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.

  • $\begingroup$ 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? $\endgroup$ Jul 11, 2020 at 23:03
  • $\begingroup$ I certainly would be satisfied with an upper/lower bound (ideally both) that is not too loose, Mateusz. $\endgroup$
    – hookah
    Jul 12, 2020 at 1:08

1 Answer 1


(This is not an answer, but too long for a comment.)

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .