sigma}\\ \Rightarrow x=z\cdot\sigma+\mu\Rightarrow dx=\sigma dz$$
2. $$u=z-\sigma t\\ \Rightarrow z=u+\sigma t \Rightarrow dz=du$$
3. See Theorems and

\begin{aligned} M_X^{(1)}(t) &= \exp\Big\{\mu t+\frac{t^2\sigma^2}{2}\Big\}\cdot(\mu+t\sigma^2) \\ \\ \\ M_X^{(2)}(t) &= \exp\Big\{\mu t+\frac{t^2\sigma^2}{2}\Big\}\cdot\sigma^2 +(\mu+t\sigma^2)\exp\Big\{\mu t+\frac{t^2\sigma^2}{2}\Big\}(\mu+t\sigma^2) \\ &= M_X^{(2)}(t)=\exp\Big\{\mu t+\frac{t^2\sigma^2}{2}\Big\}\cdot\sigma^2 +(\mu+t\sigma^2)^2\exp\Big\{\mu t+\frac{t^2\sigma^2}{2}\Big\} \\ \\ \\ E(X) &= M_X^{(1)}(0)=\exp\Big\{\mu\cdot0+\frac{0^2\sigma^2}{2}\Big\}\cdot(\mu+0\cdot\sigma) \\ &= e^0\cdot(\mu+0) \\ &= 1\cdot\mu \\ &= \mu \\ \\ \\ E(X^2) &= M_X^{(2)}(0) \\ &= \exp\Big\{\mu\cdot0+\frac{0^2\sigma^2}{2}\Big\}\cdot\sigma^2 +(\mu+0\cdot \sigma^2)^2\exp\Big\{\mu\cdot0+\frac{0^2\sigma^2}{2}\Big\} \\ &= e^0\sigma^2+(\mu+0)^2e^0 \\ &= 1\cdot\sigma^2+\mu^2\cdot1 \\ &= \sigma^2+\mu^2 \\ \\ \\ \mu &= E(X) \\ &= \mu \\ \\ \\ \sigma^2 &= E(X^2)-E(X) \\ &= \sigma^2+\mu^2-\mu^2 \\ &= \sigma^2 \end{aligned}

## 30.5 Maximum Likelihood Estimators

Let $$x_1,x_2,\ldots,x_n$$ be a random sample from a Normal distribution with mean $$\mu$$ and variance $$\sigma^2$$.

### 30.5.1 Likelihood Function

\begin{aligned} L(x_1,x_2,\ldots,x_n|\theta) &= f(x_1|\theta)f(x_2|\theta)\cdots f(x_n|\theta) \\ &= \frac{1}{\sqrt{2\pi}\sigma}\exp\bigg\{\frac{-(x_1-\mu)^2}{2\sigma^2}\bigg\} \frac{1}{\sqrt{2\pi}\sigma}\exp\bigg\{\frac{-(x_2-\mu)^2}{2\sigma^2}\bigg\} \cdots \frac{1}{\sqrt{2\pi}\sigma}\exp\bigg\{\frac{-(x_n-\mu)^2}{2\sigma^2}\bigg\} \\ &= \prod\limits_{i=1}^{n}\frac{1}{\sqrt{2\pi}\sigma}\exp\bigg\{\frac{-(x_i-\mu)^2}{2\sigma^2}\bigg\} \\ &= \bigg(\frac{1}{\sqrt{2\pi}\sigma}\bigg)^n \prod\limits_{i=1}^{n}\exp\bigg\{\frac{-(x_i-\mu)^2}{2\sigma^2} \bigg\} \\ &= \bigg(\frac{1}{\sqrt{2\pi}\sigma}\bigg)^n \exp\bigg\{ \sum\limits_{i=1}^{n}\frac{-(x_i-\mu)^2}{2\sigma^2} \bigg\} \\ &= \bigg(\frac{1}{\sqrt{2\pi}\sigma}\bigg)^n \exp\bigg\{ \frac{1}{2\sigma^2} \sum\limits_{i=1}^{n}-(x_i-\mu)^2 \bigg\} \end{aligned}

