Computer Scientists Solve Kadison-Singer Problem | Quanta Magazine

Computer Scientists Solve Kadison-Singer Problem | Quanta Magazine

On a class of random walks

$\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bR}{\mathbb{R}}$  $\newcommand{\bE}{\mathbb{E}}$ $\newcommand{\eC}{\mathscr{C}}$ $\newcommand{\bP}{\mathbb{P}}$  $\newcommand{\eF}{\mathscr{F}}$ $\newcommand{\whS}{\widehat{S}}$ $\DeclareMathOperator{\var}{\boldsymbol{var}}$ $\newcommand{\bC}{\mathbb{C}}$

I heard   of these random walks from my number-theorist colleague Andrei Jorza who  gave me a probabilistic translation of some    observations in number theory. What follows is rather a variation  on  that theme.



There is a whole family of such walks, parametrized by a complex number  $\newcommand{\ii}{\boldsymbol{i}}$ $\rho=\cos \theta +\ii \sin\theta$, $\theta\in [0,\pi]$, $\ii=\sqrt{-1}$.

This is  discrete time walk  on  the configuration space  $\newcommand{\lan}{\langle}$ $\newcommand{\ran}{\rangle}$

$$\eC_\rho:= L[\rho] \times  C(\rho,-\rho),$$

where $C(\rho,-\rho)$ is the subgroup of $S^1$ generated by $\pm\rho$, i.e.,

$$C(\rho,-\rho)=\bigl\{\; \pm \rho^k;\;\;k\in\bZ \;\bigr\},$$

and $L[\rho]\subset  \bC$ is the additive subgroup of $\bC$ generated by the elements $\rho^n$, $n\in\bZ_{\geq 0}$.

The state of the walk at time $n$ is given  by a pair of random variables $(X_n,  V_n)$, where one should think of $X_n\in L[\rho]$  as describing the "position'' and $V_n\in G(\rho,-\rho)$ as describing the ``velocity''   at time $n$.

Here is how the random walk takes place.  Suppose that at time $n$ we are in the configuration $(X_n, V_n)$. To decide where to go next, toss a fair coin.   If a  the Head shows up  then

$$V_{n+1}=\rho V_n,\;\; X_{n+1}= X_n+V_{n+1}.$$

Otherwise,

$$V_{n+1}=-\rho V_n,\;\; X_{n+1}= X_n+V_{n+1}.$$

More abstractly, we are given a probability space $(\Omega,\eF,\bP)$ and a sequence of  independent, identically distributed variables

$$S_n:\Omega\to\{-1,1\},\;\;\bP[S_n=\pm 1]=\frac{1}{2},\;\;n\in\bZ_{\geq 0}.$$

If we set
$$\whS_n:=\prod_{k=1}^nS_k,$$
then we have
$$V_n= \whS_n \rho^n,\;\;X_n=\sum_{k=1}^n V_n=\sum_{k=1}^n \whS_k\rho_k.$$
Observing that

$$\bE[\whS_k]=0,\;\;\var[\whS_k]=\prod_{j=1}^k\var[S_j]=1,$$ we deduce

$$\bE[X_n]=0,\;\;\var[S_n]=\bE[X_n\bar{X}_n]$$

$$=\sum_{j,k=1}^n \bE[ \whS_j\whS_k]\rho^{k-j}=\sum_{j=1}^n \bE[ \whS_j^2]+2\sum_{1\leq j<k\leq}^n \bE[ \whS_j\whS_k]\rho^{k-j}=n.$$

Example 1.   Suppose that $\theta=\frac{\pi}{2}$ so that $\rho=\ii$. In this case $L[\rho]$ is the lattice $\bZ^2\subset \bC$ and $G(\ii,-\ii)$ is the group  of forth  order roots of $1$.  At time $n=0$,  the moving particle is at the origin facing  East,. turns right or left  (in this case North/South) with equal probability. Once   reaches a new intersection, it turns  right/left  with equal probability.  The following MAPLE generated animation  a loop describes a 90-step   portion of one such walk.

This is somewhat typical   of  what one observes in general.


Example 2.   Suppose that  $\theta  =\frac{\pi}{3}$. In this case $G(\rho,-\rho)$ is the group of    6th order roots of  $1$ and $L[\rho]$ is a lattice in $\bC$.   The following  MAPLE generated   animation depicts a  90 step portion of such a walk and the patterns you observe are typical.

