열렬히.뛰기

이항 분포

수학 & 통계 > 수리통계1 > 3. 일변량 분포 : 예시 > 이항 분포

이항 분포

  • n번 시도 할 때 x번 성공할 확률. 복원추출
\begin{align*} P(X=x) &= \binom{n}{x}~p^x~ (1-p)^{1-x}~ I(x\in\{0, 1, \cdots, n\}) \\[15pt] M_{X}(t) &= \bigg\{(1-p)+pe^t\bigg\}^n \\[15pt] \mu &= np \\[10pt] \sigma &= np(1-p) \end{align*}
확률함수 조건 확인
\text{이항정리 : }(a + b)^n = \displaystyle\sum_{k=0}^{n} \binom{n}{k} a^k\; b^{n-k}
  1. P(X = x) \ge 0 : 만족
\begin{align*} &P(X = x) = \displaystyle\binom{n}{x} ~p^x~(1-p)^{(1-x)} ~I(x \in \{0, 1, \cdots, n\}) \end{align*} \\[30pt] \begin{align*} &\text{이항정리: } \displaystyle\binom{n}{x} = \dfrac{n!}{x!(n-x)!} \ge 0 \\[20pt] &\text{확률함수: } p^x~(1-p)^{(1-x)} \ge 0 \\[20pt] &\text{지시함수: } I(x \in \{0, 1, \cdots, n\}) \ge 0 \end{align*}
  1. \displaystyle\sum P(X = x) = 1 : 만족
\begin{align*} \displaystyle\sum P(X = x) &= \displaystyle\sum \binom{n}{x} ~p^x~(1-p)^{(1-x)} ~I(x \in \{0, 1, \cdots, n\}) \\[20pt] &= \displaystyle\sum_{x=0}^{n} \binom{n}{x} ~p^x~(1-p)^{1-x} \\[20pt] &= \{p + (1-p)\}^{n} \\[10pt] &= 1 \end{align*}
1차 적률
\begin{align*} E(X) &= \sum_{x}~xP(X=x) = np \\[15pt] &=~\sum_{x=0}^{n}~~x \cdot \binom{n}{x}~p^x~(1-p)^{n-x} \\[15pt] &=~\sum_{x=1}^{n}~~x \cdot \dfrac{n}{x}~\binom{n-1}{x-1} ~p^x~(1-p)^{n-x} \\[15pt] &=~n~\sum_{x=1}^{n} ~\binom{n-1}{x-1}~p^x~(1-p)^{n-x} \\[15pt] &= ~n~\sum_{k=0}^{n_*} ~\binom{n_*}{k}~p^x~(1-p)^{n-x} \\[15pt] &= ~np~\sum_{k=0}^{n_*} ~\binom{n_*}{k}~p^{x-1}~(1-p)^{n-x} \\[15pt] &= ~np \end{align*}
~n_* = n-1\\ ~k = x-1
2차 적률과 분산
\begin{align*} E(X^2-X) &= E\big( X(X-1) \big) \\[20pt] = \sum_{x}&~ x(x-1)~P(X=x) \\[15pt] = \sum_{x}&~ x(x-1) ~\binom {\textcolor{red}{n}} {\textcolor{red}{x}} ~p^x~(1-p)^{x-1} \\[15pt] = \sum_{x}&~ x(x-1) \cdot \dfrac {\textcolor{red}{n(n-1)}} {\textcolor{red}{x(x-1)}} \cdot \binom {\textcolor{red}{n-2}} {\textcolor{red}{x-2}} ~p^x\cdot(1-p)^{x-1} \\[15pt] =~n&(n-1)~\sum_{x} \binom {n-2} {x-2} ~p^x~(1-x)^{n-x} \\[15pt] =~n&(n-1)~\sum_{x} \binom {n_*} {k} (1-p)^{n_* - k} \\[15pt] =~n&(n-1)~p^2 \end{align*}

기댓값: 제곱

\begin{align*} E(X^2) &= E(X^2 - X + X) = E(X^2 - X) + E(X) \\[20pt] &= n(n-1)~p^2 + np \\[20pt] &= n^2p^2 - np^2 + np \end{align*}

분산

\begin{align*} Var(X) &= E(X^2) - E(X)^2 \\[10pt] &= E(X^2 - X) + E(X) - E(X)^2 \\[10pt] &= n^2p^2 - np^2 + np - n^2p^2 \\[10pt] &= np(1-p) \end{align*}
3차 적률
  • 정의
\begin{align*} E(X^3) &= np~[(n-1)p+1]~[(n-2)p+1] \\[20pt] = E \bigg(& X(X-1)(X-2) + 3X^2 -3X + X\bigg) \\[15pt] = E \bigg(& X(X-1)(X-2) + 3X^2 -2X\bigg) \\[15pt] = E \bigg(& X(X-1)(X-2) \bigg) + 3~E \bigg( X(X-1) \bigg) - E(X) \end{align*}
  • 중간과정: X(X-1)(X-2)
\begin{align*} &E\bigg( X(X-1)(X-2) \bigg) \\[20pt] &=~\sum_{x}~x~(x-1)~(x-2)~ P(X=x) \\[20pt] &=~x~(x-1)~(x-2) \cdot ~\dfrac {n(n-1)(n-2)} {x(x-1)(x-2)} \cdot \binom {n-3} {x-3} ~p^x~(1-p)^{1-x} \\[20pt] &=~n~(n-1)~(n-2)~p^3 \end{align*}
  • 다시 계산
\begin{align*} E(X^3) &= E \bigg( X(X-1)(X-2) \bigg) + 3~E \bigg( X(X-1) \bigg) - E(X) \\[20pt] = E \bigg(& X(X-1)(X-2) + 3X^2 -3X + X\bigg) \\[15pt] = E \bigg(& X(X-1)(X-2) + 3X^2 -2X\bigg) \\[15pt] = E \bigg(& X(X-1)(X-2) \bigg) + 3~E \bigg( X(X-1) \bigg) - E(X) \\[15pt] = n(n&-1)(n-2)~p^3 + 3n(n-1)p^2 + np \\[15pt] = np[&(n-1)(n-2)p^2 + 3(n-1)p + 1] \\[15pt] = np[&(n-1)p + 1][(n-2)p + 1] \end{align*}
MGF
\begin{align*} M_Y(t) &= \sum_{y}~e^{ty}~P(Y=y) = \sum_{y}~e^{ty}~\binom{n}{y}~p^y~(1-p)^{n-y} \\[20pt] &= \sum_{y}~\binom{n}{y}~(pe^t)^y~(1-p)^{n-y} \\[20pt] M_Y(t) &= \{pe^t + 1-p\}^{n} \end{align*}