Multivariate normal law, including singular covariance

For any square-integrable vector X, covariance entries ${\mathrm{\Sigma }}_{ij}=\mathbb{E}[(X_i-m_i)(X_j-m_j)]$ exist because $2\mathrm{\mid }ab\mathrm{\mid }\le a^2+b^2$. They are symmetric. Finite linearity gives $u^T\mathrm{\Sigma }u=\mathbb{E}({\sum }_i{u}_i(X_i-m_i){)}^2\ge 0$, so covariance matrices are positive semidefinite. Conversely let a symmetric positive semidefinite Sigma be given and take its nonnegative symmetric square root A by [F2], so $A{A}^T=A^2=\mathrm{\Sigma }$.

By [F3]–[F4], AC realizes d independent standard normals $Z_1,\dots ,Z_d$, each of mean zero and second moment one by [F1]. Define $X=m+AZ$. Coordinate linear combinations are measurable, so X is a Borel random vector and its pushforward is a probability by [F9]. For any u and real t, [F5] and [F1] give $\mathbb{E}{e}^{itu\cdot X}=e^{itu\cdot m}{\prod }_j{e}^{-t^2(A^{T}u{)}_j^2\mathrm{/}2}=e^{itu\cdot m-t^2{u}^T\mathrm{\Sigma }u\mathrm{/}2}$. This equals the characteristic function of $N(u\cdot m,u^T\mathrm{\Sigma }u)$ by [F1]; [F6] gives equality of scalar laws. Thus the required projection condition holds, including u=0 and all null directions.

Finite linearity gives $\mathbb{E}X=m$. By [F7], $\mathbb{E}{Z}_i{Z}_j=0$ for i different from j, and [F1] gives $\mathbb{E}{Z}_i^2=1$. Hence the covariance of AZ is $AI{A}^T=\mathrm{\Sigma }$, with every product integrable by the bound in step 1.1. If Sigma=0 then A=0 and the law is the point mass at m; no inverse or density is required even when only some directions are null. For d=1 the construction agrees with the scalar affine normal. If dimension zero is admitted, use the unique law on the singleton empty tuple instead. AC is spent in the standard-normal construction, independent-copy realization and scalar uniqueness supplier.