Example 3.    Assume   that the angle $\theta$ is small and $\frac{\theta}{\pi}$ is not rational,  say $\cos \theta=0.95$. The animation below  depicts a 90-step stretch of such a walk.

Example 4.   For comparison, I have included an animation of a 90-step segment of the usual uniform random walk on $\bZ^2$.

Clearly, I am interested to  figure out what is happening and I have a hard time asking a good question. A first  question  that comes to my mind would be to find   statistical invariants  that would distinguish the  random walk in Example 1 from the random walk in Example  4.    I will refer to them as $RW_1$ and $RW_4$

 Clear each of these two  walks   surround  many unit squares with vertices in the lattice $\bZ^2$.  For $j=1,4$ will denote by $s_j(n)$ the expected number of unit squares      surrounded by a  walk of $RW_j$ of length $n$.. Which grows faster as $n\to\infty$, $s_1(n)$ or $s_4(n)$?

I will probably update this posting as I get more ideas, but maybe this shot in the dark will initiate a conversation and will suggest other, better questions.

   

IUT Theory of Shinichi Mochizuki | Clay Mathematics Institute

IUT Theory of Shinichi Mochizuki | Clay Mathematics Institute

Random subdivisions of an interval

This is a classical problem, but I find it very appealing.

Drop $n$  points $X_1,\dotsc, X_n$  at random in the unit interval, uniformly and independently. These  points     divide the interval into $n+1$  sub-intervals.  Their lengths are called the spacings of the subdivision. What is the expected length of the longest of these intervals? What is its asymptotic behavior for $n$ large?

I was surprised to find out that the  problem  of finding  the expectation of the largest segment   was solved   in late 19th century.    What follows   is a more detailed presentation  of the arguments in Sec. 6.3 of the book  Order Statistics   by H.A. David and  H.N. Nagaraja.

If you read French, you can click here to see what P. Levy had to say about this problem back in 1939.

Relabel  by $Y_1,\dotsc, Y_n$, the points $\{X_1,\dotsc, X_n\}\subset [0,1]$ so that  $Y_1\leq \cdots \leq Y_n$.  We set

$$Y_0=0,\;\;Y_{n+1}=1,\;\;\;\;L_n= \max_{1\leq i\leq n+1} Y_i-Y_{i-1},.$$

We want to determine  the expectation $\newcommand{\bE}{\mathbb{E}}$ $E_n$ of $L_n$.

The probability distribution  of $\vec{X}=(X_1,\dotsc, X_n)$  $\newcommand{\bone}{\boldsymbol{1}}$   is

$$p_{\vec{X}}(dx)=\bone_{C_n}(x_1,\dotsc, x_n) dx_1\cdots dx_n, \;\;C_n=[0,1]^n,$$

where $\bone_A$ denotes the indicator function of a subset $A$ of a given set $S$.

The probability distribution of $\vec{Y}=(Y_1,\dotsc, Y_n)$ is

$$p_{\vec{Y}}(dy)= n! \bone_{T_n}(y_1,\dotsc, y_n) dy_1\cdots dy_n,$$

where $T_n$ is the simplex $\newcommand{\bR}{\mathbb{R}}$

$$T_n=\bigl\{ y\in\bR^n;\;\;0\leq y_1\leq \cdots \leq  y_n\leq 1\,\bigr\}.$$

The  spacings  are the lengths of the  $n+1$ segments that the    points $Y_1,\dotsc, Y_n$ determine  on $[0,1]$
$$S_i= Y_{i}-Y_{i-1},\;\; i=1,\dotsc, n+1,$$
where $Y_{n+1}:=1$. The resulting map $\vec{Y}\to\vec{S}=\vec{S}(\vec{Y})$ induces a linear bijection

$$T_n\to \Delta_n=\bigl\{\, (s_1,\dotsc, s_{n+1})\in\bR_{\geq 0}^{n+1};\;\;s_1+\cdots+s_{n+1}=1\,\bigr\}.$$

This  shows that the probability density is $\DeclareMathOperator{\vol}{vol}$

$$p_{\vec{S}}(ds) =\frac{1}{\vol(\Delta_n)} dV_{\Delta_n},$$

where $dV_{\Delta_n}$ denote the usual  Euclidean volume density on  the standard simplex $\Delta_n$. This description  shows that  the random vector of spacings  $\vec{S}$ is exchangeable, i.e., the probability  density  of $\vec{S}$ is invariant under permutation of the components. Thus the  joint probability distribution of any group  of $k$ components of $\vec{S}$,   $k\leq n$, is the same as the joint distribution of the first $k$-components $S_1,\dotsc, S_k$.

