mean_and_variance_of_binomial_distribution

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mean_and_variance_of_binomial_distribution [2024/10/09 10:27] – [For variance] hkimscilmean_and_variance_of_binomial_distribution [2025/10/06 23:50] (current) – [Proof of Binomial Expected Value, from a scratch] hkimscil
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-====== Proof of Binomial Expected Valuefrom scratch ====== +====== Proof of Binomial Expected Value and Variance (from scratch====== 
 +이항분포에서의 평균과 분산 증명
 see [[:The Binomial Theorem]] see [[:The Binomial Theorem]]
  
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 \end{eqnarray*} \end{eqnarray*}
  
-위의 식이 복잡해 보이지만 m = 3 일때의 이항정리식을 한다  +위의 식이 복잡해 보이지만 m = 3 일때 이항정리식이 아래처럼 전개됨을 한다.
 \begin{align*} \begin{align*}
-\sum^{m}_{y=0}{{m}\choose{y}} a^{y} b^{m-y} \text{m = 3} \\+\sum^{m}_{y=0}{{m}\choose{y}} a^{y} b^{m-y} \;\;\; \dots \;\;\; \text{m = 3} \\
 \end{align*} \end{align*}
  
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 ====== For Mean ====== ====== For Mean ======
 \begin{eqnarray*} \begin{eqnarray*}
-E(X) & = & \sum_{x}x p(x) \\+E(X) & = & \sum_{x}x p(x) \;\;\; \because \; p(x) = {{n}\choose{x}} p^x (1-p)^{n-x} \;\;\; \text{, binomial probability} \\
 & = & \sum_{x=0}^{n} x {{n} \choose {x}} p^x(1-p)^{n-x}  \\ & = & \sum_{x=0}^{n} x {{n} \choose {x}} p^x(1-p)^{n-x}  \\
 & = & \sum_{x=0}^{n} x \frac{n!}{x!(n-x)!} p^x(1-p)^{n-x}  \\ & = & \sum_{x=0}^{n} x \frac{n!}{x!(n-x)!} p^x(1-p)^{n-x}  \\
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 \text {we know that the underline part is} \\ \text {we know that the underline part is} \\
 +(p+(1-p))^m \\
 +\text {and, we also know that it is 1} \\
 (p+(1-p))^m = 1^m \\ (p+(1-p))^m = 1^m \\
 & = n(n-1)p^2 (p + (1-p))^m \\ & = n(n-1)p^2 (p + (1-p))^m \\
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 E[X(X - 1)] & = n(n-1)p^2 \\ E[X(X - 1)] & = n(n-1)p^2 \\
 E[X^2 - X] & = n(n-1)p^2 \\ E[X^2 - X] & = n(n-1)p^2 \\
-E[X^2]- E[X] & = n(n-1)p^2 \\ +E[X^2]- E[X] & = n(n-1)p^2 \;\;\; \because E[X] = np \\ 
-E[X^2]- np & = n(n-1)p^2 \\+E[X^2]- np & = n(n-1)p^2  \\
 E[X^2]& = n(n-1)p^2 + np \\ E[X^2]& = n(n-1)p^2 + np \\
 \\ \\
mean_and_variance_of_binomial_distribution.1728437223.txt.gz · Last modified: by hkimscil

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