Probabilistic problem on the covariance matrix of a multivariate normal distribution [closed]

Question

We have a random variable $\mathbf{X}\sim\mathcal{N}_d(\mathbf{\mu},\mathbf{\Sigma})$, where $\mathcal{N}_d(\mathbf{\mu},\mathbf{\Sigma})$ is a $d$-dimensional multivariate normal distribution with mean $\mathbf{\mu}$ and covariance matrix $\mathbf{\Sigma}$. Let $\mathbf{x}$ be the value taken by $\mathbf{X}$. We want to bet on the event $x_1\ge x_2$, where $x_1$ and $x_2$ are respectively the first and second component of $\mathbf{x}$.

Question: What are sufficient and necessary conditions for $\mathbf{\Sigma}$ that guarantee $x_1\ge x_2$ with a desired probability $p\in\left(\tfrac12,1\right)$?

Answer 1

You are just asking to compute $p=P(X>Y)=P(Z<0)$, where $(X,Y)$ has the bivariate normal distribution with given $EX=\mu_1$, $EY=\mu_2$, $Var\,X=\sigma_1^2:=\Sigma_{1,1}$, $Var\,Y=\sigma_2^2:=\Sigma_{2,2}$, and $\rho:=corr(X,Y)=\Sigma_{1,2}/(\sigma_1 \sigma_2)$, and $Z:=Y-X\sim N(\mu,\sigma^2)$, where $\mu:=\mu_2-\mu_1$ and $\sigma^2:=\sigma _1^2+\sigma _2^2-2 \rho \sigma _1 \sigma_2$. So, $$p=\Phi\Big(\frac{\mu _1-\mu _2}{ \sqrt{\sigma _1^2+\sigma _2^2-2 \rho \sigma _1 \sigma_2}}\Big)$$ or, equivalently, $$\frac{\mu _1-\mu _2}{ \sqrt{\sigma _1^2+\sigma _2^2-2 \rho \sigma _1 \sigma_2}}=\Phi^{-1}(p), \tag{1}$$ where $\Phi$ is the standard normal cdf and $\Phi^{-1}$ is the function inverse to $\Phi$. We have $p\in(\frac12,1)$ iff $\mu_1>\mu_2$.