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문서의 이전 판입니다!
기하분포 (Geometric Distribution)
표기
$$ X \sim Geo(p)$$
$$ p \in [ \ 0 \ , \ 1 \ ] $$
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[받침]
$$ x \in \{ \ 0 \ , \ 1 \ , \ 2 \ , \ \cdots \ \} $$
[확률밀도함수]
[누적분포함수]
[기대값]
$$E(X)=\frac{1-p}{p}$$
[분산]
$$Var(X)=\frac{1-p}{p^{2}}$$
[왜도]
$$ \gamma_{1} = \frac{2 - p}{\sqrt{1-p}} = \frac{2-p}{\sqrt{q}} $$
[첨도]
$$ \gamma_{2} = \frac{p^{2} - 6p + 6}{1-p} = \frac{p^{2} - 6p + 6}{q} $$
[특성함수]
$$ \phi \ (t) = \frac{p}{1 - (1 - p) \cdot e^{ \ i t}} = \frac{p}{1 - q \cdot e^{ \ i t}} $$
[적률생성함수]
$$M(t)=\frac{p}{1-(1-p) \cdot e^{t}} = \frac{p}{1-q \cdot e^{t}}$$
[원적률]
$$ \mu'_{1} = \frac{1-p}{p} $$
$$ \mu'_{2} = \frac{(2-p)(1-p)}{p^{2}} $$
$$ \mu'_{3} = \frac{(1-p) \left[ 6+(p-6)p \right] }{p^{3}} $$
$$ \mu'_{4} = \frac{(2-p)(1-p) \left[ 12+(p-12)p \right] }{p^{4}} $$
$$ \mu'_{k} = p \ \operatorname{Li}_{ \ -k} (1-p) $$
단,   $$\operatorname{Li}_{n} (z)$$ 는 ??함수(Polylogarithm)이다.
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[중심적률]
$$ \mu_{2} = \frac{1-p}{p^{2}} $$
$$ \mu_{3} = \frac{(p-1)(p-2)}{p^{3}} $$
$$ \mu_{4} = \frac{(p-1)(-p^{2} +9p -9}{p^{4}} $$
$$ \mu_{k} = p \ \Phi \left( \ 1-p \ , \ -k \ , \ \frac{p-1}{p} \ \right) $$
단,   $$\Phi ( \ z \ , \ s \ , \ a \ )$$ 는 ??함수(Lerch Transcendent)이다.
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특성
i. [무기억성]을 가진다.