자연계열 2번

(1) 모든 $\displaystyle k=1,\ldots,n$에 대하여 $\displaystyle \mathrm{E}(X_{k})=1/2$이므로

$\displaystyle \begin{aligned} \mathrm{E}(Z_{n}) &= \mathrm{E}\left(\frac{1}{2}X_{1}+\frac{1}{2^{2}}X_{2}+\cdots+\frac{1}{2^{n}}X_{n}\right) \\ &= \frac{1}{2}\mathrm{E}(X_{1})+\frac{1}{2^{2}}\mathrm{E}(X_{2})+\cdots+\frac{1}{2^{n}}\mathrm{E}(X_{n}) \\ &= \frac{1}{2}\left(\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n}}\right) \\ &= \frac{1}{2}\left(1-\left(\frac{1}{2}\right)^{n}\right) \end{aligned}$

(2) $\displaystyle Z_{n}$을 이진법으로 표시하면$\displaystyle Z_{n}=\frac{X_{1}}{2}+\frac{X_{2}}{2^{2}}+\cdots+\frac{X_{n}}{2^{n}}=\left(0.X_{1}X_{2}\cdots X_{n}\right)_{(2)}$이고,$\displaystyle \begin{aligned} \frac{1}{2}+\frac{1}{2^{m}} &= \left(0.X_{1}X_{2}\cdots X_{m-1}X_{m}X_{m+1}\cdots X_{n}\right)_{(2)} \\ &= \left(0.1\ 0\ \cdots\ 0\,1\,0\ \cdots\ 0\right)_{(2)} \end{aligned}$이다. 즉, 소수 첫 번째와 $\displaystyle m$번째 자리가 1이다.$\displaystyle Z_{n}\ge\frac{1}{2}+\frac{1}{2^{m}}$이려면 $\displaystyle X_{1}=1$이고 $\displaystyle X_{2},\cdots,X_{m}$ 중에 적어도 하나가 1 이어야 함. 따라서$\displaystyle \begin{aligned} \mathrm{P}\left(Z_{n}\ge\frac{1}{2}+\frac{1}{2^{m}}\right) &= \mathrm{P}(X_{1}=1)\left[1-\left\{\mathrm{P}(X_{2}=0)\cdots\mathrm{P}(X_{m}=0)\right\}\right] \\ &= \frac{1}{2}\left(1-\frac{1}{2^{m-1}}\right)=\frac{1}{2}-\frac{1}{2^{m}} \end{aligned}$

(3) $\displaystyle \mathrm{P}\left(\left|Z_{n}-\frac{1}{2}\right|<\frac{1}{2^{m}}\right)=1-\mathrm{P}\left(Z_{n}\ge\frac{1}{2}+\frac{1}{2^{m}}\right)-\mathrm{P}\left(Z_{n}\le\frac{1}{2}-\frac{1}{2^{m}}\right)$이므로, 먼저 확률 $\displaystyle \mathrm{P}\left(Z_{n}\le\frac{1}{2}-\frac{1}{2^{m}}\right)$을 계산하여 보자.$\displaystyle \begin{aligned} \frac{1}{2}-\frac{1}{2^{m}} &= \left(0.X_{1}X_{2}\cdots X_{m}X_{m+1}\cdots X_{n}\right)_{(2)} \\ &= \left(0.0\ 1\ \cdots\ 1\,0\ \cdots\ 0\right)_{(2)} \end{aligned}$이다. 즉, 소수 두 번째 자리부터 $\displaystyle m$번째 자리까지가 1이다. 그러므로, $\displaystyle Z_{n}\le\frac{1}{2}-\frac{1}{2^{m}}$이려면 $\displaystyle X_{1}=0$이고 $\displaystyle X_{2},\cdots,X_{m}$ 가운데 적어도 하나가 0 이거나, 혹은 $\displaystyle X_{2},\cdots,X_{m}$ 모두 1이고 $\displaystyle X_{m+1},\cdots,X_{n}$은 모두 0 이어야 함.

그러므로$\displaystyle \begin{aligned} \mathrm{P}\left(Z_{n}\le\frac{1}{2}-\frac{1}{2^{m}}\right) &= \mathrm{P}(X_{1}=0)\Big[\left(1-\left\{\mathrm{P}(X_{2}=1)\cdots\mathrm{P}(X_{m}=1)\right\}\right) \\ &\qquad+\left(\mathrm{P}(X_{2}=1)\cdots\mathrm{P}(X_{m}=1)\cdot\mathrm{P}(X_{m+1}=0)\cdots\mathrm{P}(X_{n}=0)\right)\Big] \\ &= \frac{1}{2}\left(1-\left(\frac{1}{2}\right)^{m-1}+\left(\frac{1}{2}\right)^{n-1}\right) \\ &= \frac{1}{2}-\left(\frac{1}{2}\right)^{m}+\left(\frac{1}{2}\right)^{n} \end{aligned}$따라서$\displaystyle \begin{aligned} \lim_{n\to\infty}\mathrm{P}\left(\left|Z_{n}-\frac{1}{2}\right|<\frac{1}{2^{m}}\right) &= \lim_{n\to\infty}\left\{1-\left(1-\frac{2}{2^{m}}+\frac{1}{2^{n}}\right)\right\} \\ &= \frac{1}{2^{m-1}} \end{aligned}$