Skip to content

北海道大学 情報科学院 情報科学専攻 情報理工学コース 2022年8月実施 専門科目1 問3 (確率・統計)

Author

Miyake

Description

Kai

[1]

(1)

\(x=2\) における \(F(x)\) の連続性より、

\[ \begin{aligned} k \cdot 2^3 &= 1 \\ \therefore \ \ k &= \frac{1}{8} \end{aligned} \]

がわかる。

(2)

\[ \begin{aligned} P \left( 0 \lt X \leq 1 \right) &= F(1) - F(0) \\ &= \frac{1}{8} \end{aligned} \]

(3)

\[ \begin{aligned} f(x) &= \frac{d}{dx} F(x) \\ &= \begin{cases} 0 &(x \lt 0) \\ \frac{3}{8} x^2 &(0 \leq x \lt 2) \\ 0 &(2 \leq 2) \end{cases} \end{aligned} \]

(4)

(5)

\[ \begin{aligned} \mathrm{E} \left[ X \right] &= \int_{-\infty}^\infty x f(x) dx \\ &= \frac{3}{8} \int_0^2 x^3 dx \\ &= \frac{3}{8} \left[ \frac{x^4}{4} \right]_0^2 \\ &= \frac{3}{2} \\ \mathrm{E} \left[ X^2 \right] &= \int_{-\infty}^\infty x^2 f(x) dx \ \ \ \ \ \ \ \ \left( X^2 \text{ の期待値 } \right) \\ &= \frac{3}{8} \int_0^2 x^4 dx \\ &= \frac{3}{8} \left[ \frac{x^5}{5} \right]_0^2 \\ &= \frac{12}{5} \\ \mathrm{Var} \left[ X \right] &= \mathrm{E} \left[ X^2 \right] - \mathrm{E} \left[ X \right]^2 \\ &= \frac{3}{20} \end{aligned} \]

[2]

\(X_1, X_2, \cdots, X_n\) は独立であり、

\[ \begin{aligned} \mathrm{E} \left[ X_i \right] = \mu , \ \ \mathrm{E} \left[ X_i^2 \right] = \sigma^2 + \mu^2 \ \ \ \ \ \ \ \ (i = 1, 2, \cdots, n) \end{aligned} \]

である。

(1)

\[ \begin{aligned} \mathrm{E} \left[ \bar{X} \right] &= \frac{1}{n} \sum_{i=1}^n \mathrm{E} \left[ X_i \right] \\ &= \mu \end{aligned} \]

(2)

\[ \begin{aligned} \mathrm{E} \left[ \left( \bar{X} \right)^2 \right] &= \frac{1}{n^2} \sum_{i=1}^n \sum_{j=1}^n \mathrm{E} \left[ X_i X_j \right] \\ &= \frac{1}{n^2} \left( \sum_{i=1}^n \mathrm{E} \left[ X_i^2 \right] + \sum_{i,j \ (i \ne j)} \mathrm{E} \left[ X_i X_j \right] \right) \\ &= \frac{1}{n^2} \left( \sum_{i=1}^n \mathrm{E} \left[ X_i^2 \right] + \sum_{i,j \ (i \ne j)} \mathrm{E} \left[ X_i \right] \mathrm{E} \left[ X_j \right] \right) \\ &= \frac{1}{n^2} \left( n \left( \sigma^2 + \mu^2 \right) + \left( n^2 - n \right) \mu^2 \right) \\ &= \frac{\sigma^2}{n} + \mu^2 \end{aligned} \]

(3)

\[ \begin{aligned} T_1 &= \frac{1}{n-1} \sum_{i=1}^n \left( X_i - \bar{X} \right)^2 \\ &= \frac{1}{n-1} \sum_{i=1}^n \left( X_i^2 - 2 \bar{X} X_i + \left( \bar{X} \right)^2 \right) \\ &= \frac{1}{n-1} \left( \sum_{i=1}^n X_i^2 - n \left( \bar{X} \right)^2 \right) \\ &= \frac{1}{n-1} \sum_{i=1}^n X_i^2 - \frac{n}{n-1} \left( \bar{X} \right)^2 \\ \therefore \ \ \mathrm{E} \left[ T_1 \right] &= \frac{1}{n-1} \sum_{i=1}^n \mathrm{E} \left[ X_i^2 \right] - \frac{n}{n-1} \mathrm{E} \left[ \left( \bar{X} \right)^2 \right] \\ &= \frac{n}{n-1} \left( \sigma^2 + \mu^2 \right) - \frac{n}{n-1} \left( \frac{\sigma^2}{n} + \mu^2 \right) \\ &= \sigma^2 \end{aligned} \]

(4)

\[ \begin{aligned} T_2 &= \left( \bar{X} \right)^2 - \frac{1}{n} T_1 \\ \therefore \ \ \mathrm{E} \left[ T_2 \right] &= \mathrm{E} \left[ \left( \bar{X} \right)^2 \right] - \frac{1}{n} \mathrm{E} \left[ T_1 \right] \\ &= \frac{\sigma^2}{n} + \mu^2 - \frac{1}{n} \sigma^2 \\ &= \mu^2 \end{aligned} \]

(5)