# Probability involving dependent random variables constructed from i.i.d. Gaussians

This is a problem I need to address for a certain computation in my research.

Let $$Y_1,\dots,Y_n$$ be a sequence of i.i.d. standard normal variables; and let $$I\subset[0,+\infty)$$ be an interval. In my application, if it helps one can think of $$I=[a_n,b_n]$$ where both $$a_n,b_n$$ are positive sequences that are both $$o_n(1)$$ that is $$a_n,b_n\to 0$$ as $$n\to+\infty$$.

Evaluate the following probability: $$\mathbb{P}\left(\bigcap_{1\leq i\leq n}\left\{\sum_{1\leq j\leq n,j\ne i}Y_iY_j\le 0 \right\}\cap \left\{\sum_{1\le j\le n}Y_j \in I\right\}\right).$$ In particular, (a) Is there a good way to compute this probability? (b) How does it behave as a function of the interval $$I$$, equivalently, as a function of $$a_n,b_n$$.

If we denote the sum by $$S$$, the condition $$\sum_{1\le j\le n,j\ne i}Y_iY_j\le 0$$ is equivalent to having $$Y_i^2\ge SY_i$$, which, on top of $$S\ge 0$$ (recall $$I\subset[0,\infty)$$) implies either $$Y_i\le 0$$ or $$Y_i\ge S$$. In particular, as $$I$$ gets larger, $$\mathbb{P}(S\in I)$$ gets larger, whereas $$\mathbb{P}(Y_i\in [0,S]^c)$$ gets smaller, which obviously means the size of $$I$$ brings a compromise.

I could not see a good way of computing this, and appreciate any help.

• Do you really need to evaluate the probability (i.e. give a closed-form expression), or you would be satisfied with lower or upper bounds for that quantity? Jul 11, 2020 at 23:03
• I certainly would be satisfied with an upper/lower bound (ideally both) that is not too loose, Mateusz. Jul 12, 2020 at 1:08

Let $$Z = \frac{1}{\sqrt{n}} \sum_{i=1}^n Y_i$$ and $$X_i = Y_i - \tfrac{1}{\sqrt{n}} Z$$. Then $$Z$$ and $$(X_1, \ldots, X_n)$$ are independent, $$Z$$ has standard normal distribution, and $$(X_1, \ldots, X_n)$$ has standard multivariate normal distribution on the hypersurface $$x_1 + \ldots + x_n = 0$$.
The question asks for the probability that $$Z \in \tfrac{1}{\sqrt{n}} I \qquad \text{and} \qquad X_i \notin [-\tfrac{1}{\sqrt{n}} Z, \tfrac{n - 1}{\sqrt{n}} Z] \quad \text{for every } i = 1, \ldots, n .$$ Denote by $$p(z)$$ the probability of the latter part, conditionally on $$Z = z$$. The whole story is therefore reduced to an estimate of the probability that a $$n-1$$-dimensional standard normal random vector belongs to a certain polytope.
For a fixed $$n$$, this seems doable. The general case, however, seems out of reach. Crude lower and upper bounds for $$p(z)$$ are $$p(z) \geqslant 1 - \sum_{i = 1}^n \mathbb{P}(X_i \in [-\tfrac{1}{\sqrt{n}} z, \tfrac{n - 1}{\sqrt{n}} z]) = 1 - n \mathbb{P}(N \in [-\tfrac{1}{\sqrt{n-1}} z, \tfrac{n - 1}{\sqrt{n-1}} z])$$ and \begin{aligned} p(z) & \leqslant 1 - \sum_{i = 1}^n \mathbb{P}(X_i \in [-\tfrac{1}{\sqrt{n}} z, \tfrac{n - 1}{\sqrt{n}} z]) + 2 \sum_{i = 1}^{n - 1} \sum_{j = i+1}^n \mathbb{P}(X_i \in [-\tfrac{1}{\sqrt{n}} z, \tfrac{n - 1}{\sqrt{n}} z]) \\ & = 1 - n \mathbb{P}(N \in [-\tfrac{1}{\sqrt{n-1}} z, \tfrac{n - 1}{\sqrt{n-1}} z]) + n (n - 1) \mathbb{P}((N, M - \tfrac{1}{n - 1} N) \in [-\tfrac{1}{\sqrt{n-1}} z, \tfrac{n - 1}{\sqrt{n-1}} z]^2), \end{aligned} where $$N, M$$ are auxiliary independent standard normal random variables. These estimates, of course, follow from the exclusion-inclusion principle, and are reasonably sharp when $$z = O(n^{-3/2})$$. That would require $$b_n = O(n^{-1})$$, which is likely not what is meant in the question, so I did not attempt to continue this calculation.