Normal Distribution

$\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\sigma^2}}\exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)dx = 1.$

After centering and scaling, the integrand becomes $e^{-z^2/2}/\sqrt{2\pi}$. The classical trick squares the integral and computes the resulting double integral in polar coordinates.

The factor $\sqrt{2\pi\sigma^2}$ is not optional. Drop it and the density does not normalize, every downstream probability and expectation is wrong by a constant, and the log-likelihood that drives MLE for the Normal is off by the same constant. The constant matters for likelihood ratios across $\sigma^2$ values but cancels for likelihood ratios at fixed $\sigma^2$.

The polar-coordinate trick uses the rotational symmetry of $e^{-(z_1^2+z_2^2)/2}$. No analogous trick works for non-Gaussian densities. Substituting $z = (x-\mu)/\sigma$ requires $\sigma>0$; the degenerate $\sigma = 0$ case is a point mass at $\mu$, not a Normal.

For $X\sim\mathcal{N}(\mu,\sigma^2)$ and every $s\in\mathbb{R}$, $M_X(s) = \mathbb{E}[e^{sX}] = \exp\!\left(\mu s + \frac{\sigma^2 s^2}{2}\right).$

The MGF is a quadratic in $s$ in the log scale, with linear coefficient $\mu$ and quadratic coefficient $\sigma^2/2$. The log-MGF being quadratic is the defining property of the Normal family: any random variable whose log-MGF is quadratic is Normal.

The mean and variance read directly off the MGF: $\mathbb{E}[X] = M_X'(0) = \mu$ and $\operatorname{Var}(X) = M_X''(0) - M_X'(0)^2 = \sigma^2$. The MGF is finite for every real $s$, which is stronger than the polynomial-moment condition; it forces sub-Gaussian tails. See sub-Gaussian random variables for the corresponding tail bound.

The MGF is the exponential of a quadratic only for the Normal. If you compute the MGF of a sample and find a quadratic log-MGF you have empirical evidence for Normality, but a finite-sample MGF estimate is noisy. The MGF tool is for identification of distributions from algebraic forms, not for goodness-of-fit testing; use goodness-of-fit tests for that.

1. Affine. If $X\sim\mathcal{N}(\mu,\sigma^2)$ and $a,b\in\mathbb{R}$ with $a\ne 0$, then $aX + b \sim \mathcal{N}(a\mu + b,\, a^2\sigma^2).$
2. Sum of independents. If $X\sim\mathcal{N}(\mu_X,\sigma_X^2)$ and $Y\sim\mathcal{N}(\mu_Y,\sigma_Y^2)$ are independent, then $X + Y \sim \mathcal{N}(\mu_X+\mu_Y,\, \sigma_X^2+\sigma_Y^2).$

Closure under affine maps is just a change of location and scale on the density and follows from the change-of-variables formula. Closure under independent sums is the MGF argument: by independence the MGFs multiply, and the product of two Normal MGFs is itself a Normal MGF.

Closure is what makes the sample mean of an i.i.d. Normal sample explicit: $\bar X_n = (1/n)\sum X_i \sim \mathcal{N}(\mu,\sigma^2/n)$, with no asymptotic approximation needed. Closure also drives the central limit theorem heuristic: averages of independent random variables look approximately Normal even when the summands are not Normal, because the family is closed under the operation that defines averaging.

Independence in the sum result is essential. $X + X = 2X \sim \mathcal{N}(2\mu, 4\sigma^2)$, not $\mathcal{N}(2\mu, 2\sigma^2)$. Sums of dependent Normals are still Normal if the joint law is multivariate Normal, but the variance is $\sigma_X^2 + \sigma_Y^2 + 2\operatorname{Cov}(X,Y)$, not the independent-sum variance.

Given an i.i.d. sample $X_1,\dots,X_n$ from $\mathcal{N}(\mu,\sigma^2)$, the maximum likelihood estimators are $\hat\mu = \bar X_n = \frac{1}{n}\sum_{i=1}^n X_i,\qquad \hat\sigma^2 = \frac{1}{n}\sum_{i=1}^n (X_i - \bar X_n)^2.$

The log-likelihood is quadratic in $\mu$ and concave in $\sigma^2$, so the score equations have a single explicit solution. The MLE for $\mu$ is the sample mean; the MLE for $\sigma^2$ divides the sum of squared deviations by $n$, not $n-1$.

The unbiased estimator of $\sigma^2$ is $S^2 = (1/(n-1))\sum(X_i-\bar X_n)^2$, which differs from $\hat\sigma^2$ by the factor $n/(n-1)$. Both are consistent. The MLE is biased downward by a factor $n/(n-1)$; that bias vanishes as $n\to\infty$. This is the simplest example of a finite-sample bias of an MLE that disappears asymptotically.

The MLE requires $n\ge 2$ to estimate $\sigma^2$ meaningfully; for $n=1$ the estimator $\hat\sigma^2 = 0$ is degenerate. The formula assumes the sample is i.i.d.; correlated Normal samples require the joint Normal log-likelihood and a covariance estimator.

Let $X_1,\dots,X_n$ be i.i.d. $\mathcal{N}(\mu,\sigma^2)$, $\bar X_n = (1/n)\sum X_i$, and $S^2 = (1/(n-1))\sum (X_i-\bar X_n)^2$. Then

1. $\bar X_n \sim \mathcal{N}(\mu, \sigma^2/n)$.
2. $(n-1)S^2/\sigma^2 \sim \chi^2_{n-1}$.
3. $\bar X_n$ and $S^2$ are independent.

Decompose the sample vector $(X_1,\dots,X_n)$ into its projection onto the all-ones direction (which carries $\bar X_n$) and the orthogonal complement (which carries the deviations $X_i-\bar X_n$). For i.i.d. Normal data the two projections are independent Normals, and the squared norm of an orthogonal Normal projection is a Chi-squared.

This three-part statement is the engine behind almost every classical inference for Normal samples. It identifies the law of the sample mean, the law of the sample variance (as a scaled Chi-squared), and their independence, which is what makes the t-statistic $(\bar X_n - \mu)/(S/\sqrt n)$ a Student-t random variable rather than an arbitrary ratio of dependent random variables. See Student-t distribution and t-test for the consequence.

The independence of sample mean and sample variance is a special property of the Normal distribution. For non-Normal i.i.d. samples the sample mean and sample variance are not independent in finite samples; they only become asymptotically uncorrelated. Using the Student-t exact distribution outside of Normal samples replaces an exact statement with an approximation whose accuracy depends on tail weight and sample size.