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第2講確率分布

ベルヌーイ分布

X \sim B e r (p)

$\begin{equation} X \sim Ber(p) \end{equation}$

P (X = k) = p^{k} (1 - p)^{1 - k}, k \in {0, 1}

$\begin{equation} P(X = k) = p^{k}(1-p)^{1-k},~~~~k \in \{0, 1\} \end{equation}$

E [X] = p

$\begin{equation} E[X] = p \end{equation}$

V [X] = p (1 - p)

$\begin{equation} V[X] = p(1-p) \end{equation}$

補足1: $E[X]$ の計算

\begin{aligned} E [X] & = \sum_{k \in {0, 1}} k \cdot p^{k} (1 - p)^{1 - k} \\ = 0 \cdot p^{0} (1 - p)^{1 - 0} + 1 \cdot p^{1} (1 - p)^{1 - 1} \\ = p \end{aligned}

$\begin{align} E[X] &= \sum_{k \in \{0,1\}}k \cdot p^{k}(1-p)^{1-k} \\ &= 0 \cdot p^{0}(1-p)^{1-0} + 1 \cdot p^{1}(1-p)^{1-1} \\ &= p \end{align}$

補足2: $V[X]$ の計算

\begin{aligned} E [X^{2}] & = \sum_{k = 0}^{1} k^{2} \cdot p^{k} (1 - p)^{1 - k} \\ = 0^{2} \cdot p^{0} (1 - p)^{1 - 0} + 1^{2} \cdot p^{1} (1 - p)^{1 - 1} \\ = p \end{aligned}

$\begin{align} E[X^2] &= \sum_{k=0}^{1}k^2 \cdot p^{k}(1-p)^{1-k} \\ &= 0^2 \cdot p^{0}(1-p)^{1-0} + 1^2 \cdot p^{1}(1-p)^{1-1} \\ &= p \end{align}$

\begin{aligned} V [X] & = E [X^{2}] - (E [X])^{2} \\ = p - p^{2} \\ = p (1 - p) \end{aligned}

$\begin{align} V[X] &= E[X^2] - (E[X])^2 \\ &= p - p^2 \\ &= p(1 - p) \end{align}$

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