### 30.5.2 Log-likelihood Function

\begin{aligned} \ell(\theta) &= \ln\big(L(\theta)\big) \\ &= \ln\Bigg(\bigg(\frac{1}{\sqrt{2\pi}\sigma}\bigg)^n \exp\bigg\{ \frac{1}{2\sigma^2} \sum\limits_{i=1}^{n}-(x_i-\mu)^2 \bigg\}\Bigg) \\ &= \ln\bigg(\frac{1}{\sqrt{2\pi}\sigma}\bigg)^n + \ln\bigg(\exp\bigg\{ \frac{1}{2\sigma^2} \sum\limits_{i=1}^{n}-(x_i-\mu)^2 \bigg\}\bigg) \\ &= n\ln\frac{1}{\sqrt{2\pi}\sigma} - \frac{1}{2\sigma^2} \sum\limits_{i=1}^{n}-(x_i-\mu)^2 \\ &= n\ln\frac{1}{\sqrt{2\pi}\sigma} - \frac{1}{2\sigma^2} \sum\limits_{i=1}^{n}(x_i^2 - 2\mu x_i + \mu^2) \\ &= n\ln\frac{1}{\sqrt{2\pi}\sigma} - \frac{1}{2\sigma^2} \bigg\lbrack \sum\limits_{i=1}^{n}x_i^2 - 2\mu\sum\limits_{i=1}^{n}x_i + \sum\limits_{i=1}^{n}\mu^2\bigg\rbrack \\ &= n\ln\frac{1}{\sqrt{2\pi}\sigma} - \frac{1}{2\sigma^2}\sum\limits_{i=1}^{n}x_i^2 + \frac{2\mu}{2\sigma^2}\sum\limits_{i=1}^{n}x_i - \frac{1}{2\sigma^2}n\mu^2 \\ &= n\ln\bigg(\frac{1}{\sqrt{2\pi}}\sigma^{-1}\bigg) - \frac{\sigma^{-2}}{2}\sum\limits_{i=1}^{n}x_i^2 + \mu\sigma^{-2}\sum\limits_{i=1}^{n}x_i - \frac{n\mu^2\sigma^{-2}}{2} \end{aligned}

### 30.5.3 MLE for $$\mu$$

\begin{aligned} \frac{d\ell}{d\mu} &= 0 - 0 + \sigma^{-2}\sum\limits_{i=1}^{n} - \frac{2n\mu\sigma^{-2}}{2} \\ &= \sigma^{-2}\sum\limits_{i=1}^{n} - n\mu\sigma^{-2}\\ \\ \\ 0 &= \sigma^{-2}\sum\limits_{i=1}^{n} - n\mu\sigma^{-2}\\ \Rightarrow n\mu\sigma^{-2} &= \sigma^{-2}\sum\limits_{i=1}^{n}\\ \Rightarrow n\mu\ &= \sum\limits_{i=1}^{n}\\ \Rightarrow \mu &= \frac{1}{n}\sum\limits_{i=1}^{n} \end{aligned}

So $$\hat\mu = \frac{1}{n}\sum\limits_{i=1}^{n}$$ is the maximum likelihood estimator for $$\mu$$.

### 30.5.4 MLE for $$\sigma^2$$

The work in deriving the MLE for $$\sigma^2$$ is greatly reduced if we use

\begin{aligned} \ell(\theta) &= n\ln\frac{1}{\sqrt{2\pi}\sigma} - \frac{1}{2\sigma^2} \sum\limits_{i=1}^{n}-(x_i-\mu)^2\\ &= n\ln\bigg(\frac{1}{\sqrt{2\pi}}\sigma^{-1}\bigg) - \frac{1}{2\sigma^2} \sum\limits_{i=1}^{n}-(x_i-\mu)^2 \\ \\ \\ \frac{d\ell}{d\sigma} &= n\bigg(\frac{1}{\frac{1}{\sqrt{2\pi}}\sigma^{-1}}\bigg) \bigg(\frac{-1}{\sqrt{2\pi}\sigma^{-2}}\bigg) - \frac{-2}{2}\sigma^{-3} \sum\limits_{i=1}^{n}-(x_i-\mu)^2 \\ &= \frac{-n\sqrt{2\pi}\sigma}{\sqrt{2\pi}\sigma^2} + \frac{1}{\sigma^3} \sum\limits_{i=1}^{n}-(x_i-\mu)^2\\ &= \frac{-n}{\sigma} + \frac{1}{\sigma} \sum\limits_{i=1}^{n}-(x_i-\mu)^2 \\ \\ \\ 0 &= \frac{-n}{\sigma} + \frac{1}{\sigma} \sum\limits_{i=1}^{n}-(x_i-\mu)^2 \\ \Rightarrow \frac{n}{\sigma} &= \frac{1}{\sigma} \sum\limits_{i=1}^{n}-(x_i-\mu)^2 \\ \Rightarrow \frac{\sigma^3}{\sigma} &= \frac{1}{n}\sum\limits_{i=1}^{n}-(x_i-\mu)^2 \\ \Rightarrow \sigma^2 &= \frac{1}{n}\sum\limits_{i=1}^{n}-(x_i-\mu)^2 \end{aligned}

So $$\hat\sigma^2 = \frac{1}{n}\sum\limits_{i=1}^{n}-(x_i-\mu)^2$$ is the maximum likelihood estimator for $$\sigma^2$$. Notice however that this $$MLE$$ is a biased estimator.

