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In the question I have $0<A<1$ arbitrarily fixed, and $n\in\mathbb Z^+$ goes to $+\infty$. Just in case the title doesn't parse correctly, I repeat the question: Does $$\lim_{\,n\to+\infty\,}\frac{\int_A^1(1-s^2)^n{\rm d}s}{\int_0^1(1-s^2)^n{\rm d}s}=0$$ hold? If, or if not, how does one prove it? I have some ideas but I would like to know if there is a relatively simple way of getting the result.

The background in my question above is that I am considering the possibility of getting some "reasonable" probability measures on infinite-dimensional Hilbert balls by obtaining them as a limit of probabilities on finite-dimensional subspaces. It seems that this construction will fail, and a crucial failure would be guaranteed by a positive answer to the question.

Mittens
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The-unKnowN
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  • Perhaps you would like to describe what your ideas (even if they are just that "ideas"). As it stands, your posting may attract some closing votes as it does not show context. – Mittens Apr 01 '23 at 17:09
  • @OliverDíaz The "ideas" are too complicated to be written down without writing a whole proper answer. The background, or "context" is that I am considering the possibility of getting some "reasonable" probability measures on infinite-dimensional Hilbert balls by obtaining them as a limit of probabilities on finite-dimensional subspaces. It seems that this construction will fail, and a crucial failure would be guaranteed by a positive answer to my question above. – The-unKnowN Apr 01 '23 at 18:15
  • @OliverDíaz OK, thanks for the advice. – The-unKnowN Apr 01 '23 at 18:35

1 Answers1

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Notice that $$\int^1_0(1-s^2)^n\,ds=\frac12\int^1_0u^{1/2-1}(1-u)^{n+1-1}\,du=B(1/2,n+1)$$

The sequence in the OP can be rewritten as $$\frac{\int_A^1(1-s^2)^n{\rm d}s}{\int_0^1(1-s^2)^n{\rm d}s}=\frac{1}{B(1/2,n)}\int^1_0\mathbb{1}_{(A,1]}(\sqrt{u})u^{1/2-1}(1-u)^{n+1-1}\,du$$

In terms of probability, this is $P[A<Y_n\leq 1]$, where $Y_n=X^{1/2}_n$ is a random variable with $X_n\sim \operatorname{Beta}(1/2,n+1)$.

The $\operatorname{Beta}(1/2,n)$ distribution $$\mu_n(dx)=\frac{1}{B(1/2,n+1)}x^{1/2-1}(1-x)^{n+1-1}\mathbb{1}_{(0,1)}(x)\,dx$$ converges to $\delta_{0}$ as $n\rightarrow\infty$. Thus, if $0< A\leq 1$, $$P[Y_n> A]=P[X_n> A^2]=\mu_n((A^2,1])\xrightarrow{n\rightarrow\infty}\delta_0((A^2,1])=0$$

Proof that $\mu_n\stackrel{n\rightarrow\infty}{\Longrightarrow}\delta_0$:

Let $(X_n:n\in\mathbb{N})$ be an i.i.d sequence of $\operatorname{Gamma}(1/2,\theta)$ distributions ($\theta>0$ is actually not very important). Then $$W_n=\frac{X_1}{X_1+(X_2+\ldots+X_{n+2})}$$ is a $\operatorname{Beta}(1/2,n+1)$ random variable. By the law of large numbers $\frac{1}{n+1}\sum^{n+2}_{k=2}X_k \xrightarrow{n\rightarrow\infty}E[X_2]=\frac{1/2}{\theta}$ a.s.; hence $$W_n=\frac{X_1}{X_1+(n+1)\frac{1}{n+1}\sum^{n+2}_{k=2}X_k}\xrightarrow{n\rightarrow\infty}0\qquad\text{a.s}$$ The conclusion follows.

Mittens
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  • Your solution uses quite "nonelementary machinery", and convinces me of the validity of the positive answer although as a nonprobabilist (I am a functional analyst) I do not quite follow the part of the proof that $\mu_n\to\delta_0$ holds. In fact, this is the main assertion for which I am searching a proof. The Wikipedia link you give for having $X_1/(X_1+\ldots,X_{n+2})$ a Beta$(\frac12,n+1)$ random variable only considers $X/(X+Y)$. Is there some "official" reference for the more general assertion you are using? – The-unKnowN Apr 01 '23 at 19:52
  • @The-unKnowN: the solution is based on simple Probability facts at the upper division level. the Wikipedia link I added in my posting states that $U/(U+V)$ has $Beta(\alpha,\beta)$ distribution when $U$ and $V$ are independent Gamma variables with parameters $(\alpha,\theta)$ and $(\beta,\theta)$ ($\theta$ not important) this can be shown with simple change of variables in a double integral. The sums of $n$ independent Gamma's $(\alpha,\beta)$ is a Gamma $(n\gamma,\beta)$ (just multiply the Fourier transforms)... (to be continued) – Mittens Apr 01 '23 at 19:53
  • @The-unKnowN: ...continuaation. Convergence in distribution is what in functional analysis is convergence in the weak topology $\sigma(M(\mathbb{R}),C_b(\mathbb{R})$ topology where $M(\mathbb{R})$ is the collection of all finite Borel measures in $\mathbb{R}$. – Mittens Apr 01 '23 at 19:55
  • @The-unKnowN: Here the change of variable calculation that gives you Beta from two Gammas. Like I said, is a simple calculation that involves computing a Jacobian determinant. – Mittens Apr 01 '23 at 19:59
  • @The-unKnowN: The convergence in distribution to the unit masss at $0$ ($\delta_0$) follows from the law of large numbers (from upper division courses in probability) $\frac{1}{n}(Y_2+\ldots + Y_{n+1})\xrightarrow{n\rightarrow\infty}E[X_2]=\theta$ almost surely. – Mittens Apr 01 '23 at 20:02
  • OK, thanks. There has been quite a long time since I have last considered those probability distributions. I have accepted your answer. – The-unKnowN Apr 01 '23 at 20:06