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--- mean_and_variance_of_binomial_distribution [2024/10/09 10:20] – [For variance] hkimscil
+++ mean_and_variance_of_binomial_distribution [2024/10/14 17:06] (current) – [For variance] hkimscil
@@ Line 31: / Line 31: @@
 ====== For Mean ======
-\begin{eqnarray}
+\begin{eqnarray*}
-E(X) & = & \sum_{x}x p(x) \nonumber \\
+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}  \nonumber \\
+& = & \sum_{x=0}^{n} x \frac{n!}{x!(n-x)!} p^x(1-p)^{n-x}  \\
-\text{note that   } x! = x(x-1)!  \nonumber \\
+\text{note that   } x! = x(x-1)!  \\
-& = & \sum_{x=1}^{n} \frac{n!}{(x-1)!(n-x)!} p^x(1-p)^{n-x} \nonumber  \\
+& = & \sum_{x=1}^{n} \frac{n!}{(x-1)!(n-x)!} p^x(1-p)^{n-x}  \\
-\text{cause we know that E(x) = np,} \nonumber \\
+\text{cause we know that E(x) = np,} \\
-\text{we extract np outside from summation} \nonumber \\
+\text{we extract np outside from summation} \\
-\text{note that  } p^x = p * p^{x-1} \nonumber \\
+\text{note that  } p^x = p * p^{x-1} \\
-\text{and  } n! = n * (n-1)! \nonumber \\
+\text{and  } n! = n * (n-1)! \\
-& = & \sum_{x=1}^{n} \frac{\underline{n}*(n-1)!}{(x-1)!(n-x)!} (\underline{p}*p^{x-1})(1-p)^{n-x}  \nonumber \\
+& = & \sum_{x=1}^{n} \frac{\underline{n}*(n-1)!}{(x-1)!(n-x)!} (\underline{p}*p^{x-1})(1-p)^{n-x}  \\
-\text{we take out the underlined part}  \nonumber \\
+\text{we take out the underlined part}  \\
-\text{(that is, np) out of the sigma part } \nonumber \\
+\text{(that is, np) out of the sigma part } \\
-& = & np \sum_{x=1}^{n} \frac{(n-1)!}{(x-1)!(n-x)!} p^{x-1}(1-p)^{n-x} \nonumber  \\
+& = & np \sum_{x=1}^{n} \frac{(n-1)!}{(x-1)!(n-x)!} p^{x-1}(1-p)^{n-x}  \\
-& = & np \underline{ \sum_{x=1}^{n} {\frac{(n-1)!}{(x-1)!(n-x)!} p^{x-1}(1-p)^{n-x}}} \nonumber  \\
+& = & np \underline{ \sum_{x=1}^{n} {\frac{(n-1)!}{(x-1)!(n-x)!} p^{x-1}(1-p)^{n-x}}}  \\
-\text{we want to check the underlined} \nonumber \\
+\text{we want to check the underlined} \\
-\text{part is equal to one so that np is left out} \nonumber \\
+\text{part is equal to one so that np is left out} \\
-n-x = (n-1)-(x-1) \nonumber \\
+n-x = (n-1)-(x-1) \\
-& = & np \sum_{x=1}^{n} \frac{(n-1)!}{(x-1)!((n-1)-(x-1))!} p^{x-1}(1-p)^{(n-1)-(x-1)} \nonumber  \\
+& = & np \sum_{x=1}^{n} \frac{(n-1)!}{(x-1)!((n-1)-(x-1))!} p^{x-1}(1-p)^{(n-1)-(x-1)}  \\
-m = n - 1 \nonumber \\
+m = n - 1 \\
-y = x - 1 \nonumber \\
+y = x - 1 \\
-& = & np \sum_{y=0}^{m} \frac{(m)!}{(y)!(m-y))!} p^{y}(1-p)^{(m-y)} \nonumber  \\
+& = & np \sum_{y=0}^{m} \frac{(m)!}{(y)!(m-y))!} p^{y}(1-p)^{(m-y)}  \\
-& = & np \sum_{y=0}^{m} \frac{(m)!}{(y)!(m-y))!} p^{y}(1-p)^{(m-y)} \nonumber  \\
+& = & np \sum_{y=0}^{m} \frac{(m)!}{(y)!(m-y))!} p^{y}(1-p)^{(m-y)}  \\
-& = & np \sum_{y=0}^{m} {{m}\choose {y}} p^{y}(1-p)^{(m-y)}  \nonumber \\
+& = & np \sum_{y=0}^{m} {{m}\choose {y}} p^{y}(1-p)^{(m-y)}  \\
-\text{Recall that}  \nonumber \\
+\text{Recall that}  \\
-\sum^{m}_{y=0}{{m}\choose{y}} a^{y} b^{m-y} = (a + b)^{m} \nonumber \\
+\sum^{m}_{y=0}{{m}\choose{y}} a^{y} b^{m-y} = (a + b)^{m} \\
-& = & np\;(p + (1-p))^m  \nonumber \\
+& = & np\;(p + (1-p))^m  \\
-& = & np\;(1)^m  \nonumber \\
+& = & np\;(1)^m  \\
-& = & np  \nonumber \\
+& = & np  \\
-\end{eqnarray}
+\end{eqnarray*}
 ====== For variance ======
@@ Line 72: / Line 72: @@
 \begin{align*}
 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} \\
 \end{align*}
@@ Line 80: / Line 80: @@
 \begin{align*}
 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} \\
 \text{cause} \\
@@ Line 92: / Line 92: @@
 \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 \\
 & = n(n-1)p^2 (p + (1-p))^m \\
@@ Line 100: / Line 102: @@
 E[X(X - 1)] & = 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 \\
 \\