Mark Kac - Statistical Independence in Probability, Analysis and Number Theory

計算メモ.とても楽しい本です.

以下の書籍で知りました.


【目次】
第 1 章 ヴィエトから統計的独立性の概念へ
第 2 章 ボレルとその後
第 3 章 正規法則
第 4 章 素数は賽を振る
第 5 章 気体分子運動論から連分数へ

CHAPTER1 - FROM VIETA TO THE NOTION OF STATISTICAL INDEPENDENCE

A formula of Vieta

$\cos (2\theta) = 2\cos^{2}\theta - 1$より$\cos(\pi/2^{n+1})= \dfrac{\sqrt{2 + 2\cos(\pi/2^{n})}}{2}$だから
\begin{aligned}
& \cos(\pi / 2) = 0 \\
& \cos(\pi / 2^{2}) = \frac{\sqrt{2}}{2} \\
& \cos(\pi / 2^{3}) = \frac{\sqrt{2 + \sqrt{2}}}{2} \\
& \cos(\pi / 2^{4}) = \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} \\
& \quad \vdots
\end{aligned}

Another look at Vieta's formula

$m=0,1,2,...,2^{k - 1} - 1$に対して
\begin{aligned}
\epsilon_{k} (t)
&=
\begin{cases}
\, 0 & (2m / 2^{k} \leq t < (2m + 1) / 2^{k}) \\
\, 1 & ((2m + 1) / 2^{k} \leq t < (2m + 2) / 2^{k})
\end{cases} \\
\end{aligned}
\begin{aligned}
r_{k}(t)
&= 1 - 2\epsilon_{k} (t) \\
&=
\begin{cases}
\, 1 & (2m / 2^{k} \leq t < (2m + 1) / 2^{k}) \\
\, -1 & ((2m + 1) / 2^{k} \leq t < (2m + 2) / 2^{k})
\end{cases}
\end{aligned}

\begin{aligned}
t
& = \sum_{k = 1}^{\infty} \frac{\epsilon_{k}(t)}{2^{k}} \\
1 - 2t
& = \sum_{k = 1}^{\infty} \frac{1}{2^{k}} - 2 \sum_{k = 1}^{\infty} \frac{\epsilon_{k}(t)}{2^{k}} \\
& = \sum_{k = 1}^{\infty} \frac{r_{k}(t) }{2^{k}}
\end{aligned}

\begin{aligned}
\int_{0}^{1} e^{ix(1 - 2t)} \,\mathrm{d}t
&=e^{ix} \biggl[-\frac{e^{-2ixt}}{2ix}\biggr]_{0}^{1} \\
&=\frac{e^{ix} - e^{-ix}}{2ix} \\
&=\frac{\sin x}{x}
\end{aligned}

\begin{aligned}
& \int_{0}^{1} \exp\biggl[ix\frac{r_{k}(t)}{2^{k}}\biggr] \,\mathrm{d}t \\
&= \sum_{m=0}^{2^{k - 1} - 1} (e^{ix/2^{k}} + e^{- ix/2^{k}}) \cdot \frac{1}{2^{k}} \\
&= \cos(x/2^{k})
\end{aligned}

An accident or a beginning of something deeper?

$r_{k}(t)$が$\pm 1$の値をとる区間を合計すると,それぞれ$1/2$の長さになるから
\begin{aligned}
\int_{0}^{1} e^{ic_{k} r_{k}(t)} \,\mathrm{d}t
&= \frac{e^{ic_{k}} + e^{-ic_{k}}}{2}
= \cos c_{k}
\end{aligned}

$t \in (0, 1)$に対し

\begin{aligned}
& \biggl|\sum_{k=1}^{n} \frac{r_{k}(t)}{2^{k}} - (1 - 2t) \biggr| \\
& = \biggl|\sum_{k=n + 1}^{\infty} \frac{r_{k}(t)}{2^{k}} \biggr| \\
& \leq \sum_{k=n + 1}^{\infty} \frac{1}{2^{k}} = \frac{1/2^{n + 1}}{1 - 1/2} \to 0 \quad(n \to \infty)
\end{aligned}
より,$(0, 1)$上一様収束.

