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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. Growth of log ratio

The result is similar for $n=3$. However, for $n=2$, the result is pretty striking: n=2

It should also be pointed out that there seems to a critical point for $\alpha$ when $\alpha^2$ is around 10, for all $n=2,3,4$.

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    $\begingroup$ What have you tried? Did you plot the function for a few values of $n$, and if so what did you see? $\endgroup$ – Daniel McLaury Jun 23 '19 at 16:30
  • $\begingroup$ Thank you for your comment! Please find the update with a simulation study. @DanielMcLaury $\endgroup$ – neverevernever Jun 23 '19 at 20:01
  • $\begingroup$ 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. $\endgroup$ – fedja Jun 27 '19 at 5:40
  • $\begingroup$ How do you see this? @fedja $\endgroup$ – neverevernever Jun 28 '19 at 2:22
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"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 $0<t<t_0$, 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.

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  • $\begingroup$ If $\mathbb{E}[\exp(-\beta Z)]$ is comparable to $\beta^{-q}$, the $q$ for the nominator and the denominator should be different right? $\endgroup$ – neverevernever Jun 30 '19 at 23:55
  • $\begingroup$ @neverevernever The whole point is that it is the same. $\endgroup$ – fedja Jul 1 '19 at 0:07
  • $\begingroup$ And how do we see $\mathbb{E}[\exp(-\beta Z)]$ is comparable to $\beta^{-q}$? $\endgroup$ – neverevernever Jul 1 '19 at 0:54
  • $\begingroup$ @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. $\endgroup$ – fedja Jul 1 '19 at 1:19
  • $\begingroup$ 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? $\endgroup$ – neverevernever Jul 1 '19 at 1:43

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