Joint distribution from multivariate Normal distribution Consider two random variables $\{Y_1, Y_2\}$, which follow a joint multivariate normal distribution: $Y = [Y_1, Y_2]^T,$
\begin{equation}
Y_1 \sim \mathcal{N}(\mu_1,\sigma_1),\; Y_2\sim\mathcal{N}(\mu_2,\sigma_2).
\end{equation}
If they are non-negatively correlated, i.e., $\operatorname{Cov}(Y_1, Y_2)\ge 0$, does the following heuristic inequality statement generally hold? Why?
$$
P(Y_1 \le c, Y_2\le c) \ge P(Y_1 \le c )\,P(Y_2\le c),
$$
where $c$ is a constant.
 A: Partial Answer:
Following whuber: In the joint distribution function, we have
\begin{align*}
f(y_1,y_2)&=\frac{e^{-Q/2}}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}},\quad\text{where}\\
-\frac{Q}{2}&=-\frac{1}{2(1-\rho^2)}\left[
\frac{(y_1-\mu_1)^2}{\sigma_1^2}
-2\rho\,\frac{(y_1-\mu_1)(y_2-\mu_2)}{\sigma_1\sigma_2}
+\frac{(y_2-\mu_2)^2}{\sigma_2^2}
\right].\\
\text{Let}\quad T&=\frac{(y_1-\mu_1)^2}{\sigma_1^2}
-2\rho\,\frac{(y_1-\mu_1)(y_2-\mu_2)}{\sigma_1\sigma_2}
+\frac{(y_2-\mu_2)^2}{\sigma_2^2}.
\end{align*}
We seek an affine transformation so that $T$ has no cross-term between the new variables. We can obtain this by diagonalizing the quadratic form given. First, we simplify to regular $z$ terms:
\begin{align*}
Z_1&=\frac{Y_1-\mu_1}{\sigma_1}\\
Z_2&=\frac{Y_2-\mu_2}{\sigma_2}\\
T&=z_1^2-2\rho z_1z_2+z_2^2\\
&=
\left[\begin{matrix}z_1 &z_2\end{matrix}\right]
\left[\begin{matrix}1 &-\rho\\-\rho &1\end{matrix}\right]
\left[\begin{matrix}z_1\\z_2\end{matrix}\right]
\end{align*}
Diagonalizing the matrix $\left[\begin{matrix}1 &-\rho\\-\rho &1\end{matrix}\right]$ yields
$$\left[\begin{matrix}1 &-\rho\\-\rho &1\end{matrix}\right]=
\frac{1}{\sqrt{2}}\left[\begin{matrix}1 &-1\\1 &1\end{matrix}\right]
\left[\begin{matrix}1-\rho &0\\0 &1+\rho\end{matrix}\right]
\left[\begin{matrix}1 &1\\-1 &1\end{matrix}\right]\frac{1}{\sqrt{2}}$$
So the new transformation is
\begin{align*}
\hat{Z}_1&=\frac{Z_1+Z_2}{\sqrt{2}}\\
\hat{Z}_2&=\frac{-Z_1+Z_2}{\sqrt{2}}.
\end{align*}
You can show that
$$\operatorname{Cov}\!\left(\hat{Z}_1,\hat{Z}_2\right)=0.$$
Because the transformations are affine, the new distributions are also normal, and hence independent. Note that $\hat{\mu}_1=0=\hat{\mu}_2,$ and $\hat{\sigma}_1=1=\hat{\sigma}_2.$ Let
\begin{align*}
c_1&=\frac{c-\mu_1}{\sigma_1}\\
c_2&=\frac{c-\mu_2}{\sigma_2}\\
\hat{c}_1&=\sqrt{2}\,c_1\\
\hat{c}_2&=\sqrt{2}\,c_2.
\end{align*}
Now we translate the original probability problem: we want to show
\begin{align*}
P(Y_1\le c,Y_2\le c)&\ge P(Y_1\le c)\,P(Y_2\le c)\\
P\!\left(Z_1\le c_1,Z_2\le c_2\right)
&\ge P\!\left(Z_1\le c_1\right) P\!\left(Z_2\le c_2\right)\\
P\!\left(\hat{Z}_1-\hat{Z}_2\le \hat{c}_1,\hat{Z}_1+\hat{Z}_2\le \hat{c}_2\right)
&\ge P\!\left(\hat{Z}_1-\hat{Z}_2\le \hat{c}_1\right) P\!\left(\hat{Z}_1+\hat{Z}_2\le \hat{c}_2\right).
\end{align*}
This is saying that the "quadrant" defined by
$\hat{Z}_1-\hat{Z}_2\le \hat{c}_1,\hat{Z}_1+\hat{Z}_2\le \hat{c}_2$ has a probability greater than the product of the two half-planes separately.
This seems intuitively correct to me, but I'm not sure how to finish.
