東大数理院試2025年度専門B問13解答

(1)
$$ E[Y_r] = \sum_{k \geq 1} kp(1 - p)^{k - 1} = \frac{p}{(1 - (1 - p))^2} = \frac{1}{p} $$
である．また
$$ E\bigg[ \bigg( \sum_{i = 1}^k X_i \bigg)^2 \bigg] = \sum_{1 \leq i, j \leq k} E[X_i X_j] = \sum_{i = 1}^k E[X_i^2] = k $$
より
$$ E[U_r^2] = \sum_{k \geq 1} E\bigg[ \bigg( \sum_{i = 1}^k X_i \bigg)^2 \bigg] P(Y_r = k) = \sum_{k \geq 1} kP(Y_r = k) = E[Y_r] = \frac{1}{p}. $$
(2)
$$ \sum_{\substack{1 \leq i \leq n \\ 1 \leq j \leq m}} E[X_i X_j] = \sum_{i = 1}^{\min\{ n, m\}} E[X_i^2] = \min\{ n, m\} $$
より
\begin{align*} E[U_r U_s] &= \sum_{n, m \geq 1} \min\{ n, m\} P(Y_r = n)P(Y_s = m) \\ &= \sum_{n \geq 1} nP(Y_r = n)^2 + 2\sum_{n < m} nP(Y_r = n)P(Y_s = m). \end{align*}
ここで右辺第$2$項は
$$ 2\sum_{n \geq 1} nP(Y_r = n)P(Y_s > n) = 2\sum_{n \geq 1} np(1 - p)^{n - 1} (1 - p)^n = 2p(1 - p)\sum_{n \geq 1} n(1 - p)^{2(n - 1)} $$
だから，
\begin{align*} E[U_r U_s] &= \sum_{n \geq 1} np^2 (1 - p)^{2(n - 1)} + 2p(1 - p)\sum_{n \geq 1} n(1 - p)^{2(n - 1)} \\ &= p(2 - p)\sum_{n \geq 1} n(1 - p)^{2(n - 1)} = \frac{p(2 - p)}{(1 - (1 - p)^2)^2} = \frac{1}{p(2 - p)}. \end{align*}
(3)
$$ S_n = \frac{1}{n} \sum_{r = 1}^n \sum_{i = 1}^{Y_r} X_i = \frac{1}{n} \sum_{i \geq 1} \# \{ r \in [1, n] \, ; \, Y_r \geq i\} X_i = \frac{1}{n} \sum_{i \geq 1} \sum_{r = 1}^n 1_{\{ Y_r \geq i\}} X_i $$
である．$\dis{T_{n, i} = \frac{1}{n} \sum_{r = 1}^n 1_{\{ Y_r \geq i\}}}$とおく．
$$ E[T_{n, i}] = E[1_{\{ Y_r \geq i\}}] = P(Y_r \geq i) = (1 - p)^{i - 1} $$
より
\begin{align*} E[|T_{n, i} - (1 - p)^{i - 1}|^2] &= V[T_{n, i}] = \frac{1}{n^2} \sum_{r = 1}^n V[1_{\{ Y_r \geq i\}}] \leq \frac{1}{n} E[1_{\{ Y_r \geq i\}}^2] \\ &= \frac{1}{n} E[1_{\{ Y_r \geq i\}}] = \frac{1}{n} P(Y_r \geq i) = \frac{1}{n} (1 - p)^{i - 1} \end{align*}
であるから，
\begin{align*} E\bigg[ \bigg| S_n - \sum_{i \geq 1} (1 - p)^{i - 1} X_i \bigg|^2 \bigg] &= \sum_{i \geq 1} E[|T_{n, i} - (1 - p)^{i - 1}|^2] \\ &\leq \sum_{i \geq 1} \frac{1}{n} (1 - p)^{i - 1} = \frac{1}{np} \to 0 \quad (n \to \infty). \end{align*}
(4)
確率変数$X$の特性関数を$\phi_X(t)$と書く．また$\phi(t) = \phi_{X_1}(t)$とする．