The joint  probability distribution of the first $n$ components is

$$p_n(ds)= n! \bone_{D_n}(s_1,\dotsc,s_n) ds,$$

where $D_n$ is the simplex

$$D_n=\{\, (s_1,\dotsc, s_n)\in\bR^n_{\geq 0};\;\;s_1+\cdots +s_n\leq 1\,\}.$$

Indeed, $(S_1,\dotsc, S_n)\in D_n$ and

$$ds=|ds_1\cdots  ds_n|=|d y_1\wedge d(y_2-y_{1})\wedge  \cdots \wedge d(y_{n}-y_{n-1})|=|dy_1\cdots  d y_n|.$$

The  joint density of the first $n-1$ components $(S_1, \dotsc, S_{n-1})$ is

$$p_{n-1}(s_1,\dotsc, s_{n-1}) =  n! \int_0^\infty \bone_{D_n}(s_1,\dotsc,  s_n)  ds_n$$

$$=  n! \bone_{D_{n-1}}(s_1,\dotsc,  s_{n-1})\int_0^{1-s+1-\cdots-s_{n-1}}  ds_n= n! \bone_{D_{n-1}}(s_1,\dotsc,  s_{n-1})(1-s_1-\cdots-s_{n-1})$$

The joint density of the first $(n-2)$ components is

$$p_{n-2}(s_1,\dots, s_{n-2})=\int_0^\infty p_{n-1}(s_1,\dotsc, s_{n-1}) ds_{n-1}$$

$$=  n!\bone_{D_{n-2}}(s_1,\dotsc, s_{n-2})\int_0^{1-s_1-\cdots-s_{n-2}} (1-s_1-\cdots -s_{n-1}) d s_{n-1} =  \frac{n!}{2!} \bone_{D_{n-2}}(s_1,\dotsc, s_{n-2})(1-s_1-\cdots -s_{n-2})^2.$$

Iterating, we deduce that the joint probability density of the the first $k$ variables is

$$p_k(s_1,\dotsc, s_k) =\frac{n!}{(n-k)!}p_{D_k}(s_1,\dotsc, s_k) (1-s_1-\cdots-s_k)^{n-k}.$$

If $1\leq k\leq n$,  then$\newcommand{\bP}{\mathbb{P}}$

$$\bP[S_1>s,\dotsc, S_k>s]=\frac{n!}{(n-k)!}\int_s^\infty\cdots\int_s^\infty p_{D_k}(s_1,\dotsc, s_k) (1-s_1-\cdots-s_k)^{n-k} ds_1\dotsc ds_k$$

$$=\frac{n!}{(n-k)!}\int_s^1ds_1\int_s^{1-s_1}ds_2\cdots\int_s^{1-s_1-\cdots-s_{k-1}}  (1-s_1-\cdots-s_k)_+^{n-k} ds_k =(1-ks)_+^n,\;\;x_+:=\max(x,0).$$

Also

$$\bP[S_1>s,\dotsc, S_n>s, S_{n+1}>s] = \bP[S_1>s,\dotsc, S_n>s, S_1+S_2+\cdots +S_n\leq 1-s]$$

$$=n!\int_{\substack{s_1,\dotsc s_{n}>s\\ s_1+\cdots +s_{n}\leq 1-s}}ds_1\cdots d s_{n}$$

$$=n!\int_{\substack{s_1,\dotsc s_{n-1}>s\\ s_1+\cdots +s_{n-1}\leq 1-2s}}ds_1\cdots d s_{n-1}\int_s^{1-s-s_1-\cdots -s_{n-1}} d s_n$$

$$= n!\int_{\substack{s_1,\dotsc s_{n-1}>s\\ s_1+\cdots +s_{n-1}\leq 1-2s}} (1-2s-s_1-\cdots -s_{n-1}) ds_1\cdots  ds_{n-1}$$

$$=n!\int_{\substack{s_1,\dotsc s_{n-2}>s\\ s_1+\cdots +s_{n-2}\leq 1-3s}}ds_1\cdots ds_{n-2}\int_s^{1-2s-s_1-\cdots -s_{n-2}} (1-2s-s_1-\cdots -s_{n-1})  d s_{n-1}$$


$$=\frac{n!}{2!} \int_{\substack{s_1,\dotsc s_{n-2}>s\\ s_1+\cdots +s_{n-2}\leq 1-3s}} (1-3s-s_1-\cdots -s_{n-2})^2 ds_1\cdots ds_{n-2}$$

