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 $p\in\left(\tfrac12,1\right)$?



Probabilistic problem on the covariance matrix of a multivariate normal distribution
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 $\mu\_1>\mu\_2$.