$(\frac{1}{2})^{n} = \frac{1}{2}\cdots \frac{1}{2}$ ($n$ times)

同じ$k$同士が隣り合うように並び替えれば
\begin{aligned}
& \prod_{k=1}^{n} e^{ic_{k}\delta_{k}} \cdot \prod_{k=1}^{n} \mu\{r_{k} (t) = \delta_{k}\} \\
&=\prod_{k=1}^{n} e^{ic_{k}\delta_{k}} \mu\{r_{k} (t) = \delta_{k}\}
\end{aligned}


次もよくやる式変形.因数分解したものと展開したものの関係.

\begin{aligned}
& \sum_{x(1)\in X_{1}, ..., x(n) \in X_{n}} \prod_{k=1}^{n} A_{x(k)} \\
& =\sum_{x(1)\in X_{1}, ..., x(n) \in X_{n}} A_{x(1)} A_{x(2)} \cdots A_{x(n)} \\
&= \biggl( \sum_{x(1)\in X_{1}} A_{x(1)} \biggr) \cdots \biggl( \sum_{x(n)\in X_{n}} A_{x(n)} \biggr) \\
&=\prod_{k=1}^{n} \biggl(\sum_{x(k)\in X_{k}} A_{x(k)} \biggr)
\end{aligned}

Heads or tails

$r_{1}(t), ..., r_{n}(t)$のうち$l$個が$+1$をとるとき,残りの$(n - l)$個は$-1$をとるから,
\begin{aligned}
& r_{1}(t)+ \cdots + r_{n}(t) \\
& = (+1) \cdot l + (- 1) \cdot (n - l) \\
& = 2l - n.
\end{aligned}

$\cos x = \dfrac{e^{ix} + e^{-ix}}{2}$より

\begin{aligned}
\cos^{n} x
&= \frac{1}{2^{n}} \sum_{k = 0}^{n} \binom{n}{k} e^{ikx} e^{-i(n - k)x} \\
&= \frac{1}{2^{n}} \sum_{k = 0}^{n} \binom{n}{k} e^{i(2k - n)x}
\end{aligned}
だから,
\begin{aligned}
& \frac{1}{2\pi} \int_{0}^{2\pi} e^{-i(2l-n)x} \cos^{n} x \,\mathrm{d}x \\
&= \frac{1}{2^{n}} \sum_{k = 0}^{n} \binom{n}{k}
\cdot \underbrace{\frac{1}{2\pi} \int_{0}^{2\pi} e^{i2(k - l)x} \,\mathrm{d}x}_{=\delta_{kl}} \\
&= \frac{1}{2^{n}} \binom{n}{l}
\end{aligned}

Independence and "Independence" Problems

Problems

1

$m=0,1,2,...,3^{k - 1} - 1$に対して
\begin{aligned}
\eta_{k} (t)
&=
\begin{cases}
\, 0 & (3m / 3^{k} \leq t < (3m + 1) / 3^{k}) \\
\, 1 & ((3m + 1) / 3^{k} \leq t < (3m + 2) / 3^{k}) \\
\, 2 & ((3m + 2) / 3^{k} \leq t < (3m + 3) / 3^{k})
\end{cases} \\
\end{aligned}

2

\begin{aligned}
\frac{\sin x}{x}
= \prod_{k = 1}^{\infty} \frac{1 + 2\cos\frac{2x}{3^{k}}}{3}
\end{aligned}
を示す.

\begin{aligned}
s_{k}(t)
& = 1 - \eta_{k} \\
&=
\begin{cases}
\, 1 & (3m / 3^{k} \leq t < (3m + 1) / 3^{k}) \\
\, 0 & ((3m + 1) / 3^{k} \leq t < (3m + 2) / 3^{k}) \\
\, -1 & ((3m + 2) / 3^{k} \leq t < (3m + 3) / 3^{k})
\end{cases}
\end{aligned}

\begin{aligned}
t
& = \sum_{k = 1}^{\infty} \frac{\eta_{k}(t)}{3^{k}} \\
1 - 2t
& = \sum_{k = 1}^{\infty} \frac{2}{3^{k}} - 2\sum_{k = 1}^{\infty} \frac{\eta_{k}(t)}{3^{k}} \\
& = 2\sum_{k = 1}^{\infty} \frac{s_{k}(t) }{3^{k}}
\end{aligned}