$$ \phi_{V_1 + V_r}(t) = E[e^{i(V_1 + V_r)t} | Y_r]P(Y_r < r) + E[e^{i(V_1 + V_r)t} | Y_r]P(Y_r \geq r) $$
の右辺第$2$項の絶対値は$\leq P(Y_r \geq r) = (1 - p)^{r - 1} \to 0 \, (r \to \infty)$であり，第$1$項は
\begin{align*} &\sum_{n < r, m \geq 1} E[e^{i(Y_1 + Y_r)t} | Y_1 = n, Y_r = m]P(Y_1 = n)P(Y_r = m) \\ =& \, \sum_{n < r, m \geq 1} \phi(t)^{n + m} p^2 (1 - p)^{n + m - 2} \to \sum_{n, m \geq 1} \phi(t)^{n + m} p^2 (1 - p)^{n + m - 2} \quad (r \to \infty) \\ =& \, \bigg( \sum_{n \geq 1} \phi(t)^n p(1 - p)^{n - 1} \bigg)^2 = \bigg( \frac{p\phi(t)}{1 - (1 - p)\phi(t)}\bigg)^2 \tag{$\ast$} \end{align*}
となる．
一方
\begin{align*} \phi_{U_1 + U_r}(t) &= \sum_{n, m \geq 1} E[e^{i(U_1 + U_r)t} | Y_1 = n, Y_r = m]P(Y_1 = n)P(Y_r = m) \\ &= \sum_{n, m \geq 1} \phi(2t)^{\min\{ n, m\}} \phi(t)^{\max\{n, m\} - \min\{ n, m\}} p^2 (1 - p)^{n + m - 2} \\ &= p^2 (1 - p)^{-2} \frac{(1 - p)^2 \phi(2t)}{1 - (1 - p)^2 \phi(2t)} \frac{1 + (1 - p)\phi(t)}{1 - (1 - p)\phi(t)} \\ &= \frac{p^2 \phi(2t)}{1 - (1 - p)^2 \phi(2t)} \frac{1 + (1 - p)\phi(t)}{1 - (1 - p)\phi(t)} \tag{$\ast'$} \end{align*}
である．ただし$a, b, c \in \CC \, (|a|, |b|, |c| < 1)$に対し
\begin{align*} &\sum_{n, m \geq 1} a^{\min\{ n, m\}} b^{\max\{ n, m\} - \min\{ n, m\}} c^{n + m} \\ =& \, \sum_{n \geq 1} (ac^2)^n + 2\sum_{n \geq 1} \sum_{m > n} a^n b^{m - n} c^{n + m} = \frac{ac^2}{1 - ac^2} + 2\sum_{n \geq 1} \frac{(bc)^{n + 1}}{1 - bc} (ab^{-1}c)^n \\ =& \, \frac{ac^2}{1 - ac^2} + \frac{2bc}{1 - bc} \frac{ac^2}{1 - ac^2} = \frac{ac^2}{1 - ac^2} \frac{1 + bc}{1 - bc} \end{align*}
となることを用いた．
今十分大きな任意の$r$に対し$V_1 + V_r$の分布と$U_1 + U_r$の分布が等しいとすると，任意の$t \in \RR$に対し$(\ast)$と$(\ast')$が等しいから
$$ \frac{\phi(t)^2}{1 - (1 - p)^2 \phi(t)^2} = \frac{\phi(2t)}{1 - (1 - p)^2 \phi(2t)} $$
となる．
ここで$\dis{\frac{x}{1 - (1 - p)^2 x}}$は一次分数変換ゆえ単射なので，$\phi(2t) = \phi(t)^2$を得る．これより
$$ \phi(t) = \phi(t / 2)^2 = \cdots = \phi(t / 2^n)^{2^n} = \bigg( 1 + \phi'(\theta_n)\frac{t}{2^n} \bigg)^{2^n} \to e^{\phi'(0)t} = 1 \quad (n \to \infty) $$
となる．ここで$|\theta_n| < |t| / 2^n.$よって$X_1 = 0$ a.s. となるが，これは$V[X_1] = 1$に矛盾．