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In one paper I read, the author claimed that the Gaussian measure of the $B_n$ ball tends to 1/2 when n tends to infinity. Here $$B_n = \{x \in \mathbb{R}^n : \|x \|^2 \le n \},$$ and the Gaussian measure is the probability measure on $\mathbb{R}^n$ with density $$\frac{1}{(2\pi)^{n/2}} e^{\frac{-\|x\|^2}{2}}.$$

Can any one suggest me a proof of this statement? I am aware that the Gaussian measure concentrates on the "flatten set" $\{\sqrt{n} - \epsilon \le \|x\| \le \sqrt{n} + \epsilon\}$, but could not find a proof anywhere for this case (when $x$ is only on one side of $\sqrt{n}$).

The question made explicitly: $$\lim_{n \to \infty} \int_{\|x \|^2 \le n} \frac{1}{(2\pi)^{n/2}} e^{\frac{-\|x\|^2}{2}} dx = \frac{1}{2} ?$$

Qui-vous
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5 Answers5

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This is like integrating the density you mentioned from in the range $(0,\infty)$ since the radius of ball can't be negative. But since the function is symmetrical about 0 the integration in the aforementioned range should be $1/2$ i.e $$\lim_{n\rightarrow \infty}\int_0^{\sqrt{n}}\frac{1}{(2\pi)^{n/2}} e^{\frac{-\|x\|^2}{2}}=\frac{1}{2}\lim_{n\rightarrow \infty}\int_{-\sqrt{n}}^{\sqrt{n}}\frac{1}{(2\pi)^{n/2}} e^{\frac{-\|x\|^2}{2}}$$

dohmatob
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Sonal_sqrt
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Let $Z_1,Z_2,\dots$ be i.i.d $\chi^2(1)$ variables (squares of independent normal variables). By the central limit theorem,

$$\mathbb P\left [\frac{Z_1+\dots+Z_n}n - \mathbf E[Z_1] \leq 0\right] \to \tfrac 1 2$$

which is a restatement of the claim.

Dap
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  • You seem to mix strong law and CLT but CLT does work as you see in my post. – Kavi Rama Murthy Sep 19 '17 at 07:49
  • Your statement of the central limit theorem is not correct. It should actually say $$P \bigg( Z_1+...+Z_n \leq n \bigg) = P\bigg( \frac{Z_1+...+Z_n - n}{\sigma \sqrt n} \leq 0 \bigg) \to 1/2$$ where $\sigma = E[(Z_1-1)^2]^{1/2}$. – shalop Sep 22 '17 at 06:10
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    @Shalop: "By the central limit theorem" does not mean "Here is a statement of the central limit theorem". It just means the statement obviously follows. – Dap Sep 22 '17 at 09:34
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A quick proof when $n$ is even: Spherical coordinates change yields

\begin{align*} \int_{B_n} \frac{1}{(2\pi)^{n/2}} e^{-\|x\|^2/2} \, dx &= \int_{0}^{\sqrt{n}} \frac{1}{(2\pi)^{n/2}} \underbrace{ \frac{2\pi^{n/2}}{\Gamma(\frac{n}{2})} r^{n-1} }_{=|\partial B_r|} e^{-r^2/2} \, dr \\ (r = \sqrt{2u}) \quad &= \int_{0}^{\frac{n}{2}} \frac{u^{\frac{n}{2}-1}}{\Gamma(\frac{n}{2})} e^{-u} \, du. \end{align*}

Now assume that $n = 2k$ is even and $T_1, T_2, \cdots$ are i.i.d. exponential random variables of unit rate. Then the integral above can be written as

$$ \int_{0}^{k} \frac{u^{k-1}}{(k-1)!} e^{-u} \, du = \mathbb{P}(T_1+\cdots+T_k \leq k) = \mathbb{P}\left( \frac{T_1+\cdots+T_k - k}{\sqrt{k}} \leq 0 \right)$$

By the classical CLT, we have $\frac{T_1+\cdots+T_k - k}{\sqrt{k}} \Rightarrow Z \sim \mathcal{N}(0,1)$ as $k\to\infty$. So the above probability converges to $\mathbb{P}(Z \leq 0) = \frac{1}{2}$.