\begin{aligned}
\frac{\sin x}{x}
&= \int_{0}^{1} e^{ix(1 - 2t)} \,\mathrm{d}t \\
&= \int_{0}^{1} \exp\biggl(i 2x \sum_{k = 1}^{\infty} \frac{s_{k}(t) }{3^{k}} \biggr) \,\mathrm{d}t \\
&=\prod_{k = 1}^{\infty}
\underbrace{\int_{0}^{1} \exp\biggl(i 2x \frac{s_{k}(t) }{3^{k}} \biggr) \,\mathrm{d}t}_{=\frac{1}{3} (e^{-i2x/3^{k}} + 1 + e^{i2x/3^{k}} )} \\
&= \prod_{k = 1}^{\infty} \frac{1 + 2\cos(2x/3^{k})}{3}
\end{aligned}


一般化する.
$m=0,1,2,...,p^{k - 1} - 1$に対して
\begin{aligned}
&\xi_{k} (t) \\
&=
\begin{cases}
\, 0 & (pm / p^{k} \leq t < (pm + 1) / p^{k}) \\
\, 1 & ((pm + 1) / p^{k} \leq t < (pm + 2) / p^{k}) \\
\, \vdots \\
\, p - 1 & ((pm + p - 1) / p^{k} \leq t < (pm + p) / p^{k})
\end{cases} \\
\end{aligned}

$\sin x/ x$を出すために$1 - 2t $は変えられないから,先に$1 - 2t $の計算をして$u_{k}$を決めている.

\begin{aligned}
&u_{k}(t) = p - 1 - 2\xi_{k} (t) \\
&=
\begin{cases}
\, p - 1 & (pm / p^{k} \leq t < (pm + 1) / p^{k}) \\
\, p - 3 & ((pm + 1) / p^{k} \leq t < (pm + 2) / p^{k}) \\
\, \vdots \\
\, -(p - 1) & ((pm + p - 1) / p^{k} \leq t < (pm + p) / p^{k})
\end{cases} \\
\end{aligned}

\begin{aligned}
t
& = \sum_{k = 1}^{\infty} \frac{\xi_{k}(t)}{p^{k}} \\
1 - 2t
& = \sum_{k = 1}^{\infty} \frac{p-1}{p^{k}} - 2\sum_{k = 1}^{\infty} \frac{\xi_{k}(t)}{p^{k}} \\
& = \sum_{k = 1}^{\infty} \frac{u_{k}(t) }{p^{k}}
\end{aligned}

\begin{aligned}
\frac{\sin x}{x}
&= \int_{0}^{1} e^{ix(1 - 2t)} \,\mathrm{d}t \\
&= \int_{0}^{1} \exp\biggl(ix \sum_{k = 1}^{\infty} \frac{u_{k}(t) }{p^{k}} \biggr) \,\mathrm{d}t \\
&=\prod_{k = 1}^{\infty}
\int_{0}^{1} \exp\biggl(i x \frac{u_{k}(t) }{p^{k}} \biggr) \,\mathrm{d}t
\end{aligned}

\begin{aligned}
&\int_{0}^{1} \exp\biggl(i x \frac{u_{k}(t) }{p^{k}} \biggr) \,\mathrm{d}t \\
&=
\begin{cases}
\, \displaystyle
\frac{2}{p} \sum_{l = 0}^{p/2 - 1} \cos [x (2l + 1) / p^{k}]& (p: \text{even}) \\
\, \displaystyle
\frac{1}{p} \biggl\{ 1 + 2 \sum_{l = 1}^{(p - 1)/2} \cos [x \cdot 2l / p^{k}] \biggr\} & (p: \text{odd})
\end{cases}
\end{aligned}

