====== Proof of Binomial Expected Value, from a scratch ======

see [[:The Binomial Theorem]]

\begin{eqnarray*}
\text{The binomial theorem} &  & \\
(a + b)^{m} & = & \sum^{m}_{y=0}{{m}\choose{y}} a^{y} b^{m-y} \\
\end{eqnarray*}

위의 식이 복잡해 보이지만 m = 3 일때의 이항정리식을 말한다  
\begin{align*}
\sum^{m}_{y=0}{{m}\choose{y}} a^{y} b^{m-y} \text{, m = 3} \\
\end{align*}

\begin{align*}
\sum^{3}_{y=0}{{3}\choose{y}} a^{y} b^{3-y} 
& = {{3}\choose{0}} a^{0} b^{3-0} 
+ {{3}\choose{1}} a^{1} b^{3-1} 
+ {{3}\choose{2}} a^{2} b^{3-2} 
+ {{3}\choose{3}} a^{y} b^{3-3}  \\
& = 1*a^0*b^3 
+ 3*a^1*b^2 
+ 3*a^2*b^1  
+ 1*a^3*b^0  \\
& = a^3 
+ 3 a^2 b^1 
+ 3 a^1 b^2  
+ b^3  \\
\end{align*}


====== For Mean ======
\begin{eqnarray*}
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 \frac{n!}{x!(n-x)!} p^x(1-p)^{n-x}  \\
\text{note that   } x! = x(x-1)!  \\
& = & \sum_{x=1}^{n} \frac{n!}{(x-1)!(n-x)!} p^x(1-p)^{n-x}  \\
\text{cause we know that E(x) = np,} \\
\text{we extract np outside from summation} \\
\text{note that  } p^x = p * p^{x-1} \\
\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}  \\
\text{we take out the underlined part}  \\
\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}  \\
& = & 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} \\
\text{part is equal to one so that np is left out} \\
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)}  \\
m = n - 1 \\
y = x - 1 \\
& = & 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)}  \\
& = & np \sum_{y=0}^{m} {{m}\choose {y}} p^{y}(1-p)^{(m-y)}  \\
\text{Recall that}  \\
\sum^{m}_{y=0}{{m}\choose{y}} a^{y} b^{m-y} = (a + b)^{m} \\
& = & np\;(p + (1-p))^m  \\
& = & np\;(1)^m  \\
& = & np  \\
\end{eqnarray*}

====== For variance ======
\begin{align*}
Var(X) & = E[(X-\mu)^2] \\
& = \sum_{x}(x-\mu)^2p(x) \\
\text{We also know that: } \\
E[(X-\mu)^2] & = E(X^2) - [E(X)]^2 \\
\end{align*}

\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} \;\;\; \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*}

그런데 $E(X^2)$ 대신 $E[X(X-1)]$을 생각해보면

\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} \;\; \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} \\
n - x = (n - 2)-(x - 2) \\
& = n(n-1)p^2 \sum_{x=2}^{n} \frac{(n-2)!}{(x-2)!((n-2)-(x-2))!} p^{x-2} (1-p)^{(n-2)-(x-2)} \\
m = n - 2 \\
y = x - 2 \\
& = n(n-1)p^2 \sum_{y=0}^{m} \frac{m!}{y!((m-y))!} p^{y} (1-p)^{(m-y)} \\
& = n(n-1)p^2 \underline {\sum_{y=0}^{m} \frac{m!}{y!((m-y))!} p^{y} (1-p)^{m-y} } \\
& = n(n-1)p^2 \underline {\sum_{y=0}^{m} {{m}\choose{y}} p^{y} (1-p)^{m-y} } \\

\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 \\
& = n(n-1)p^2 \\
\end{align*}
. . . . 
\begin{align*}
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 \;\;\; \because E[X] = np \\
E[X^2]- np & = n(n-1)p^2  \\
E[X^2]& = n(n-1)p^2 + np \\
\\
\end{align*}

\begin{align*}
Var(X) & = E[X^2] - [E(X)]^2  \\
& = \left[n(n-1)p^2 + np \right] - \left[np \right]^2  \\
& = np[(n-1)p + 1 - np]  \\
& = np[np - p + 1 - np]  \\
& = np(1 - p)  \\
& = npq  \\
\\
\end{align*}