# Normal distribution by successive approximation?

$$\newcommand\R{\mathbb R}\newcommand\la\lambda$$It is well known and easy to see that the rotationally invariant product of two probability measures on $$\R$$ has to be a Gaussian (or Dirac) measure; see e.g. this answer.

This appears to make the conjecture below somewhat plausible.

Let $$\mu$$ be any probability measure on $$\R^2$$ with a finite nonzero covariance matrix. Let $$\mu_1:=\mu$$. For each natural $$n$$, consider the following three-step procedure:

1. let $$\la_n:=\mu_n^{(1)}\otimes\mu_n^{(2)},$$ where $$\mu_n^{(1)}$$ and $$\mu_n^{(2)}$$ are the marginals of $$\mu_n$$;
2. let $$\nu_n:=\frac1{2\pi}\int_0^{2\pi}\la_n R_t\,dt,$$ where $$\la_n R_t$$ is the pushforward measure obtained from $$\la_n$$ by the rotation about the origin through angle $$-t$$;
3. let $$\mu_{n+1}$$ be obtained by rescaling the probability measure $$\nu_n$$ so that the covariance matrix of $$\mu_{n+1}$$ be the unit matrix.

Conjecture: $$\mu_n$$ converges weakly (as $$n\to\infty$$) to the standard Gaussian measure on $$\R^2$$.

Is this conjecture true?

Comment: Perhaps Step 3, the rescaling, is not essential. Of course, if we have the convergence to a (nondegenerate) Gaussian measure without Step 3, then we have such a convergence with Step 3 as well.

We can restate the problem (without rescaling) analytically as follows. Let $$f_n$$ denote the characteristic function of $$\nu_n$$, so that $$f_n(u,v)=\int_{\R^2}\nu_n(dx\times dy)e^{i(ux+uy)} \\ =\int_{\R^2}\nu_n(dx\times dy)\cos(ux+uy)\quad \text{(by symmetry)}$$ for all real $$u$$ and $$v$$. Then $$f_n(u,v)=g_n\big(\sqrt{u^2+v^2}\big)$$ for some function $$g_n\colon[0,\infty)\to\mathbb R$$ and all real $$u$$ and $$v$$. Then for all natural $$n$$ and all real $$r\ge0$$ $$g_{n+1}(r)=\frac2\pi\int_0^{\pi/2} dt\,g_n(r\cos t)g_n(r\sin t). \tag{1}$$ We want to show that $$g_n(r)\to e^{-c^2 r^2/2}$$ for some $$c\in(0,\infty)$$ and all real $$r\ge0$$.

So, analytically, the problem may be viewed as one of stability of a (nonlinear) integral equation or as one of solving such an integral equation by iterations.

Using substitutions $$g_n(r)=h_n(r^2)$$ and $$r^2=s$$, we can rewrite (1) as $$h_{n+1}(s)=\frac1\pi\int_0^s du\,\frac{h_n(s-u)}{\sqrt{s-u}}\frac{h_n(u)}{\sqrt{u}}$$ and then as $$\pi H_{n+1}(s)\sqrt s=[(H_n*H_n)(s)=]\int_0^s du\,H_n(s-u)H_n(u)$$ for all natural $$n$$ and all real $$s\ge0$$, where $$H_n(u):=h_n(u)/\sqrt u$$. We want to show that $$H_n(u)\to e^{-c^2 u/2}/\sqrt u$$ for some $$c\in(0,\infty)$$ and all real $$u>0$$.

• Commented May 31, 2021 at 14:46
• @AbdelmalekAbdesselam : Thank you for this reference. Both settings indeed involve iterations. I think that other setting is a case of the central limit theorem for iid summands and $2^k$ summands, for natural $k$. Here the setting involves a different kind of iterations. But presumably/hopefully in this setting too one has normality in the limit. Commented May 31, 2021 at 15:03
• Of course the iterations are different, but the methods used for one might work for the other. It would take me some time to work it out, but the renormalization group inspired strategy I would use for your questions is: 1) write $g$ as the Gaussian times $(1+h)$ and write the iteration for the function $h$, 2) write the linearization of the transformation at the fixed point $h=0$, 3) see if you can diagonalize it explicitly using suitable orthogonal polynomials for the integral over $t$. Ultimately, one would need a Lyapunov function like some kind of entropy. Does Stein's method help? Commented May 31, 2021 at 15:11
• @AbdelmalekAbdesselam : Thank you for your suggestions. I see your point better now, and will have these suggestions in mind. Commented May 31, 2021 at 15:50
• I just did a quick computation which gives support to your conjecture. Functions $h_k(r)=r^k$ are eigenfunctions of the linearization (no need for orthogonal polynomials like Hermite etc.) with eigenvalues $c_k=(4/\pi)\times W_k$ in terms of Wallis integrals in the notations of en.wikipedia.org/wiki/Wallis%27_integrals The only expanding/relevant directions are for $k=0,1$, while $k=2$ is neutral/marginal. Finally for $k>2$ , the corresponding directions are contracting. This is exactly the same as in the RG link I mentioned. One can mimick Koralov-Sinai and make this into a proof... Commented May 31, 2021 at 16:05

In the rewritten form it doesn't require any ingenuity at all. For brevity, introduce the notation $$(Th)(s)=\frac 1\pi\int_0^s\frac{h(s-u)h(u)}{\sqrt{u(s-u)}}\,du=\frac 1\pi\int_0^1\frac{h(s(1-v))h(sv)}{\sqrt{v(1-v)}}\,dv$$ Suppose that $$h_0$$ is any real valued Lipschitz function on $$[0,+\infty)$$ (the finiteness of the second moment guarantees Lipschitzness for your $$h_0$$ and, if you really want to discuss it, I can show how this particular condition can be dropped) such that $$\|h_0\|_\infty=h_0(0)=1$$ and there exists $$h_0'(0)=-c< 0$$ (the latter two conditions are essential). Then the iterations $$h_{n+1}=Th_n$$ converge to $$e^{-cx}$$ uniformly on compact subsets of $$[0,+\infty)$$.