3

$\delta_{i} \in \{-1, 1\}$とする.
\begin{aligned}
&\int_{0}^{1} r_{k_{1}} r_{k_{2}} \cdots r_{k_{s}} \,\mathrm{d}t \\
&= \sum_{\delta_{1},..., \delta_{s}}
\biggl[
\prod_{i = 1}^{s} \delta_{i}
\cdot \mu\{r_{k_{1}}(t) = \delta_{1}, \cdots , r_{k_{s}}(t) = \delta_{s} \} \biggr] \\
&= \sum_{\delta_{1},..., \delta_{s}}
\biggl[
\prod_{i = 1}^{s} \delta_{i}
\cdot \prod_{j = 1}^{s} \mu\{r_{k_{j}}(t) = \delta_{j}\} \biggr] \\
&= \sum_{\delta_{1},..., \delta_{s}}
\prod_{i = 1}^{s} \delta_{i} \cdot \mu\{r_{k_{i}}(t) = \delta_{i}\} \\
&= \prod_{i = 1}^{s}
\underbrace{\sum_{\delta_{i}} \delta_{i} \cdot \mu\{r_{k_{i}}(t) = \delta_{i}\}}_{=0} \\
&=0
\end{aligned}


4

\begin{aligned}
\int_{0}^{1} w_{0}(t) w_{0}(t) \,\mathrm{d}t
& =1.
\end{aligned}
$n \geq 1$のとき,$r_{k}^{2}(t) \equiv 1$より
\begin{aligned}
\int_{0}^{1} w_{n}(t) w_{n}(t) \,\mathrm{d}t
&=1.
\end{aligned}
$m \neq n$なら,$w_{m}(t) w_{n}(t) $を$r_{k}$の積で表したとき,各$r_{k}$は高々2回現れる.また,$m \neq n$だから,1回しか現れないものが必ず存在する.2回現れるものについては$r_{k}^{2}(t) \equiv 1$より
\begin{aligned}
& w_{m}(t) w_{n}(t)
= r_{k_{1}}(t) \cdots r_{k_{s}}(t) \\
& (k_{1} < \cdots < k_{s})
\end{aligned}
の形になる.よって,問題3. より
\begin{aligned}
\int_{0}^{1} w_{m}(t) w_{n}(t) \,\mathrm{d}t
&=0.
\end{aligned}


5


CHAPTER2 - BOREL AND AFTER

"Laws of large numbers"

\begin{aligned}
& \lim_{n \to \infty}
\sum_{|2l - n| > \epsilon n} \frac{1}{2} \binom{n}{l} \\
&=
\end{aligned}


$r_{i}^{2}(t) \equiv 1$より

\begin{aligned}
& \int_{0}^{1} \biggl[ \sum_{i = 1}^{n} r_{i}(t) \biggr]^{2} \,\mathrm{d}t \\
&= \sum_{i = 1}^{n} \int_{0}^{1} r_{i}^{2} (t) \,\mathrm{d}t
+ 2 \sum_{i < j} \underbrace{\int_{0}^{1} r_{i} (t) r_{j}(t) \,\mathrm{d}t}_{=0} \\
&=n
\end{aligned}

Borel and "normal numbers"

ゼロにならないのは$r_{i}^{4}$と$r_{i}^{2}r_{j}^{2}$の項だけだから,
\begin{aligned}
& \int_{0}^{1} \biggl[ \sum_{i = 1}^{n} r_{i}(t) \biggr]^{4} \,\mathrm{d}t \\
&= \sum_{i = 1}^{n} \int_{0}^{1} r_{i}^{4} (t) \,\mathrm{d}t
+ \sum_{i \neq j} \int_{0}^{1} r_{i}^{2} (t) r_{j}^{2} (t) \,\mathrm{d}t .
\end{aligned}
2項目の選び方は,($n$から2コの添字を選ぶ方法)×(4つの積から2コ選ぶ方法)だから
\begin{aligned}
\binom{n}{2} \binom{4}{2}
&= \frac{n(n-1)}{2} \frac{4\cdot 3}{2}
\end{aligned}
通り.

4乗ではなく2乗だと,$1/n$が出てきて収束が言えない.3乗だと$r_{i}$の奇数次しか現れないので積分がゼロになってしまう.

almost every number $t$ has (asymptotically!) the same number of zeros and ones in its binary expansion!

Mark Kac, "Statistical Independence in Probability, Analysis and Number Theory"



Problems

"Heads or Tails"—a more abstract formulation

What price abstraction?

Example 1. Convergence of series with random signs

Example 2 . Divergence of series with random signs

Problems