# The probability that the minimum of a multivariate Gaussian exceeds zero

Suppose $$X \sim \mathcal N_n(\text{diag}(\Sigma), \sigma^2 \Sigma)$$ where $$\Sigma$$ may be allowed to be low rank, and let $$Y = \min_i > X_i$$.

What can be said about $$P\left(Y \geq 0\right)$$?

In general, I know that the exact distributions of Gaussian order statistics can be intractable, such as this math.se Q&A and the discussion here, but I'm hoping that the relationship between the mean and covariance matrix may lead to some simplification, or how I don't need the distribution of $$Y$$ but rather just the probability that it is greater than zero. The $$X_i$$ not being iid prevents me from using the usual things I know for examining minima and maxima, but I'm still hoping something can be done aside from numerical integration/simulation given values of $$\Sigma$$ and $$\sigma$$. I'd be very interested in approximations too.

The context on this and the unusual mean vector come from a now-deleted question on stats.se that essentially asked the following:

If we have $$X\sim\mathcal N_k(\mathbf 0, \sigma^2 I)$$ and nonrandom nonzero vectors $$z_1,\dots,z_n\in\mathbb R^k$$, what is the probability that $$\|X\|^2 \leq \|X-z_i\|^2 \text{ for all } i?$$

$$\|X-z_i\|^2 = \|X\|^2 - 2 X^Tz_i + \|z_i\|^2$$

so the question is equivalent to $$P(\|z_i\|^2- 2 X^Tz_i \geq 0 \text{ for all }i)$$. I collected the $$z_i$$ into the columns of a $$k\times n$$ matrix $$Z$$ so I can write the random variables in question as an affine transformation of $$X$$ via $$\text{diag}(Z^TZ) - 2 Z^TX \sim \mathcal N_n(\text{diag}(Z^TZ), 4\sigma^2 Z^TZ)$$ and I want the probability that this random vector is all non-negative, so this led me to the question I asked. The factored form of $$\Sigma$$ here is why I want to allow for possibly low rank covariance matrices since I could have $$k \leq n$$.

• Do you really intend that the mean of the multinormal should be $\text{diag}(\Sigma)$ or is that a typo? Looks strange ... – kjetil b halvorsen Aug 23 '20 at 4:24
• @kjetilbhalvorsen yeah I do mean that – jld Aug 23 '20 at 15:50
• Interesting --- out of curiosity, what is the context of the problem? – kjetil b halvorsen Aug 23 '20 at 19:33
• @kjetilbhalvorsen I've just updated with the context – jld Aug 24 '20 at 13:14

At least in the simplest non-trivial case, there are some tractable formulas. Let $$\Sigma=\begin{pmatrix}a & b\\b & d\end{pmatrix}, \text{ where }a>0,\ d>0,\ ad-b^2>0$$ The probability is given by a messy integral of the form $$P(Y\ge0)=\int_0^\infty\!\!\!\int_0^\infty f_{a,b,d,\sigma}(x_1,x_2)\,dx_1\, dx_2$$ However, the integral turns out nicely when $$b=0$$, giving an expression with the regularized incomplete gamma function $$Q$$: $$P(Y\ge0)|_{b=0}=\frac14 \left(Q\left(\frac12,\frac{a}{2\sigma^2}\right)-2\right)\! \left(Q\left(\frac12,\frac{d}{2\sigma^2}\right)-2\right)$$ The integral also turns out nicely when we differentiate by $$b$$ under the integral sign: $$\frac{dP(Y\ge0)}{db}= \int_0^\infty\!\!\!\int_0^\infty \frac{df_{a,b,d,\sigma}(x_1,x_2)}{db}\,dx_1\, dx_2 = \frac{ \exp\left(\frac{-ad(a+d-2b)}{2\sigma^2(ad-b^2)}\right)}{2\pi\sqrt{ad-b^2}}$$ This lets us calculate the probability for any $$a,B,d,\sigma$$ as $$P(Y\ge0) = P(Y\ge0)|_{b=0}+\int_{b=0}^B \frac{dP(Y\ge0)}{db} db$$ Then:
• We can check, e.g., that this correctly gives $$P(Y\ge0)=0.4794$$ when $$a=1,\, b=1/2,\, d=2,\, \sigma=3$$.
• The formula allows asymptotic analysis, e.g. for calculating a limiting probability as $$b$$ approaches $$\sqrt{ad}$$, where the messy approach of $$\iint\lim_{b\to\sqrt{ad}}f_{a,b,d,\sigma}$$ is ill-defined.
Some variant of this approach may also work for $$n>2$$, especially in the low-rank situation mentioned in the post. Feynman was famous for solving problems like this.