The proof consists of several easy observations:

1. $$\|Th\|_\infty\le\|h\|_\infty^2$$ and, thereby, $$\|h_n\|_\infty\le 1$$ for all $$n$$.

2. If $$h$$ is bounded and $$L$$-Lipschitz, then $$Th$$ is $$L\|h\|_\infty$$-Lipschitz.

Indeed, $$Th(s)-Th(S)=\frac1\pi\int_0^1\frac{dv}{\sqrt{v(1-v)}}[h(s(1-v))h(sv)-h(S(1-v))h(Sv)]$$ and $$|h(s(1-v))h(sv)-h(S(1-v))h(Sv)| \\ \le\|h\|_\infty[|h(s(1-v))-h(S(1-v))|+|h(sv)-h(Sv)|] \\ \le\|h\|_\infty L[|s-S|(1-v+v)]=\|h\|_\infty L|s-S|\,.$$ Thus all $$h_n$$ are Lipschitz with the same Lipschitz constant $$L$$ as $$h_0$$.

1. If $$A\in\mathbb R$$ and $$h(s)\ge e^{-As}$$ on $$[0,s_0]$$, then $$(Th)(s)\ge e^{-As}$$ on $$[0,s_0]$$. 3') If $$a\in\mathbb R$$ and $$0\le h(s)\le e^{-as}$$ on $$[0,s_0]$$, then $$(Th)(s)\le e^{-as}$$ on $$[0,s_0]$$.

(all exponential functions are fixed points of $$T$$ and we have monotonicity as long as $$h$$ stays non-negative)

1. Now choose any $$0 and choose $$s_0>0$$ so that $$e^{-As}\le h_0(s)\le e^{-as}$$ on $$[0,s_0]$$ (by the derivative at zero condition such $$s_0$$ exists).

Consider the largest $$S_n$$ such that $$e^{-Bs}\le h_n(s)\le e^{-bs}$$ on $$[0,S_n]$$. We have $$S_0\ge s_0$$ and $$S_{n+1}\ge S_n$$. We want to improve the latter trivial inequality to some quantitative advance $$S_{n+1}\ge S_n+\delta(S_n)$$ where $$\delta>0$$ is separated from $$0$$ on any compact subinterval of $$[s_0,+\infty)$$. That is quite easy: $$h_{n+1}(S_n)=(Th_n)(S_n)=\frac1\pi\int_0^{S_n}\frac{h_n(S_n-u)h_n(u)}{\sqrt{u(S_n-u)}}\,du\ge \\ \frac1\pi\int_0^{S_n}\frac{e^{-B(S_n-u)}e^{-Bu}}{\sqrt{u(S_n-u)}}\,du +\frac1\pi\int_0^{s_0}\frac{e^{-B(S_n-u)}[e^{-Au}-e^{-Bu}]}{\sqrt{u(S_n-u)}}\,du \\ \ge e^{-BS_n}+e^{-BS_n}S_n^{-1}\int_0^{s_0}[e^{-Au}-e^{-Bu}]\,du=e^{-BS_n}+\Delta(S_n)$$ Then, by the Lipschitz property of both $$h_{n+1}$$ and $$e^{-Bs}$$, the inequality $$h_{n+1}(s)\ge e^{-Bs}$$ persists on $$[S_n,S_n+\delta(S_n)]$$ with $$\delta(S_n)=\frac{\Delta(S_n)}{B+L}$$, say. The extension of the upper bound is similar.

The outcome is that the double inequality $$e^{-Bs}\le h_n(s)\le e^{-bs}$$ propagates from $$[0,s_0]$$ to the entire real line. Since $$0 were arbitrary, we conclude that $$h_n$$ tend to $$e^{-cs}$$ uniformly on compact intervals, finishing the story.

• Thank you for your answer. The solution is indeed when it's done. :-) I have fixed some typos. Commented Jan 17, 2023 at 2:09
• Do you want to try the problem at mathoverflow.net/q/438390/36721 ? Commented Jan 17, 2023 at 2:11
• @IosifPinelis Yes, if you or somebody else translates it into the language I can understand (I know neither what copula is, nor what the Frechet-Hoeffding upper bound refers to and google search is definitely not my favorite thing) Commented Jan 17, 2023 at 2:16
• A copula is the joint cdf of two uniformly distributed random variables. The Frechet--Hoeffding upper bound is, I think, the copula $C(u,v)=\min(u,v)$ for $u,v$ in $[0,1]$. Commented Jan 17, 2023 at 3:32
• I meant to say, a copula is the joint cdf of two random variables each uniformly distributed on $[0,1]$. Commented Jan 17, 2023 at 4:08