Note that
\({\boldsymbol x}=\mathbf {S}{\boldsymbol z}+{\boldsymbol \mu }\), where
\({\boldsymbol z}\) is standard multinormal, and the symmetric square root
\(\mathbf {S}={\boldsymbol \Sigma }^\frac{1}{2}\) may be regarded as the multidimensional sd, since it linearly scales the normal (like
\(\sigma\) in one dimension), and its eigenvectors and values are the axes of the 1 sd error ellipsoid. We first invert this transform to standardize the normal:
\({\boldsymbol z}=\mathbf {S}^{-1} ({\boldsymbol x}-{\boldsymbol \mu })\). This decorrelates or “whitens” the variables and transforms the integration domain to a different quadratic:
\begin{equation}
\begin{array}{ll}\tilde{q}({\boldsymbol z}) &\;= {\boldsymbol z}^{\prime } \mathbf {\widetilde{Q}}_2 {\boldsymbol z} + \tilde{{\boldsymbol q}}_1^{\prime } {\boldsymbol z} + \tilde{q}_0 \gt 0, \text{with} \\
\mathbf {\widetilde{Q}}_2 &\;= \mathbf {S} \mathbf {Q}_2 \mathbf {S}, \\
\tilde{{\boldsymbol q}}_1 &\;= 2\mathbf {S}\mathbf {Q}_2 {\boldsymbol \mu }+\mathbf {S}{\boldsymbol q}_1, \\
\tilde{q}_0 &\;= q({\boldsymbol \mu }). \end{array}
\end{equation}
Now the problem is to find the probability of the standard normal
\({\boldsymbol z}\) in this domain. If there is no quadratic term
\(\mathbf {\widetilde{Q}}_2\),
\(\tilde{q}({\boldsymbol z})\) is normally distributed, the integration domain boundary is a flat, and the probability is
\(\Phi (\frac{\tilde{q}_0}{\Vert \tilde{{\boldsymbol q}}_1 \Vert })\), where
\(\Phi\) is the standard normal cdf (
Ruben, 1960). Otherwise, say
\(\mathbf {\widetilde{Q}}_2=\mathbf {RDR}^{\prime }\) is its eigen-decomposition, where
\(\mathbf {R}\) is orthogonal (i.e., a rotoreflection). So
\({\boldsymbol y}=\mathbf {R}^{\prime }{\boldsymbol z}\) is also standard normal, and in this space the quadratic is
\begin{eqnarray*}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\hat{q}({\boldsymbol y}) &\;=&{\boldsymbol y}^{\prime } \mathbf {D}{\boldsymbol y} + {\boldsymbol b}^{\prime } {\boldsymbol y} + \tilde{q}_0 \quad \quad \left({\boldsymbol b}=\mathbf {R}^{\prime }\tilde{{\boldsymbol q}}_1\right) \\
&\;=&\sum _i \left(D_i y_i^2 + b_i y_i \right) + \sum _{i^{\prime }} b_{i^{\prime }} y_{i^{\prime }} + \tilde{q}_0 \\
&&\!\!\text{($i$ and $i^{\prime }$ index the nonzero and zero eigenvalues)} \\
&\;=&\sum _i D_i \left(y_i + \frac{b_i}{2D_i} \right)^2 + \sum _{i^{\prime }} b_{i^{\prime }} y_{i^{\prime }} + \tilde{q}_0 \\
&& -\, \sum _i \left(\frac{b_i}{2D_i} \right)^2 \\
&\;=&\sum _i D_i \, \chi ^{\prime 2}_{1,(b_i/2D_i)^2} + x,
\end{eqnarray*}
a weighted sum of noncentral chi-square variables
\(\chi ^{\prime 2}\), each with 1 degree of freedom, and a normal variable
\(x \sim N(m,s)\). So this is a generalized chi-square variable
\(\tilde{\chi }^2_{{\boldsymbol w}, {\boldsymbol k}, {\boldsymbol \lambda },m,s}\), where we merge the noncentral chi-squares with the same weights, so that the vector of their weights
\({\boldsymbol w}\) are the
unique nonzero eigenvalues among
\(D_i\), their degrees of freedom
\({\boldsymbol k}\) are the numbers of times the eigenvalues occur and their noncentralities, and normal parameters are
\begin{eqnarray*}
&&\lambda _j =\frac{1}{4 w_j^2} \sum _{i: D_i=w_j} b_i^2, \quad m =q({\boldsymbol \mu })- {\boldsymbol w.\lambda }, \\
&& s = \sqrt{\sum _{i^{\prime }} b_{i^{\prime }}^2}.
\end{eqnarray*}