mean_and_variance_of_binomial_distribution

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mean_and_variance_of_binomial_distribution [2024/10/09 10:22] – [For Mean] hkimscilmean_and_variance_of_binomial_distribution [2024/10/14 17:06] (current) – [For variance] hkimscil
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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} \\
 & = & \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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 \begin{align*} \begin{align*}
 E(X^2) & = \sum_{x=0}^{n} x^2 p(x) \\ E(X^2) & = \sum_{x=0}^{n} x^2 p(x) \\
-& = \sum_{x=0}^{n} x^2 {{n}\choose{x}} p^x (1-p)^{n-x} \;\;\; \text{see equation (1) }\\+& = \sum_{x=0}^{n} x^2 {{n}\choose{x}} p^x (1-p)^{n-x} \;\;\; \because \; p(x) = {{n}\choose{x}} p^x (1-p)^{n-x} \\
 & = \sum_{x=0}^{n} x^2 \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x} \\ & = \sum_{x=0}^{n} x^2 \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x} \\
 \end{align*} \end{align*}
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 \begin{align*} \begin{align*}
 E[X(X-1)] & = \sum_{x=0}^{n} x(x-1) \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x} \\ E[X(X-1)] & = \sum_{x=0}^{n} x(x-1) \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x} \\
-& = \sum_{x=2}^{n} \frac{n!}{(x-2)!(n-x)!} p^x (1-p)^{n-x} \\+& = \sum_{x=2}^{n} \frac{n!}{(x-2)!(n-x)!} p^x (1-p)^{n-x} \;\; \because \; x! = x (x-1) (x-2)!  \\
 & = n(n-1)p^2 \sum_{x=2}^{n} \frac{(n-2)!}{(x-2)!(n-x)!} p^{x-2} (1-p)^{n-x} \\ & = n(n-1)p^2 \sum_{x=2}^{n} \frac{(n-2)!}{(x-2)!(n-x)!} p^{x-2} (1-p)^{n-x} \\
 \text{cause} \\ \text{cause} \\
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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.1728436922.txt.gz · Last modified: 2024/10/09 10:22 by hkimscil

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