# Ratio of expectation involving random unit vectors

Let $$u=(u_1,...,u_n), v=(v_1,...,v_n)$$ be two random vectors independently and uniformly distributed on the unit sphere in $$\mathbb{R}^n$$. Define two other random variables $$X=\sum_{i=1}^nu_i^2v_i^2$$, $$Y=u_1^2v_1^2$$. Consider the following ratio of expectation: $$r_n(\alpha)=\frac{\mathbb{E}\{\exp[-\alpha^2\frac{1-X}{2}]\}}{\mathbb{E}\{\exp[-\alpha^2(1-Y)]\}}$$ Does there exist a finite upper bound for $$r_n(\alpha)$$, independent of $$\alpha$$? I'm interested in the behavior with fixed $$n$$ and large $$\alpha$$.

Update:

I did some simulation, the answer seems to be negative. Below is a plot for $$n=4$$ using Monte Carlo simulation by averaging 20000 samples on the numerator and denominator respectively. The result is similar for $$n=3$$. However, for $$n=2$$, the result is pretty striking: It should also be pointed out that there seems to a critical point for $$\alpha$$ when $$\alpha^2$$ is around 10, for all $$n=2,3,4$$.

• What have you tried? Did you plot the function for a few values of $n$, and if so what did you see? – Daniel McLaury Jun 23 '19 at 16:30
• Thank you for your comment! Please find the update with a simulation study. @DanielMcLaury – neverevernever Jun 23 '19 at 20:01
• Actually, the ratio is bounded as $\alpha\to\infty$ and your mysterious $10$ is just the cutoff for which 20000 trials is any good. Raise the number of trials to $10^{20}$ at least if you want to simulate in $\mathbb R^4$ in the range of your plot. – fedja Jun 27 '19 at 5:40
• How do you see this? @fedja – neverevernever Jun 28 '19 at 2:22

"How do you see this?" It is quite simple, actually. For a positive random variable $$Z$$, we have $$E[e^{-\beta Z}]=\beta\int_0^\infty e^{-\beta t}P[Z\le t]\,dt$$. Thus, if $$P[Z\le t]\asymp t^q$$ for $$0, our expectation is comparable to $$\beta^{-q}$$.

Now if $$t>0$$ is small and $$\frac {1-X}2\le t$$, then $$X\ge 1-2t$$ which means that there exists $$i$$ such that $$u_i^2\ge 1-2t$$ and there is $$j$$ such that $$v_j^2\ge 1-2t$$ (if there is no such $$i$$, then $$X=\sum_i u_i^2v_i^2<(1-2t)\sum_iv_i^2=1-2t$$ and similarly for $$j$$. Note that such $$i,j$$ are unique (provided that $$t$$ is small enough), all other entries of $$u$$ and $$v$$ squared are at most $$2t$$ and if $$i\ne j$$, then we still have $$X\le 2\cdot 2t(1-2t)+4t^2(n-2)< 1-2t$$ if $$t$$ is small. Thus, we must have an index $$i$$ such that $$u_i^2,v_i^2\ge 1-2t$$. Conversely, if there is an index $$i$$ such that $$u_i^2,v_i^2\ge 1-t$$, then $$X\ge 1-2t$$. By symmetry and independence, up to a factor of $$n$$, we just need to look at the probability that $$u_1^2\ge 1-t$$ and establish a power rate of decay for it as $$t\to 0$$. This amounts to the estimate of the corresponding portion of the sphere and gives something like $$t^{(n-1)/2}$$. The computation for $$Y$$ is the same.

Now, the main part of the integral comes from $$t\approx \beta^{-1}$$, which means that you should detect events of probability about $$\beta^{-(n-1)}$$ in your trials. If $$\beta=\alpha^2=10$$ and $$n=4$$, that is $$10^{-3}$$, so the life is still fair (you have about $$20$$ hits out of $$20000$$) but when you try $$\beta=2000$$, you get an event of probability $$<10^{-10}$$ to be noticed so fewer than $$10^{10}$$ trials are useless (I wrote $$10^{20}$$ because I thought you had $$\alpha$$ rather than $$\alpha^2$$ as your parameter).

That's it.

• If $\mathbb{E}[\exp(-\beta Z)]$ is comparable to $\beta^{-q}$, the $q$ for the nominator and the denominator should be different right? – neverevernever Jun 30 '19 at 23:55
• @neverevernever The whole point is that it is the same. – fedja Jul 1 '19 at 0:07
• And how do we see $\mathbb{E}[\exp(-\beta Z)]$ is comparable to $\beta^{-q}$? – neverevernever Jul 1 '19 at 0:54
• @neverevernever By the integral identity I wrote. It does not matter what to integrate from $t_0$ to $\infty$: with either $t^q$ or some probability, the integral is exponentially small in $\beta$. And on $[0,t_0]$ the quantities are comparable. – fedja Jul 1 '19 at 1:19
• I understand that we only need to look at $[0,t_0]$. I also know that on this interval, $\beta^{-q}$ is the upper bound of the integral. But why is $\beta^{-q}$ also the lower bound? – neverevernever Jul 1 '19 at 1:43