Starting from a multivariate density of $R$, i.e. $f(R,\check{R},\sum)=(2\pi)^{-\frac{M}{2}}|\sum|^{-\frac{1}{2}}e^{-\frac{1}{2}[(R-\check{R})^T\sum^{-1}(R-\check{R})]}$ with
- $\sum$ covariance matrix
- $\check{R}$ the expectation of $R$,
I observe that the argument of exponential form in $M_2\operatorname{d}$imensional case a pencil of ellipsis $(\operatorname{x}-\mu)^T \sum^{-1}(\operatorname{x}-\mu)=c$. Then I write $\sum^{-1}=C^TC$ (with $C$ lower triangular matrix) for Cholesky, so $(R-\check{R})^T\sum^{-1}(R-\check{R})=(R-\check{R})^T C^TC(R-\check{R})$.
Knowing this is a quadratic form $x^TAx$,
- how can I prove that $(R-\check{R})^T C^TC(R-\check{R})=\begin{Vmatrix} C(R-\check{R}) \end{Vmatrix}_2$ ?
After I assume that this quantity is $\leq \Gamma$ for $\Gamma \geq 0$. So
- how can I prove that $\begin{Vmatrix} C(R-\check{R}) \end{Vmatrix}_2\leq \Gamma \Rightarrow \begin{matrix} \sum_{i=1}^MC_{m,i}\cdot R^i-s^m\leq \sum_{i=1}^M C_{m,i}\cdot \check{R}^i,\forall m=1,...,M\\ -\sum_{i=1}^MC_{m,i}\cdot R^i-s^m\leq -\sum_{i=1}^M C_{m,i}\cdot \check{R}^i,\forall m=1,...,M\\ \sum_{m=1}^M (s^m)^2 \leq (\Gamma)^2 \end{matrix}$ ?
My book says that $\begin{Vmatrix} x-u \end{Vmatrix}_2\leq \Gamma \Rightarrow \begin{Vmatrix} x-u \end{Vmatrix}_2^2=\sum_{n=v+1}^{+\infty}|x_n|^2<(\Gamma)^2$ but I don't understand the meaning of this condition.