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