Proof of Standardization and Cumulative Distribution Function of a Gaussian Random Variable

Fix a Gaussian representation $\bigl(m,(\mu_1),(a_j),(Z_j)\bigr)$ of $(X)$ and write $W=\mu_1+\sum_{j=1}^{m}a_jZ_j$, so $P(X=W)=1$; when $m=0$ the sum is empty and equal to $0$, in accordance with Gaussian Random Vectors and Jointly Gaussian Random Variables. By Claim 2 of Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector, $\mu=\mathbb{E}[X]=\mu_1$ and $\sigma^{2}=\operatorname{Var}(X)=\operatorname{Cov}(X,X)=\sum_{j=1}^{m}a_j^{2}$, using Covariance of Square-Integrable Random Variables.

Claim 1. If $\sigma^{2}=0$, then $\sum_j a_j^{2}=0$ with every term nonnegative, so $a_j=0$ for every $j$ (vacuously when $m=0$) and $W=\mu$ at every point of $\Omega$. Hence $P(X=\mu)\ge P(X=W)=1$.

Claim 2. Assume $\sigma^{2}>0$ and let $J=\{j:a_j\ne0\}$, which is nonempty. For $j\in J$ define $Z'_j=Z_j$ if $a_j>0$ and $Z'_j=-Z_j$ if $a_j<0$. Each $Z'_j$ is standard normal, by Claim 3 of Reflection Invariance of Lebesgue Measure and Symmetry of the Standard Normal Distribution in the reflected case. The family $(Z'_j)_{j\in J}$ is independent: for Borel sets $B_j$, each event $\{Z'_j\in B_j\}$ equals $\{Z_j\in B_j\}$ or $\{Z_j\in-B_j\}$, and $-B_j$ is a Borel set by Claim 1 of Reflection Invariance of Lebesgue Measure and Symmetry of the Standard Normal Distribution, so the required product identities follow from the corresponding identities for the independent subfamily $(Z_j)_{j\in J}$. Moreover $a_jZ_j=|a_j|Z'_j$ at every point, so

Enumerate $J=\{j_1,\dots,j_r\}$ and set $c_i=|a_{j_i}|>0$ and $s_k=c_1^{2}+\dots+c_k^{2}$ for $1\le k\le r$, so $s_r=\sigma^{2}$. We prove by induction on $k$ that

is standard normal, with $\sqrt{\,\cdot\,}$ the positive square root. For $k=1$, $Y_1=Z'_{j_1}$. For the step from $k$ to $k+1$, write

where $a,b>0$ and $a^{2}+b^{2}=(s_k+c_{k+1}^{2})/s_{k+1}=1$. The random variables $Y_k$ and $Z'_{j_{k+1}}$ are independent: by the grouping lemma applied to the independent family $(Z'_j)_{j\in J}$ with the disjoint blocks $\{j_1,\dots,j_k\}$ and $\{j_{k+1}\}$, the $\sigma$-algebras $\sigma(Z'_{j_1},\dots,Z'_{j_k})$ and $\sigma(Z'_{j_{k+1}})$ are independent; $Y_k$ is measurable with respect to the former (sums and scalar multiples of measurable functions are measurable with respect to any $\sigma$-algebra making the summands measurable, by the preliminaries of Square-Integrable Random Variables and the Mean-Square Inner Product), and $Z'_{j_{k+1}}$ with respect to the latter, so every pair of events $\{Y_k\in B\}$, $\{Z'_{j_{k+1}}\in B'\}$ with $B,B'$ Borel multiplies, which is independence in the sense of Independence of Events and of Random Variables (using the identification of independence of random variables with independence of their generated $\sigma$-algebras recorded in Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras). By Claim 2 of Moments and Stability of the Standard Normal Distribution, $Y_{k+1}$ is standard normal, completing the induction.

For $k=r$ this says $(W-\mu)/\sigma=Y_r$ is standard normal. Now $(X-\mu)/\sigma$ is a random variable by the preliminaries of Square-Integrable Random Variables and the Mean-Square Inner Product, and for every Borel set $B$ the events $\bigl\{(X-\mu)/\sigma\in B\bigr\}$ and $\bigl\{(W-\mu)/\sigma\in B\bigr\}$ agree off the event $\{X\ne W\}$ of probability $0$; each is contained in the union of the other with $\{X\ne W\}$, so by the monotonicity and additivity of the probability measure $P$ their probabilities are equal. Hence the distribution of $(X-\mu)/\sigma$ equals that of $(W-\mu)/\sigma$, namely the standard normal distribution; so $(X-\mu)/\sigma$ is standard normal.

Finally, since $\sigma>0$, for every real $t$ the inequality $X\le t$ holds exactly when $(X-\mu)/\sigma\le(t-\mu)/\sigma$, so

by the definition of $\Phi$ in Standard Normal Distribution. The final sentence of the statement follows because in the degenerate case Claim 1 forces $F_X(t)=0$ for $t<\mu$ and $F_X(t)=1$ for $t\ge\mu$, again determined by $(\mu,\sigma^{2})$. $\blacksquare$