기하분포 (Geometric Distribution)

• 여기서는 확률변수 $X$를 Fail의 횟수로 정의한다.

$$ X \sim Geo(p)$$

• $$ p \in [ \ 0 \ , \ 1 \ ] $$

$$ x \in \{ \ 0 \ , \ 1 \ , \ 2 \ , \ \cdots \ \} $$

$$ p(x) = p \ (1-p)^{x} = p \ q^{x} $$

$$ F(x) = 1 - (1-p)^{x+1} = 1 - q^{x+1} $$

$$E(X)=\frac{1-p}{p}$$

$$Var(X)=\frac{1-p}{p^{2}}$$

$$ \gamma_{1} = \frac{2 - p}{\sqrt{1-p}} = \frac{2-p}{\sqrt{q}} $$

$$ \gamma_{2} = \frac{p^{2} - 6p + 6}{1-p} = \frac{p^{2} - 6p + 6}{q} $$

$$ \phi \ (t) = \frac{p}{1 - (1 - p) \cdot e^{ \ i t}} = \frac{p}{1 - q \cdot e^{ \ i t}} $$

$$M(t)=\frac{p}{1-(1-p) \cdot e^{t}} = \frac{p}{1-q \cdot e^{t}}$$

$$ \mu'_{1} = \frac{1-p}{p} $$ $$ \mu'_{2} = \frac{(2-p)(1-p)}{p^{2}} $$ $$ \mu'_{3} = \frac{(1-p) \left[ 6+(p-6)p \right] }{p^{3}} $$ $$ \mu'_{4} = \frac{(2-p)(1-p) \left[ 12+(p-12)p \right] }{p^{4}} $$ $$ \mu'_{k} = p \ \operatorname{Li}_{ \ -k} (1-p) $$

단, $\mathrm{Li}_{n} (z)$는 ??함수(Polylogarithm)이다.

$ \mu_{2} = \frac{1-p}{p^{2}} $ $ \mu_{3} = \frac{(p-1)(p-2)}{p^{3}} $ $ \mu_{4} = \frac{(p-1)(-p^{2} +9p -9}{p^{4}} $ $ \mu_{k} = p \ \Phi \left( \ 1-p \ , \ -k \ , \ \frac{p-1}{p} \ \right) $

단, $\Phi ( \ z \ , \ s \ , \ a \ )$는 ??함수(Lerch Transcendent)이다.

1. 무기억성을 가진다.

• 분포