Suppose that $\boldsymbol{t}\sim \mathcal{N}(\boldsymbol{u};\boldsymbol{0},\boldsymbol{M})=f_{\boldsymbol{t}}(\boldsymbol{u})$, where $\boldsymbol{t}$ is a $N$-dimensional gaussian random vector, and
\begin{equation}
\boldsymbol{M}=
\left[
\begin{matrix}
1&m_{12}&\cdots& m_{1N}\\
m_{21}&1&\cdots& m_{2N}\\
\vdots&\vdots&&\vdots\\
m_{N1}&m_{N2}&\cdots& 1
\end{matrix}
\right]
\end{equation}
where $m_{i,j}=m_{j,i}\leq1$, $i=1,2,\cdots,N$, and $j=1,2,\cdots,N$;
Let
\begin{equation}
F=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\boldsymbol{\gamma})} f_{\boldsymbol{t}}(\boldsymbol{u}) \,d\boldsymbol{u}
\end{equation}
where $\mathcal{Y}(\boldsymbol{\gamma})=\left(-\infty,\gamma_1\right)\times\cdots\times \left(-\infty,\gamma_N\right)$
Now, if let the $(i,j)\text{th}$ element of $\boldsymbol{M}$ be an independent variable $a,$ $i\neq j$, i.e., we have $m_{i,j}=m_{j,i}=a<1$.
Then we let $F(a)=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\gamma)} f_{\boldsymbol{t}}(\boldsymbol{u}) \, d\boldsymbol{u}$.
My question is:
Is $F(a)$ a monotonically increasing function of $a$?
Best Answer
Yes, this is a special case of Slepian's inequality.