Joint distribution from multivariate Normal distribution

Consider two random variables $$\{Y_1, Y_2\}$$, which follow a joint multivariate normal distribution: $$Y = [Y_1, Y_2]^T,$$

$$$$Y_1 \sim \mathcal{N}(\mu_1,\sigma_1),\; Y_2\sim\mathcal{N}(\mu_2,\sigma_2).$$$$

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.

• If they are positively correlated, then surely $\operatorname{Cov}(X,Y)$ should be strictly greater than zero? – Adrian Keister Jul 9 '20 at 15:37
• @AdrianKeister, thanks for the point out, I have edited the description. – Naofumi Jul 9 '20 at 15:47
• Do the $Y_1$ and $Y_2$ variables have different means and variances? – Adrian Keister Jul 9 '20 at 15:54
• I believe that if you perform an affine transformation from $(Y_1,Y_2)$ to standard Normal (and uncorrelated) variables $(Z_1,Z_2)$ and rephrase the probability inequality in terms of the $Z_i,$ the answer becomes clear. The positive correlation implies the event on the left hand side is an acute-angled wedge whereas the lack of correlation of the $Z_i$ can be used to show the right hand side is the probability associated with a right-angled wedge. – whuber Jul 9 '20 at 16:21
• @Adrian Yes, that's right--and in the diagonalization process, the angle made by the images of the lines $y_1=c$ and $y_2=c$ changes to reflect the covariance. – whuber Jul 9 '20 at 22:32

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.