## 30.6 Theorems for the Normal Distribution

### 30.6.1 Validity of the Distribution (Polar Coordinates)

$\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^\frac{-(x-\mu)^2}{2\sigma^2}dx=1$

Proof:

Let $$A = \int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^\frac{-(x-\mu)^2}{2\sigma^2}dx$$.

By using the identity transformation $$y=x$$, we may also write

\begin{aligned} A &= \int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^\frac{-(y-\mu)^2}{2\sigma^2}dy \\ \Rightarrow A^2 &= \int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma} e^\frac{-(x-\mu)^2}{2\sigma^2}dx\cdot \int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma} e^\frac{-(y-\mu)^2}{2\sigma^2}dy \\ &= \frac{1}{2\pi\sigma^2}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} e^\frac{-(x-\mu)^2}{2\sigma^2}e^\frac{-(y-\mu)^2}{2\sigma^2}dx\ dy \\ &= \frac{1}{2\pi\sigma^2}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} e^{\frac{-(x-\mu)^2}{2\sigma^2}-\frac{(y-\mu)^2}{2\sigma^2}} \\ &= \frac{1}{2\pi\sigma^2}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} e^{-\frac{1}{2\sigma^2}[(x-\mu)^2+(y-\mu)^2]}dx\ dy \\ ^{[1]} &= \frac{1}{2\pi\sigma^2}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} e^{-\frac{1}{2\sigma^2}t^2+u^2}dt\ du \\ ^{[2]} &= \frac{1}{2\pi\sigma^2}\int\limits_{0}^{2\pi}\int\limits_{0}^{\infty} e^{-\frac{r^2}{2\sigma^2}}r\ dr\ d\theta \\ &=\frac{1}{2\pi\sigma^2}\int\limits_{0}^{2\pi} -\sigma^2e^{-\frac{r^2}{2\sigma^2}}\Big|_0^\infty d\theta \\ &=\frac{1}{2\pi\sigma^2}\int\limits_{0}^{2\pi}0+\sigma^2\ d\theta \\ &=\frac{1}{2\pi\sigma^2}\int\limits_{0}^{2\pi}\sigma^2\ d\theta \\ &=\frac{1}{2\pi\sigma^2}\cdot \Big(\sigma^2\theta\Big|_0^{2\pi}\Big) \\ &=\frac{1}{2\pi\sigma^2}[2\pi\sigma^2-0] \\ &=\frac{2\pi\sigma^2}{2\pi\sigma^2} \\ &=1\\ \Rightarrow A^2 &= 1 \\ \Rightarrow A &= 1 \end{aligned}

1. Substitute $$t=x-\mu$$ and $$u=y-\mu$$
2. $$t=r\cos\theta$$ and $$u=r\sin\theta$$.\ Thus $$t^2+u^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2(\cos^2\theta+\sin^2\theta)=r^2$$

Thus $$\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^\frac{-(x-\mu)^2}{2\sigma^2}dx=1$$

### 30.6.2 Validity of the Distribution (Gamma Function)

$\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^\frac{-(x-\mu)^2}{2\sigma^2}dx=1$

Proof:

