Gaussian random variable ($X\sim N(\mu ,{\sigma }^2)$)

A Gaussian random variable is a random variable whose probability distribution is a normal distribution. For $\sigma >0$, write $X=\mu +\sigma Z$ with a standard normal random variable $Z$; conversely, standardization of a normal random variable recovers $Z=(X-\mu )/\sigma$. For example, $X\sim N(100,15^2)$ has mean $100$, standard deviation $15$, and variance $225$. The outcome $x=115$ corresponds to standard score $z=1$, one standard deviation above the mean.

Some definitions also allow a Gaussian random variable with $\sigma =0$. This means $X=\mu$ almost surely, with variance zero and probability distribution equal to the Dirac measure ${\delta }_{\mu }$. It has no probability density function with respect to Lebesgue measure: $\mathbb{P}(X=\mu )=1$. The positive-scale density formula and division by $\sigma$ apply only when $\sigma >0$. MIT's constructive definition explicitly includes the constant case. ^[ref].

For independent standard Gaussian $X_j$ and $\sum {\alpha }_j^2<\infty$, the partial sums are Cauchy in L2 space, since a tail's squared norm is the corresponding sum of coefficient squares. Completeness gives an $L^2$ limit. The finite-sum characteristic functions converge to $\exp [-t^2\sum {\alpha }_j^2/2]$, identifying the limit as Gaussian, including the zero-variance case. The identification needs convergence of distributions, not just agreement of the first two moments.

For independent standard Gaussian random variables, the partial sums have characteristic functions $\exp (-u^2{v}_n/2)$, where $v_n={\sum }_{k\le n}{a}_k^2$. A finite almost-sure limit implies convergence of these characteristic functions to one continuous at zero; $v_n\to \infty$ would instead give zero at every nonzero $u$. Conversely, bounded $v_n$ makes the partial sums an martingale bounded in L2 space, which converges almost surely and in L2 space.

Let $M_n={\sum }_{k\le n}{a}_k{Z}_k$, and stop on first reaching a fixed positive threshold. If the coefficient squares are summable, uniform integrability gives the stopped mean zero, with the terminal limit used on $T=\infty$. If the coefficient squares diverge, $\mathbb{P}(M_n\ge R)\to 1/2$ for each fixed $R$. Thus $\mathbb{P}({\sup }_n{M}_n=\infty )\ge 1/2$. This event is unaffected by changing finitely many of the independent variables, so the Kolmogorov zero-one law makes its probability one. Hence $T<\infty$ and $M_T$ is at least the positive threshold; its expectation, finite or infinite, cannot be zero.

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 standardization of a normal random variable $(X-\mu )/\sigma$ is a standard normal random variable. Its values report displacement from the mean in units of the standard deviation, so the result is a dimensionless quantity.