기하분포
- 성공할 때 까지 독립적으로 실시한 베르누이 시행의 횟수
\begin{align*}
X &\sim geo(p)
\\[10pt]
P(X=x) &= pq^{x-1}~~I(x \in \mathbb{N})
\\[20pt]
E(X) &= \dfrac{1}{p}
\\[10pt]
Var(X) &= \dfrac{1-p}{p^2}
\end{align*}
확률함수 조건 확인
- P(X = x) ≥ 0
p \ge 0,~
g \ge 0,~
I(x \in \mathbb{N}) \ge 0
- \textstyle\sum P(X = x) = 1
\sum_{x}~P(X=x) =
\sum_{1}^{\infty}~pq^{x-1} =
\dfrac{p}{1-q} =
\dfrac{p}{p} = 1
1차 적률
E(X) = \dfrac{1}{p}
\begin{align*}
\because~
\sum_{x=1}^{\infty}
~x~P(X=x)
&= \sum_{x=1}^{\infty}
~x\cdot p\cdot q^{x-1}
= ~p~\sum_{x=1}^{\infty}~x\cdot q^{x-1}
\\[20pt]
&=~p~\cdot~\sum_{x=1}^{\infty}
\bigg( \dfrac{d}{dq}~q^x\bigg)
=~p\cdot\dfrac{d}{dq}~ \sum_{x=1}^{\infty}~q^x
\\[20pt]
&=~p\cdot\dfrac{d}{dq}
\bigg(\dfrac{q}{1-q}\bigg)
= p\cdot\dfrac{d}{dq}
\bigg(\dfrac{q-1+1}{1-q}\bigg)
\\[20pt]
&=~p\cdot\dfrac{d}{dq}~
\bigg(-1 + (1-q)^{-1}\bigg)
\\[20pt]
&=~p\cdot
\bigg\{(-1)(1-q)^2\cdot(-1)
\bigg\}
\\[20pt]
&=~p\cdot\dfrac{1}{(1-q)^2}
= \dfrac{p}{p^2}
= \dfrac{1}{p}
\end{align*}
2차 적률
- E(X^2)
\begin{align*}
E(X^2) &=~E(X(X-1)) + E(X)
\\[10pt]
&=~\dfrac{2q}{p^2}~+~\dfrac{1}{p}
\end{align*}
- E\{X(X-1)\} 구하기 1
\begin{align*}
E(X(X-1))
&=~\sum_{x}~x(x-1)\cdot P(X=x)
\\[15pt]
&=~\sum_{x}~x(x-1)\cdot pq^{x-1}
\\[15pt]
&=~pq~\cdot~\sum_{x=2}^{\infty}
~x(x-1)\cdot q^{x-2}
\end{align*}
- 중간 과정
\sum_{x=2}^{\infty}
~x(x-1)\cdot q^{x-2}
= \sum_{x=2}^{\infty}
~\dfrac{d^2}{dq^2}~q^x
= \sum_{x=2}^{\infty}
~\dfrac{q^2}{1-q}
\\[20pt]
\begin{align*}
&= \dfrac{d^2}{dq^2}
~\{-(1+q)+(1+q)^{-1}\}~
\\[20pt]
&= \dfrac{d^2}{dq^2}
\{-(1+q)\}
+ \dfrac{d^2}{dq^2}
\{(1-q)^{-1}\}
\\[20pt]
&= 0 + \dfrac{d}{dq}
(-1)\cdot(1-q)^2\cdot(-1)
\\[20pt]
&= (-2)\cdot(1-q)^3~\cdot(-1)
\\[20pt]
&= (-2)\cdot(1-q)^3~\cdot(-1)
\end{align*}
- E\{X(X-1)\} 구하기 2
\begin{align*}
E(X(X-1))
&=~pq\cdot2\cdot(1-q)^3
\\[10pt]
&=~\dfrac{pq\cdot 2}{p^3}
\\[10pt]
&=~\dfrac{2q}{p^2}
\end{align*}
- 분산
\begin{align*}
Var(X) &=~E(X^2)-E(X)^2 =
\dfrac{1-p}{p^2}
\\[20pt]
&=~\dfrac{2q}{p^2}~+~\dfrac{1}{p}
~-~\dfrac{1}{p^2}~~
=~\dfrac{1}{p^2}
\big(2q + p - 1\big)
\\[20pt]
&=~\dfrac{1}{p^2}
~\big(2q - q\big)
\\[20pt]
&=~\dfrac{q}{p^2}
\end{align*}
MGF
\begin{align*}
M_X(t) &= \sum_{x}~e^{tx}~P(X=x)
=~
\sum_{x}~e^{tx}\cdot p\cdot q^{x-1}
\\[15pt]
&=~\sum_{x=1}^{\infty}
~p\cdot\big(qe^t\big)^x\cdot q^{-1}
\\[15pt]
&=~\dfrac{p}{q}~\sum_{x=1}^{\infty}
~\big(qe^t\big)^x
=~\dfrac{p}{q}~\cdot
\dfrac{qe^t}{1-qe^t}
\\[15pt]
&=~\dfrac{pe^t}{1-qe^t}
\quad\bigg(~|qe^t|<1~\bigg)
\\[15pt]
\end{align*}