Normal distribution ($N(\mu ,{\sigma }^2)$)

The normal distribution has density proportional to $\exp (-(x-\mu {)}^2/(2{\sigma }^2))$.

Here take $X,Y$ iid, centered with variance one, and with finite moment-generating functions at all real arguments. Independence of their sum and difference gives $M(2t)=M(t{)}^{3}M(-t)$. Dividing the positive and negative argument identities shows $M(t)/M(-t)$ satisfies the symmetry step of dyadic rigidity of a moment-generating function, hence $M$ is even. The identity becomes $M(2t)=M(t{)}^4$, and the unit-variance expansion then forces $M(t)=e^{t^2/2}$. The uniqueness theorem for moment-generating functions identifies the standard normal distribution. This is a precise sufficient-hypothesis version; the displayed implication is understood with those stated assumptions.

The standard normal distribution has probability density function $\varphi (z)=(2\pi {)}^{-1/2}{e}^{-z^2/2}$. Its rapid decay explains both concentration at one location and the logarithmic growth of a Gaussian maximum over many locations.

For independent standard normal random variables $Z_1,\dots ,Z_N$, the maximum of $|Z_j|$ has scale $\sqrt{\log N}$. A union bound and the Gaussian tail bound give the upper scale. Integrating the normal density over $[r,r+1/r]$ gives a lower tail bound $c{r}^{-1}{e}^{-r^2/2}$, implying $\mathbb{P}({\max }_j|Z_j|\le \sqrt{\log N})\to 0$.

For $N$ variables with the standard normal distribution, the expected maximum of their absolute values is at most $\sqrt{2\log (2N)}$, even when they are dependent. Bound the exponential of the maximum by the sum of the $2N$ signed exponentials, use the normal moment-generating function, then apply Jensen inequality and minimize over the exponential parameter. The resulting logarithmic growth is useful in simultaneous estimation.

For a centered normal random variable $Z\sim N(0,v)$, $\mathbb{E}{Z}^4=3v^2$ and $\mathrm{Var}(Z^2)=2v^2$. This converts squared normal increments into exact variance calculations, including the degenerate case $v=0$.

If $X_i\sim N({\mu }_i,{\sigma }_i^2)$ are independent random variables, thenThe result follows by multiplying their characteristic functions.

A multivariate Gaussian distribution has density proportional to $\exp [-(x-\mu {)}^T{\Sigma }^{-1}(x-\mu )/2]$. Its mean is $\mu$ and its covariance matrix is $\Sigma$.

For centered jointly Gaussian random variables, an odd product has zero expectation, and an even product's expectation is the sum over all pairings of the products of the pair covariances. In particular,For a proof, the moment-generating function is $\exp (t^T\Sigma t/2)$. Differentiating once in each variable at zero leaves precisely products of covariance entries indexed by complete pairings. No independence assumption between the coordinates is needed.

Every linear combination $a^{T}X$ of the coordinates of a jointly normal random vector is a normal random variable. Its mean is $a^T\mu$ and its variance is $a^T\Sigma a$.

For the standard normal density $\varphi$ and distribution function $\Phi$, the upper-tail Mills ratio is $(1-\Phi (x))/\varphi (x)$. Ratios involving a Gaussian density and tail probability control truncated-normal moments and Gaussian-tail asymptotics.

Two successive integrations by parts of the standard normal density tail give positive remainder integrals establishing these strict inequalities for every $x>0$. The lower bound can be negative at small $x$ without invalidating the inequality.

An inverse Mills ratio divides a Gaussian density by a Gaussian tail probability. For lower truncation at zero, the conditional mean correction is $\sigma \varphi (\mu /\sigma )/\Phi (\mu /\sigma )$.

A Gaussian random variable is a random variable whose probability distribution is a normal distribution.

A standard normal random variable has the standard normal distribution, mean zero and variance one. If $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, the standardized variable $(X-\mu )/\sigma$ is standard normal.

If $X\sim N(\mu ,{\sigma }^2)$, completing the square in its density gives

Gaussian noise is an additive random error with a normal distribution, often assumed independent across observations and centered at zero.

If $Z_1,Z_2$ are independent $N(0,1)$ random variables, their product has characteristic function $(1+t^2{)}^{-1/2}$ and density $K_0(|x|)/\pi$, where $K_0$ is a modified Bessel function. A scale multiple $a{Z}_1{Z}_2$ has the corresponding scaled distribution.

The absolute value of a standard normal random variable has density

A positive random variable is log-normal when its logarithm has a normal distribution. If $Z\sim N(0,1)$ and $X=e^Z$, then $\mathbb{E}[X^n]=e^{n^2/2}$.

The support is $x>\theta$ and the probability density function isThe change of variables $z=\log (x-\theta )$ proves normalization. The threshold $\theta$ is a lower endpoint, distinct from the log-mean $m$ and log-standard-deviation $\sigma$. In measured sea-ice pressure ridge spacings, finite ridge widths and detection rules can create such a lower endpoint.

If $\log S=m+sZ$ with a standard normal $Z$, then $\mathbb{E}[S^p{1}_{\{S>K\}}]=e^{pm+p^2{s}^2/2}\Phi ((m+p{s}^2-\log K)/s)$. Completing the square proves the formula for every real $p$, including the inverse-price moment $p=-1$.

For $Z\sim N(0,1)$ and a sufficiently regular function $g$,It follows by integrating $zg(z)e^{-z^2/2}$ by parts.

The standard normal distribution is $N(0,1)$ and has cumulative distribution function $\Phi$.

The standard normal density is

Completing the square in the standard normal density gives $\mathbb{E}[e^{tZ}]=e^{t^2/2}$. Differentiating at zero yields $\mathbb{E}[Z]=0$, $\mathbb{E}[Z^2]=1$, and $\mathbb{E}[Z^4]=3$.

For $Z$ with a standard normal distribution and $a\ge 0$, split the Gaussian integral at zero and complete the square to obtain $\mathbb{E}{e}^{a|Z|}=2e^{a^2/2}\Phi (a)\le 2e^{a^2/2}$, where $\Phi$ is the standard normal distribution function. This converts a bound on an absolute Gaussian random variable into a two-sided Gaussian tail bound using Markov inequality.

The standard normal distribution function is

The standard normal quantile $z_q$ is the unique real number satisfying $\mathbb{P}(Z\le z_q)=q$ for $Z\sim N(0,1)$ and $0<q<1$.

If $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, then $(X-\mu )/\sigma$ has the standard normal distribution.