Probability that quadratic polynomial in independent gaussian variables is negative

Question

Let $a,b \in \mathbb R^m$ be fixed (deterministic vectors), and let $w=(w_1,\ldots,w_m)$ be a random vector in $\mathbb R^m$ with iid coordinates from $N(0,1)$. Define the random variable $$ h(w) := w^Tab^Tw = \sum_{i=1}^m\sum_{j=1}^ma_ib_jw_iw_j. $$ Note that $h(w)$ can also be thought of as a random polynomial of degree 2 (or else, is identically equal to zero if $a_i b_j = 0$ for all $i,j$).

Question. Is there an analytic formula for $p:=\mathbb P(h(w) \le 0)$ ?

Solution to special case whenr $a$ and $b$ are parallel

If $b = ra$, for some $r \in \mathbb R$, then $h(w) = rw^Taa^Tw$, and so one computes $$ p = \begin{cases}1,&\mbox{ if }r \le 0 \text{ or }a = 0,\\0,&\mbox{otherwise.}\end{cases} $$

Observations

Using standard formula for moments of quadratic forms, one notes that

• $\mathbb E[h(w;a,b)] = a^Tb$,
• $\mbox{Var}(h(w;a,b)) = (a^Tb)^2$.

Answer 1

We can rewrite the expression as $(w\cdot a)(w\cdot b)$. This is positive when the dot products of $w$ with $a$ and $b$ are both positive or both negative. We can normalize $a$ and $b$ to unit vectors. Next we can assume $w$ lies in the plane of $a$ and $b$, since the Gaussian is spherically symmetric, and the perpendicular component does not contribute to the dot products. Thus we can reduce the problem to $a$, $b$, and $w$ unit vectors in the plane. Then simple geometry shows that if $a$ and $b$ make an angle of $\theta$, the probability of the expression being positive is $1 - \theta/\pi$. Note that for $\theta = 0$ or $\theta=\pi$, this reduces to the case $a = rb$ with $r$ positive or negative respectively.

Answer 2

This would be a very verbose comment, so I'm posting it here as a separate answer...

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Solution via links with kernels (statistics)

It turns out that the sought-for quantity bares an unespected link with kernels. Indeed, let $\Theta(u) := (1 + \mbox{sign}(u)) / 2$ define the Heaverside function. Then, one can rewrite

$$ 1 - p = \mathbb E[\Theta(w\cdot a)\Theta(w\cdot b)(w\cdot a)^0(w\cdot b)^0] = \kappa_0(a,b), \tag{1} $$ where $(x,x') \mapsto \kappa_n(x,x)$ is the arc-cosine kernel of order $n$. See section 2.1 of this paper for terminology.

The good news: its is well-known that if $\theta(x,x') := \arccos\left(\frac{x\cdot x'}{\|x\|\|x'\|}\right)$, then we have the representation $\kappa_n(x,x') = \frac{1}{\pi}\|x\|^n\|x'\|^nJ_n(\theta(x,x'))$, where

$$ J_n(\theta) := (-1)^n(\sin \theta)^{2n+1}\left(\frac{1}{\sin \theta}\frac{\partial}{\partial \theta}\right)^n\left(\frac{\pi-\theta}{\sin\theta}\right). $$

In particular, $J_0(\theta) = \pi-\theta$, and so returning to (1), we get the analytic formula

$$ 1-p = \kappa_0(a,b) = \frac{\pi-\theta(a,b)}{\pi} = 1 - \frac{\theta(a,b)}{\pi}, $$

which is precisely the formula proposed by @Anand by way of planar geometry!