Find $\mathbb{P}[ U+V \in S, V \in S], $ where $U,V$ are standard normal vectors, $S=\{ x \in \mathbb{R}^k: x_1 \le x_2 \le ... \le x_k \}$

Question

Let $U \in \mathbb{R}^k$ and $V\in \mathbb{R}^k$ be two independent standard normal vectors (i.e., $U \sim \mathcal{N}(0,I)$ and $U \sim \mathcal{N}(0,I)$ ). Define a set $S$ as \begin{align} S=\{ x \in \mathbb{R}^k: x_1 \le x_2 \le x_3 \le ... \le x_k \} \end{align}

Our goal is to find an order tight bound on $\mathbb{P}[ U+V \in S, V \in S]$.

Here are some preliminary bounds.

Upper Bound: Using monotonicty of probability measure \begin{align} \mathbb{P}[ U+V \in S, V \in S] \le \min (\mathbb{P}[ U+V \in S],\mathbb{P}[ V \in S])=\min \left(\frac{1}{k!} ,\frac{1}{k!} \right)=\frac{1}{k!} . \end{align}

Lower Bound: \begin{align} \mathbb{P}[ U+V \in S, V \in S] \ge \mathbb{P}[ U \in S, V \in S]= \left( \frac{1}{k!} \right)^2 \end{align} where we have used that $ U \in S, V \in S \Rightarrow U+V \in S, V \in S$.

Note that the orders are very different here. This question is inspired by something that I asked earlier here.

Answer 1

Sorry for the delay.

We have to figure out when $(U,U+V)$ lies in the cone $C=S\times S\subset \mathbb R^{2k}$ of angular measure $\frac 1{(k!)^2}$. Note that then $(U,V)$ lies in the cone $TC$ where $T$ is the linear transformation given by $(x,y)\mapsto (x,y-x)$. So we need to find or estimate the angular measure of $TC$ in $\mathbb R^{2k}$. The determinant of $T$ is $1$, so it preserves volumes. The norm of $T^{-1}$, however, is $\alpha=\sqrt{\frac{3+\sqrt 5}2}$, so if $TC$ intersects the unit ball by volume $V$, then $C$ intersects the ball of radius $\alpha$ by volume at least $V$. Passing to the angular measures, we see that the angular measure of $TC$ is at most the angular measure of $C$ times $\alpha^{2k}$, so we get an upper bound $\frac{1}{(k!)^2}(\frac{3+\sqrt 5}{2})^k$. That is not sharp, of course, but it gives you a general idea of what to expect.