\begin{aligned} \int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^\frac{-(x-\mu)^2}{2\sigma^2}dx \\ &=\frac{1}{\sqrt{2\pi}\sigma}\int\limits_{-\infty}^{\infty}e^\frac{-(x-\mu)^2}{2\sigma^2}dx \\ ^{[1]} &= \frac{1}{\sqrt{2\pi}\sigma} \int\limits_{-\infty}^{\infty}e^\frac{-(y)^2}{2\sigma^2}dy\\ &=\frac{1}{\sqrt{2\pi}\sigma} \int\limits_{-\infty}^{\infty}e^{-\frac{1}{2}\frac{y^2}{\sigma^2}}dy \\ ^{[2]} &= \frac{1}{\sqrt{2\pi}\sigma} \int\limits_{-\infty}^{\infty} e^{-\frac{u}{2}}\frac{1}{2}\sigma u^{-\frac{1}{2}}du \\ &=\frac{\sigma}{2\sqrt{2\pi}\sigma} \int\limits_{-\infty}^{\infty}u^{-\frac{1}{2}}e^{-\frac{u}{2}}du \\ ^{[3]} &= \frac{2\sigma}{2\sqrt{2\pi}\sigma} \int\limits_{0}^{\infty}u^{\frac{1}{2}-1}e^{-\frac{u}{2}}du\\ ^{[4]} &= \frac{1}{\sqrt{2\pi}}2^\frac{1}{2}\Gamma\big(\frac{1}{2}\big) \\ ^{[5]} &= \frac{\sqrt{2\pi}}{\sqrt{2\pi}} \\ &= 1 \end{aligned}

1. $$y=x-\mu\ \Rightarrow x=y+\mu\ \Rightarrow dx=dy$$
2. $$u=(\frac{y}{\sigma})^2\ \Rightarrow y=\sigma u^\frac{1}{2}\ \Rightarrow dy=\frac{1}{2}\sigma u^{-\frac{1}{2}}$$
3. If $$f(x)$$ is an even function then $$\int\limits_{-\infty}^{\infty}f(x)dx =2\int\limits_{0}^{\infty}f(x)dx$$ (Theorem ).
4. $$\int\limits_{0}^{\infty}x^{\alpha-1}e^{-\frac{x}{\beta}}dx =\beta^\alpha\Gamma(\alpha)$$
5. $$\Gamma(\frac{1}{2})=\sqrt{\pi}$$

### 30.6.3 Multiple of a Normal Random Variable

Let $$X$$ be a Normal random variable with parameters $$\mu$$ and $$\sigma^2$$, and let $$c$$ be a constant. If $$Y=cX$$, then $$Y\sim Normal(c\mu,\ c^2\sigma^2)$$.

Proof:

\begin{aligned} M_Y(t) &= E(e^{tY}) \\ &= E(e^{tcX}) \\ &= \exp\Big\{\mu tc+\frac{t^2c^2\sigma^2}{2}\Big\} \\ &= \exp\Big\{c\mu t+\frac{t^2c^2\sigma^2}{2}\Big\} \end{aligned}

Which is the Moment Generating Function of a Normal random variable with mean $$c\mu$$ and variance $$c^2\sigma^2$$. Thus $$Y\sim Normal(c\mu,\ c^2\sigma^2)$$.

### 30.6.4 Weighted Sum of Normal Random Variables

Let $$X_1,X_2,\ldots,X_n$$ be independent random variables from Normal distributions with parameters $$\mu_i$$ and $$\sigma_i^2$$, i.e. $$X_i\sim Normal(\mu_i,\sigma_i^2),\ i=1,2,\ldots,n$$, and let $$a_1,a_2,\ldots,a_n$$ be constants.

If $$Y=\sum\limits_{i=1}^{n}a_iX_i$$, then $$Y\sim Normal\big(\sum\limits_{i=1}^{n}a_i\mu_i,\ \sum\limits_{i=1}^{n}a_i^2\sigma_i^2\big)$$.

Proof:

\begin{aligned} M_Y(t) &=E(e^{tY}) \\ &=E(e^{t(a_1X_1+a_2X_2+\cdots+a_nX_n}) \\ &=E(e^{ta_1X_1}e^{ta_2X_2}\cdots e^{ta_nX_n}) \\ &=E(e^{ta_1X_1})E(e^{ta_2X_2})\cdots E(e^{ta_nX_n}) \\ &=\exp\Big\{ta_1\mu_1+\frac{ta_1^2\sigma_1^2}{2}\Big\}\cdot \exp\Big\{ta_2\mu_2+\frac{ta_2^2\sigma_2^2}{2}\Big\} \cdot \cdots \cdot \exp\Big\{ta_n\mu_n+\frac{ta_n^2\sigma_n^2}{2}\Big\} \\ &=\exp\Big\{ta_1\mu_1+\frac{ta_1^2\sigma_1^2}{2}+ta_2\mu_2+\frac{ta_2^2\sigma_2^2}{2} +\cdots+ta_n\mu_n+\frac{ta_n^2\sigma_n^2}{2}\Big\} \\ &=\exp\Big\{t\sum\limits_{i=1}^{n}a_i\mu_i + \frac{t\sum\limits_{i=1}^{n}a_i^2\sigma_i^2}{2}\Big\} \end{aligned}

Which is the mgf of a Normal random variable with parameters $$\sum\limits_{i=1}^{n}a_i\mu_i$$ and $$\sum\limits_{i=1}^{n}a_i^2\sigma_i^2$$. Thus $$Y\sim Normal\big(\sum\limits_{i=1}^{n}a_i\mu_i,\ \sum\limits_{i=1}^{n}a_i^2\sigma_i^2\big)$$.

