第2講確率分布

指数分布

X \sim E x p (λ)

$\begin{equation} X \sim Exp(\lambda) \end{equation}$

指数分布はガンマ分布で形状母数 $\alpha = 1$ としたものである．

f (x) = {\begin{cases} λ e^{- λ x} & (x \geq 0) \\ 0 & (x < 0) \end{cases}

$\begin{equation} f(x) = \begin{cases} \lambda e^{{-\lambda x}} & (x \ge 0) \\ 0 & (x < 0) \end{cases} \end{equation}$

F (x) = {\begin{cases} 1 - e^{- λ x} & (x \geq 0) \\ 0 & (x < 0) \end{cases}

$\begin{equation} F(x) = \begin{cases} 1 - e^{{-\lambda x}} & (x \ge 0) \\ 0 & (x < 0) \end{cases} \end{equation}$

E [X] = \frac{1}{λ}

$\begin{equation} E[X] = \frac{1}{\lambda} \end{equation}$

V [X] = \frac{1}{λ^{2}}

$\begin{equation} V[X] = \frac{1}{\lambda^{2}} \end{equation}$

M_{X} (t) = \frac{λ}{λ - t}

$\begin{equation} M_{X}(t) = \frac{\lambda}{\lambda - t} \end{equation}$

無記憶性

$t > 0,~s > 0$ として，

\begin{aligned} P (X > t + s | X > s) = P (X > t) = e^{- λ t} \end{aligned}

$\begin{align} P(X > t + s | X > s) = P(X > t) = e^{-\lambda t} \end{align}$

待ち行列

単位時間当たりの事象の発生回数 $X$ が，平均 $\lambda$ のポアソン分布に従うとき，事象の発生間隔 $T$ は指数分布に従う． $t$ 時間当たりの発生回数 $X_t$ は，

P (X_{t} = k) = \frac{(λ t)^{k}}{k!} e^{- λ t}

$\begin{equation} P(X_t = k) = \frac{(\lambda t)^{k}}{k!}e^{-\lambda t} \end{equation}$

\begin{aligned} F (t) & = P (T \leq t) \\ = 1 - P (T > t) \\ = 1 - P (X_{t} = 0) \dots \dots P (X_{t} > 0) のこと \\ = 1 - e^{- λ t} \end{aligned}

$\begin{align} F(t) &= P(T \le t) \\ &= 1 - P(T > t) \\ &= 1 - P(X_t = 0)~~\cdots\cdots~~P(X_t > 0)\,\text{のこと} \\ &= 1 - e^{-\lambda t} \end{align}$

f (x) = F^{'} (x) = λ e^{- λ x}

$\begin{equation} f(x) = F'(x) = \lambda e^{-\lambda x} \end{equation}$

「1回目の事象が生起した時点が

t

$t$ 以前である」という事象は，「時点

t

$t$ まで事象は発生していない」という事象と等しいことを示している．

生存関数

S (x) = 1 - F (x) = P (T > t)

$\begin{equation} S(x) = 1 - F(x) = P(T > t) \end{equation}$

ハザード関数

\begin{aligned} h (x) & = lim_{Δ t \to 0} \frac{1}{Δ t} P (t < T < t + Δ t ∣ T > t) \\ = lim_{Δ t \to 0} \frac{1}{Δ t} \frac{P (t < T < t + Δ t \cap T > t)}{P (T > t)} \\ = lim_{Δ t \to 0} \frac{1}{Δ t} \frac{P (t < T < t + Δ t)}{P (T > t)} \\ = \frac{1}{P (T > t)} lim_{Δ t \to 0} \frac{P (t < T < t + Δ t)}{Δ t} \\ = \frac{1}{1 - F (x)} lim_{Δ t \to 0} \frac{F (t + Δ t) - F (t)}{Δ t} \\ = \frac{1}{1 - F (x)} f (x) \\ = \frac{f (x)}{1 - F (x)} \end{aligned}

$\begin{align} h(x) &= \lim_{\varDelta t \to 0}\frac{1}{\varDelta t}P(t < T < t + \varDelta t \mid T > t) \\ &= \lim_{\varDelta t \to 0}\frac{1}{\varDelta t}\frac{P(t < T < t + \varDelta t~~\cap~~T > t)}{P(T > t)} \\ &= \lim_{\varDelta t \to 0}\frac{1}{\varDelta t}\frac{P(t < T < t + \varDelta t)}{P(T > t)} \\ &= \frac{1}{P(T > t)}\lim_{\varDelta t \to 0}\frac{P(t < T < t + \varDelta t)}{\varDelta t} \\ &= \frac{1}{1 - F(x)}\lim_{\varDelta t \to 0}\frac{F(t + \varDelta t) - F(t)}{\varDelta t} \\ &= \frac{1}{1 - F(x)}f(x) \\ &= \frac{f(x)}{1 - F(x)} \end{align}$

補足1: $E[X]$ の計算

f_{X} (x) = λ e^{- λ x}

$\begin{equation} f_X(x) = \lambda e^{{-\lambda x}} \end{equation}$

\begin{aligned} E [X] & = \int_{0}^{\infty} x λ e^{- λ x} d x \\ = λ \int_{0}^{\infty} x^{2 - 1} e^{- λ x} d x \\ = λ \frac{Γ (2)}{λ^{2}} \\ = λ \frac{(2 - 1)!}{λ^{2}} \\ = \frac{1}{λ} \end{aligned}

$\begin{align} E[X] &= \int_{0}^{\infty}x\,\lambda e^{{-\lambda x}}\,dx \\ &= \lambda \int_{0}^{\infty}x^{2-1}\,e^{{-\lambda x}}\,dx \\ &= \lambda \frac{\varGamma(2)}{\lambda^2} \\ &= \lambda \frac{(2-1)!}{\lambda^2} \\ &= \frac{1}{\lambda} \end{align}$

補足2: $V[X]$ の計算

\begin{aligned} E [X^{2}] & = \int_{0}^{\infty} x^{2} λ e^{- λ x} d x \\ = λ \int_{0}^{\infty} x^{3 - 1} e^{- λ x} d x \\ = λ \frac{Γ (3)}{λ^{3}} \\ = λ \frac{(3 - 1)!}{λ^{3}} \\ = \frac{2}{λ^{2}} \end{aligned}

$\begin{align} E[X^2] &= \int_{0}^{\infty}x^2\,\lambda e^{{-\lambda x}}\,dx \\ &= \lambda \int_{0}^{\infty}x^{3-1}\,e^{{-\lambda x}}\,dx \\ &= \lambda \frac{\varGamma(3)}{\lambda^3} \\ &= \lambda \frac{(3-1)!}{\lambda^3} \\ &= \frac{2}{\lambda^2} \end{align}$

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

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

第2講 確率分布