統計検定準1級例題集解答/解答例と解説

論述問題問1 [1]

設問の要約

$K = 2$
$X$ : 2種類のシールの両方を手に入れるまでに購入する菓子の個数
$P(X = x)$ を求めよ．
$E[X]$ を求めよ．

解答例

$K = 2$ 種類のシールすべてを手にいれるまでに購入する菓子の個数を $X$ とする． $P(X = 0) = P(X = 1) = 0$ であり，また， $x \ge 2$ のとき，

P (X = x) = {(\frac{1}{2})}^{x - 1}

$\begin{equation} P(X = x) = \left(\frac{1}{2}\right)^{x-1} \end{equation}$

となることがわかる．期待値は，

\begin{aligned} E [X] & = \sum_{x = 0}^{\infty} x P (X = x) \\ = 0 \cdot P (X = 0) + 1 \cdot P (X = 1) + \sum_{x = 2}^{\infty} x P (X = x) \\ = \sum_{x = 2}^{\infty} x {(\frac{1}{2})}^{x - 1} \\ = \sum_{x = 0}^{\infty} x {(\frac{1}{2})}^{x - 1} - 0 {(\frac{1}{2})}^{0 - 1} - 1 {(\frac{1}{2})}^{1 - 1} \\ = \frac{1}{{(1 - \frac{1}{2})}^{2}} - 0 {(\frac{1}{2})}^{0 - 1} - 1 {(\frac{1}{2})}^{1 - 1} \\ = 4 - 0 - 1 \\ = 3 \end{aligned}

$\begin{align} E[X] &= \sum_{x=0}^{\infty}x\,P(X = x) \\ &= 0 \cdot P(X = 0) + 1 \cdot P(X = 1) + \sum_{x=2}^{\infty}x\,P(X = x) \\ &= \sum_{x=2}^{\infty}x\,\left(\frac{1}{2}\right)^{x-1} \\ &= \sum_{x=0}^{\infty}x\,\left(\frac{1}{2}\right)^{x-1} - 0\,\left(\frac{1}{2}\right)^{0-1} - 1\,\left(\frac{1}{2}\right)^{1-1} \\ &= \frac{1}{\left(1 - \frac{1}{2}\right)^2} - 0\,\left(\frac{1}{2}\right)^{0-1} - 1\,\left(\frac{1}{2}\right)^{1-1} \\ &= 4 - 0 - 1 \\ &= 3 \end{align}$

補足1: $\ds \sum_{k=0}^{\infty}k\,r^{k-1} = \frac{1}{(1 - r)^2}$

$|r| < 1$ とする．

\begin{aligned} \sum_{k = 0}^{\infty} a r^{k - 1} ＝ lim_{K \to \infty} \sum_{k = 0}^{K} a r^{k - 1} ＝ lim_{K \to \infty} \frac{a (1 - r^{K})}{1 - r} ＝ \frac{a}{1 - r} \end{aligned}

$\begin{align} \sum_{k=0}^{\infty}a\,r^{k-1} ＝ \lim_{K \to \infty}\sum_{k=0}^{K}a\,r^{k-1} ＝ \lim_{K \to \infty}\frac{a(1 - r^K)}{1 - r} ＝ \frac{a}{1 - r} \end{align}$

\begin{aligned} S & = \sum_{k = 0}^{\infty} k r^{k - 1} = 1 \cdot r^{0} + 2 \cdot r^{1} + 3 \cdot r^{2} + \dots \\ r S & = r \sum_{k = 0}^{\infty} k r^{k - 1} = 1 \cdot r^{1} + 2 \cdot r^{2} + 3 \cdot r^{3} + \dots \end{aligned}

$\begin{align} S &= \sum_{k=0}^{\infty}k\,r^{k-1} = 1 \cdot r^{0} + 2 \cdot r^{1} + 3 \cdot r^{2} + \cdots \\ r\,S &= r\sum_{k=0}^{\infty}k\,r^{k-1} = 1 \cdot r^{1} + 2 \cdot r^{2} + 3 \cdot r^{3} + \cdots \end{align}$

\begin{aligned} S - r S = r^{0} + r^{1} + r^{2} + r^{3} + \dots = \frac{1}{1 - r} \\ ⟹ (1 - r) S = \frac{1}{1 - r} \\ ⟹ S = \frac{1}{(1 - r)^{2}} \\ ⟹ \sum_{k = 0}^{\infty} k r^{k - 1} = \frac{1}{(1 - r)^{2}} \end{aligned}

$\begin{align} & S - r\,S = r^{0} + r^{1} + r^{2} + r^{3} + \cdots = \frac{1}{1 - r} \\ &\Longrightarrow (1 - r)\,S = \frac{1}{1 - r} \\ &\Longrightarrow S = \frac{1}{(1 - r)^2} \\ &\Longrightarrow \sum_{k=0}^{\infty}k\,r^{k-1} = \frac{1}{(1 - r)^2} \end{align}$

統計検定 準1級 例題集 解答/解答例と解説

論述問題 問1 [1]

設問の要約

解答例

補足1: ∑k=0∞krk−1=1(1−r)2\ds \sum_{k=0}^{\infty}k\,r^{k-1} = \frac{1}{(1 - r)^2}

統計検定準1級例題集解答/解答例と解説

論述問題問1 [1]

補足1: $\ds \sum_{k=0}^{\infty}k\,r^{k-1} = \frac{1}{(1 - r)^2}$