For a more general proof, let $\nu = n/2$ and apply the change of variables $u = \nu - \sqrt{\nu}t$. Then

$$ \int_{0}^{\nu} \frac{u^{\nu-1}}{\Gamma(\nu)} e^{-u} \, du = \frac{\nu^{\nu-\frac{1}{2}}e^{-\nu}}{\Gamma(\nu)} \int_{0}^{1} \left(1 - \frac{t}{\sqrt{\nu}} \right)^{\nu - 1} e^{\sqrt{\nu}t} \mathbf{1}_{[0,\sqrt{\nu}]}(t)\, dt. $$

Now using the inequality $\log(1-x) \leq -x - \frac{x^2}{2}$ for $0 < x < 1$ and the quantitative form of the Stirling's approximation, we know that the integrand is uniformly bounded by

$$ \frac{\nu^{\nu-\frac{1}{2}}e^{-\nu}}{\Gamma(\nu)} \left(1 - \frac{t}{\sqrt{\nu}} \right)^{\nu - 1} e^{\sqrt{\nu}t} \mathbf{1}_{[0,\sqrt{\nu}]}(t) \leq e^{\frac{3}{2}} \sqrt{2\pi} e^{-t^2/2}, $$

which is integrable. So by the dominated convergence theorem and the Stirling's approximation, we have

$$ \lim_{\nu\to\infty} \frac{\nu^{\nu-\frac{1}{2}}e^{-\nu}}{\Gamma(\nu)} \int_{0}^{1} \left(1 - \frac{t}{\sqrt{\nu}} \right)^{\nu - 1} e^{\sqrt{\nu}t} \mathbf{1}_{[0,\sqrt{\nu}]}(t)\, dt = \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} e^{-t^2/2} \, dt = \frac{1}{2}. $$

dohmatob
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Sangchul Lee
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  • Note that $\int_0^k \frac{u^{k-1}e^{-u}}{(k-1)!}du = 1-e^{-k}e_{k-1}(k)$, where $e_n(x):=\sum_{s=0}^n \frac{x^s}{s!}$ is the exponential sum function http://mathworld.wolfram.com/IncompleteGammaFunction.html – dohmatob Jan 03 '20 at 09:46
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Let $I_n := (2\pi)^{-n/2}\int_{\|x\| \le \sqrt{n}}e^{-\|x\|^2/2}dx$. By equation (*) of this post https://math.stackexchange.com/a/3496441/168758, we have the identity $I_n = \frac{\gamma(n/2,n/2)}{\Gamma(n/2)}$, where $\gamma(a,x) := \int_{0}^xt^{a-1}e^{-t}dt$ is the incomplete gamma function. Now, use standard asymptotics results (Stirling's formulae for $\gamma$), to get that $I_n \longrightarrow 1/2$.

dohmatob
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$$ \begin{align} \lim_{n\to\infty}\int_{\|x\|^2\le n}\frac1{(2\pi)^{n/2}}e^{-\|x\|^2/2}\,\mathrm{d}x &=\lim_{n\to\infty}\frac1{(2\pi)^{n/2}}\int_0^{n^{1/2}}e^{-r^2/2}\omega_{n-1}r^{n-1}\,\mathrm{d}r\tag1\\ &=\lim_{n\to\infty}\frac1{2^{n/2}\Gamma(n/2)}\int_0^{n^{1/2}}e^{-r^2/2}r^{n-2}\,\mathrm{d}r^2\tag2\\ &=\lim_{n\to\infty}\frac1{\Gamma(n/2)}\int_0^{n/2}e^{-r}r^{n/2-1}\,\mathrm{d}r\tag3\\[3pt] &=\lim_{n\to\infty}\frac1{n!}\int_0^ne^{-r}r^n\,\mathrm{d}r\tag4\\[3pt] &=\lim_{n\to\infty}\left(\frac12-\frac{2/3}{\sqrt{2\pi n}}+O\!\left(\frac1n\right)\right)\tag5\\[3pt] &=\frac12\tag6 \end{align} $$ Explanation:
$(1)$: convert to polar coordinates
$(2)$: $\omega_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$
$(3)$: substitute $r\mapsto(2r)^{1/2}$
$(4)$: substitute $n\mapsto2n+2$
$(5)$: Equation $(11)$ from this answer
$(6)$: evaluate the limit

robjohn
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