### 30.6.5 Sum of Normal Random Variables

Let $$X_1,X_2,\ldots,X_n$$ be independent random variables from Normal distributions with parameters $$\mu_i$$ and $$\sigma_i^2$$, i.e. $$X_i\sim Normal(\mu_i, \sigma_i^2),\ i=1,2,\ldots,n$$. Let $$Y=\sum\limits_{i=1}^{n}X_i$$. Then $$Y\sim Normal\big(\sum\limits_{i=1}^{n}\mu_i,\ \sum\limits_{i=1}^{n}\sigma_i^2\big)$$.

Proof:

$$Y=\sum\limits_{i=1}^{n}X_i$$ is a special case of the Weighted Sum of Normal Random Variables where each of the weights is equal to 1. It follows then that $$Y\sim Normal\big(\sum\limits_{i=1}^{n}1\mu_i,\ \sum\limits_{i=1}^{n}1^2\sigma_i^2\big)$$ which is equivalent to saying $$Y\sim Normal\big(\sum\limits_{i=1}^{n}\mu_i,\ \sum\limits_{i=1}^{n}\sigma_i^2\big)$$.

### 30.6.6 Standard Normal Transformation

Let $$X$$ be a Normal random variable with mean $$\mu$$ and variance $$\sigma^2$$. That is, $$X\sim$$Normal($$\mu,\sigma^2$$). If $$Z=\frac{X-\mu}{\sigma}$$, then $$Z\sim$$Normal(0,1), and we say $$Z$$ has a Standard Normal distribution.

Furthermore, if $$Z\sim$$Normal(0,1), and $$X=Z\cdot\sigma+\mu$$, then $$X\sim Normal(\mu,\sigma^2$$).

Proof:

First, since $$X\sim Normal(\mu,\sigma^2$$) and $$Z=\frac{X-\mu}{\sigma}$$

\begin{aligned} E(e^{tZ}) &=E\Big(\exp\Big\{t\frac{x-\mu}{\sigma}\Big\}\Big) \\ &=E\Big(\exp\Big\{\frac{tx-t\mu}{\sigma}\Big\}\Big) \\ &=E\Big(\exp\Big\{\frac{tx}{\sigma}-\frac{t\mu}{\sigma}\Big\}\Big) \\ &=E\Big(\exp\Big\{\frac{tx}{\sigma}\Big\}\exp\Big\{-\frac{t\mu}{\sigma}\Big\}\Big) \\ &=\exp\Big\{\frac{-t\mu}{\sigma}\Big\}E\Big(\exp\Big\{\frac{t}{\sigma}x\Big\} \\ &=\exp\Big\{\frac{-t\mu}{\sigma}\Big\}\exp\Big\{\frac{t\mu}{\sigma}+\frac{t^2\sigma^2}{2\sigma^2}\Big\} \\ &=\exp\Big\{\frac{-t\mu}{\sigma}-\frac{t\mu}{\sigma}+\frac{t^2\sigma^2}{2\sigma^2}\Big\} \\ &=\exp\Big\{\frac{t^2\sigma^2}{2\sigma^2}\Big\} \end{aligned}

Which is the Moment Generating Function of a Normal random variable with $$\mu=0$$ and $$\sigma^2=1$$. Thus $$Z\sim Normal(0,1)$$.

To complete the proof, we start with $$Z\sim$$Normal(0,1) and let $$X=Z\cdot\sigma+\mu$$.

\begin{aligned} E(e^{tX}) = E(e^{t(z\sigma+\mu)}) \\ = E(e^{tz\sigma}e^{t\mu}) \\ = e^{t\mu}E(e^{t\sigma z}) \\ = \exp\{t\mu\}\exp\Big\{\frac{t^2\sigma^2}{2}\Big\} \\ = \exp\Big\{\mu t+\frac{t^2\sigma^2}{2}\Big\} \end{aligned}

Which is the Moment Generating Funciton of a Normal random variable with mean $$\mu$$ and variance $$\sigma^2$$. Thus $$X\sim Normal(\mu,\sigma^2)$$.

