Let $Y = (Y\_1, Y\_2) \sim N(0, 11^T + I)$, be a bivariate normal random variable with non-isotropic covariance.


Define $y = (y\_1, y\_2)$ and let
\begin{align}
F\_{\delta}(y) = \Pr[Y\_1 > y\_1 - \delta, Y\_2 > y\_2 - \delta].
\end{align}


Consider
\begin{align}
G\_{\delta}(y) = \frac{F\_{\delta}(y)
}{
F\_0(y)
},
\end{align}


We wish to show that $G\_{\delta}(y)$ is monotonically increasing in $y$ for $1 > \delta > 0$ and $y \in \mathbb{R}^2$.



The monotonicity of the bivariate normal with non-isotropic covariance
For real $u,v$, let
\begin{equation\*}
 Q(u,v):=P(Y\_1>u\sqrt2,Y\_2>v\sqrt2)=\int\_u^\infty dz\,\varphi(z)\Big(1-\Phi\Big(\frac{2v-z}{\sqrt3}\Big)\Big),
\end{equation\*}
where $\varphi$ and $\Phi$ are the standard normal pdf and cdf, respectively.


The question in the OP can be restated as follows: show that
\begin{equation\*}
 r(u,v):=r\_t(u,v):=\frac{Q(u-t,v-t)}{Q(u,v)} \tag{10}\label{10}
\end{equation\*}
is increasing in real $u$ and in real $v$ for each $t\in(0,1/\sqrt2)$. Since $u$ and $v$ are interchangeable, it is enough to shown that
\begin{equation\*}
 r(v):=r(u,v)
\end{equation\*}
is increasing in real $v$ for each $t\in(0,1)$.


This will be done by using so-called l'Hospital-type rules for monotonicity. Indeed, consider the "derivative ratio" for the ratio in \eqref{10} with the extra constant factor $e^{t^2/2}$:
\begin{equation\*}
 r\_1(v):=e^{t^2/2}\frac{Q\_v(u-t,v-t)}{Q\_v(u,v)}, 
\end{equation\*}
where the subscript $\_v$ denotes the partial derivative in $v$. We have
\begin{equation\*}
 Q\_v(u,v)=-\int\_u^\infty dz\,\varphi(z)\,\frac2{\sqrt3}\,\varphi\Big(\frac{2v-z}{\sqrt3}\Big)
 =-\frac{\varphi(v)}2\, \left(\text{erf}\left(\frac{v-2 u}{\sqrt{6}}\right)+1\right). 
\end{equation\*}
So,
\begin{equation\*}
 r\_1(v)=\frac{f\_1(v)}{g\_1(v)}, \tag{20}\label{20}
\end{equation\*}
where
\begin{equation\*}
 f\_1(v):=e^{t v} \left(\text{erf}\left(\frac{t-2 u+v}{\sqrt{6}}\right)+1\right),\quad 
 g\_1(v):=\text{erf}\left(\frac{v-2 u}{\sqrt{6}}\right)+1. 
\end{equation\*}
Next, consider the "derivative ratio" for the ratio in \eqref{20}:
\begin{equation\*}
 r\_2(v):=\frac{f'\_1(v)}{g'\_1(v)}=\frac{f\_2(v)}{g\_2(v)}, \tag{30}\label{30}
\end{equation\*}
where
\begin{equation\*}
 f\_2(v):=\sqrt{6 \pi } t \left(\text{erf}\left(\frac{t-2 u+v}{\sqrt{6}}\right)+1\right)+2
 e^{-(t-2 u+v)^2/6},\quad 
 g\_2(v):=2 e^{-t v-(v-2 u)^2/6}. 
\end{equation\*}
Further, consider the third (and final) "derivative ratio" for the ratio in \eqref{30}:
\begin{equation\*}
 r\_3(v):=\frac{f'\_2(v)}{g'\_2(v)}=\frac{e^{-t (t-4 (u+v))/6}\, (-2 t-2 u+v)}{3 t-2 u+v}. \tag{40}\label{40}
\end{equation\*}
Note that for all real $z$
\begin{equation}
 r'\_3(2u+z)=\frac{t \left(2 z^2+2 t z+15-12 t^2\right) e^{-t (t-4 (3 u+z))/6}}{3 (3
 t+z)^2},
\end{equation}
which is obviously $>0$ for all $t\in(0,1)$ if $z\ne-3t$.


So, $r\_3$ is increasing on $(-\infty,2u-3t)$ and on $(2u-3t,\infty)$. Also, $g\_2>0$. Also, $g'\_2>0$ on $(-\infty,2u-3t)$ and $g'\_2<0$ on $(2u-3t,\infty)$.


So, by [lines 1 and 3 of Table 1.1](https://www.emis.de/journals/JIPAM/images/157_05_JIPAM/157_05.pdf), (the continuous function) $r\_2$ is down-up on $(-\infty,2u-3t]$ and up-down on $[2u-3t,\infty)$ -- that is, (i) there is some $a\in[-\infty,2u-3t]$ such that $r\_2$ is decreasing on $(-\infty,a]$ and increasing on $[a,2u-3t]$ and (ii) there is some $b\in[2u-3t,\infty]$ such that $r\_2$ is increasing on $[2u-3t,b]$ and decreasing on $[b,\infty)$.


So, $r\_2$ is down-up-down on $(-\infty,\infty)$. But $r\_2$ is a positive function such that $r\_2(v)\to0$ as $v\to-\infty$ and $r\_2(v)\to\infty$ as $v\to\infty$. So, $r\_2$ is increasing on $(-\infty,\infty)$.


But $f\_1(v)\to0$ and $g\_1(v)\to0$ as $v\to-\infty$. So, by Proposition 4.1 of the cited paper, $r\_1$ is increasing on $(-\infty,\infty)$.


Also, clearly $Q(u,v)\to0$ as $v\to\infty$. So, again by Proposition 4.1 of the cited paper, $r(u,v)$ is increasing in $v\in (-\infty,\infty)$. $\quad\Box$