$$=\frac{n!}{2!} \int_{\substack{s_1,\dotsc s_{n-3}>s\\ s_1+\cdots +s_{n-3}\leq 1-4s}}  ds_1\cdots ds_{n-3} \int_s^{1-3s -s_1-\cdots -s_{n-3}} (1-3s-s_1-\cdots -s_{n-2})^2ds_{n-2}$$

$$=\frac{n!}{3!} \int_{\substack{s_1,\dotsc s_{n-3}>s\\ s_1+\cdots +s_{n-3}\leq 1-4s}}  (1-4s-s_1-\cdots -s_{n-3})^3ds_1\cdots ds_{n-3}$$

$$=\cdots = \bigl(\, 1-(n+1)s\,\bigr)_+^n.$$

With

$$L_n:=\max_{1\leq i\leq n+1} S_i$$

 we have


$$\bP[L_n>s]=\bP\left[\,\bigcup_{i=1}^{n+1}\bigl\lbrace\, S_i>s\,\bigr\rbrace\,\right]$$

(use inclusion-exclusion  exchangeability)

$$= \sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k} \bP\bigl[\, S_1>s,\dotsc, S_k>s\,\bigr] =\sum_{k=1}^{n+1} \binom{n+1}{k}(1-ks)_+^n .$$

The last equality is sometime known as Whitworth  formula and it goes all the way back to 1897. We have

$$E_n=\bE[L_n]=\int_0^\infty \bP[L_n>s] ds =\sum_{k=1}^{n+1}(-1)^{k+1} \binom{n+1}{k}\int_0^\infty (1-ks)_+^n  ds = \sum_{k=1}^{n+1}(-1)^{k+1} \binom{n+1}{k}\int_0^{1/k}(1-ks)^n  ds$$

$$=\frac{1}{n+1} \sum_{k=1}^{n+1}(-1)^{k+1} \frac{1}{k} \binom{n+1}{k}=\sum_{k=1}^{n+1}(-1)^{k+1} \frac{1}{k^2} \binom{n}{k-1}.$$

Here  is  $E_n$ answers for  small $n$ (thank you MAPLE)


$$E_1=\frac{3}{4},\;\;E_2=\frac{11}{18},\;\;E_3=\frac{25}{48},\;\;E_4=\frac{137}{300},\;\; E_5=\frac{49}{120}.$$


It turn out that $L_n$ is highly concentrated around its mean. In the work of P. Levy I mentioned earlier he showed  that the random variables   $nL_n-\log n$ converges in probability   to a random  variable $X_\infty$   with probability  distribution  $\mu$ given by

$$\mu\bigl(\,(-\infty,x])= \exp\bigl(\,-e^{-x}\,\bigr),\;\;x\geq 0.$$

 Moreover

$$\lim_{n\to \infty} \bE[\, nL_n-\log n\,]=\gamma=\bE[X_\infty],\;\;\lim_{n\to \infty}\mathrm{Var}[ nL_n\,]=\frac{\pi^2}{6}=\mathrm{Var}[X_\infty].$$

$$\frac{n}{\log n} L_n \to 1\;\; \mbox{a.s.}$$

Note that this shows that

$$E_n\sim\frac{\log n}{n}\;\; \mbox{as $n\to \infty$}.$$

I'll stop here, but you can learn more from this paper of Luc Devroye and its copious   list of references.

Denote by $A_n$ the smallest spacing

$$A_n:=\min_{1\leq k\leq n+1} S_k.$$ .

Note that 

$$\bP[ A_n>a]=\bP[ S_1,S_2,\dotsc, S_{n+1}>a]=\big( 1-(n+1)a\big)^n_+.$$

Hence 

$$\bE[ A_n]=\int_0^\infty\bP[A_n>a]  da= \int_0^\infty \big( 1-(n+1)a\big)^n_+ da=\int_0^{\frac{1}{n+1}} (1-(n+1)a)^n  da$$

$$= \frac{1}{n+1}\int_0^1 (1-t)^n dt= \frac{1}{(n+1)^2}.$$

The Luzin affair

I was not aware that Luzin was persecuted by Stalin and some of his detractors were big names like Kolmogorov and Sobolev.



"Aleksandrov, Kolmogorov and some other students of Luzin accused him of plagiarism and various forms of misconduct. Sergei Sobolev and Otto Schmidt incriminated him with charges of disloyalty to Soviet power. "



Nikolai Luzin - Wikipedia, the free encyclopedia