### 30.6.7 Distribution of $$\bar X$$

Let $$X_1,X_2,\ldots,X_n$$ be independent and identically distributed random variables from a Normal distribution with parameters $$\mu$$ and $$\sigma^2$$, i.e. $$X_i\sim Normal(\mu,\ \sigma^2)$$. Let $$\bar X=\frac{1}{n}\sum\limits_{i=1}^{n}X_i$$. Then $$\bar X\sim Normal\big(\mu,\ \frac{\sigma^2}{n}\big)$$.

Proof:

$$\bar X=\frac{1}{n}\sum\limits_{i=1}^{n}X_i$$ is a special case of the Weighted Sum of Normal Random Variables where $$a_i=\frac{1}{n},\ i=1,2,\ldots,n$$. It follows then that

\begin{aligned} M_{\bar X}(t) &=exp\Big(t\sum\limits_{i=1}^{n}\frac{1}{n}\mu +t\frac{\sum\limits_{i=1}^{n}\frac{1}{n}\sigma^2}{2}\Big) \\ &=exp\big(t\frac{n\mu}{n}+\frac{t\frac{n\sigma^2}{n}}{2}\big) \\ &=exp\big(t\mu+\frac{t\frac{\sigma^2}{n}}{2}\big) \end{aligned}

Which is the mgf of a Normal random variable with parameters $$\mu$$ and $$\frac{\sigma^2}{n}$$. Thus $$\bar X\sim Normal\big(\mu,\ \frac{\sigma^2}{n}\big)$$.

### 30.6.8 Square of a Standard Normal Random Variable

If $$Z\sim N(0,1)$$, then $$Z^2\sim\chi^2(1)$$.

Proof: \begin{aligned} M_{Z^2}(t) &= E(e^{tZ^2}) \\ &= \int\limits_{-\infty}^{\infty}e^{tz^2}\frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}}dz \\ &= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}e^{tz^2} e^{-\frac{z^2}{2}}dz \\ &= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty} e^{-\frac{z^2}{2}(-2t+1)}dz \\ &= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty} e^{-\frac{z^2}{2}(1-2t)}dz \\ ^{[1]} &= \frac{2}{\sqrt{2\pi}}\int\limits_{0}^{\infty} e^{-\frac{z^2}{2}(1-2t)}dz \\ ^{[2]} &= \frac{2}{\sqrt{2\pi}}\int\limits_{0}^{\infty}e^{-u} \frac{\sqrt{2}u^{-\frac{1}{2}}}{2(1-2t)^{\frac{1}{2}}}du \\ &= \frac{2\sqrt{2}}{2\sqrt{2\pi}(1-2t)^{\frac{1}{2}}} \int\limits_{0}^{\infty}e^{-u}u^{-\frac{1}{2}}du \\ &= \frac{2\sqrt{2}}{2\sqrt{2\pi}(1-2t)^{\frac{1}{2}}} \int\limits_{0}^{\infty}u^{\frac{1}{2}-1}e^{-u}du \\ ^{[3]} &= \frac{1}{\sqrt{\pi}(1-2t)^{\frac{1}{2}}}\Gamma(\frac{1}{2}) \\ &= \frac{\sqrt{\pi}}{\sqrt{\pi}(1-2t)^{\frac{1}{2}}} \\ &= \frac{1}{(1-2t)^{\frac{1}{2}}}=(1-2t)^{-\frac{1}{2}} \\ \end{aligned}

1. $$\int\limits_{-\infty}^{\infty}f(x)dx = 2\int\limits_{0}^{\infty}f(x)dx$$ when f(x) is an even function ()
2. Let $$u=\frac{z^2}{2}(1-2t) \ \ \ \ \Rightarrow z=\frac{\sqrt{2}u^{\frac{1}{2}}}{(1-2t)^{\frac{1}{2}}}$$
So $$dz=\frac{\sqrt{2}u^{-\frac{1}{2}}} {2(1-2t)^{\frac{1}{2}}}$$
3. $$\int\limits_{0}^{\infty}x^{\alpha-1}e^{-\frac{x}{\beta}}dx =\beta^\alpha\Gamma(\alpha)$$

Which is the mgf of a Chi-Square random variable with 1 degree of freedom. Thus $$Z^2\sim\chi^2(1)$$.