====== Mean and Variance of Poisson Distribution ======
====== Mean ======
Mean Poisson distribution = $\lambda$

Poisson Distribution
\begin{eqnarray*}
P(X=x) = \frac{e^{-\lambda} \cdot \lambda^x}{x!} \\
\end{eqnarray*}
혹은
\begin{eqnarray*}
P(x) = \frac{e^{-\lambda} \cdot \lambda^x}{x!} \\
\end{eqnarray*}

우선 Taylor series을 이용하면
\begin{eqnarray*}
e^{a} = \sum_{y=0}^{\infty} \frac{a^y}{y!} \\
\end{eqnarray*}
임을 알고 있다. 

\begin{eqnarray*}
E(X) & = & \sum_{x} xp(X=x) \\
\text{or }  \\
E(X) & = & \sum_{x} xp(x) \\
\end{eqnarray*}
Poisson distribution 을 다루고 있으므로  
\begin{eqnarray*}
E(X) & = & \sum_{x=0}^{\infty} x \cdot  \frac{e^{-\lambda} \cdot \lambda^x}{x!} \\
& = & e^{-\lambda} \cdot \sum_{x=0}^{\infty} x \cdot  \frac{\lambda^x}{x!} \\
& = & e^{-\lambda} \cdot \sum_{x=0}^{\infty} x \cdot  \frac{\lambda^x}{x(x-1)!} \\
& = & e^{-\lambda} \cdot \sum_{x=1}^{\infty} x \cdot  \frac{\lambda \cdot \lambda^{x-1}}{x(x-1)!} \\
& = & \lambda \cdot e^{-\lambda} \cdot \sum_{x=1}^{\infty} \frac{\lambda^{x-1}}{(x-1)!} \\
\text{let y = x-1} & \\
& = & \lambda \cdot e^{-\lambda} \cdot \sum_{y=0}^{\infty} \frac{\lambda^{y}}{(y)!} \\
& = & \lambda \cdot e^{-\lambda} \cdot \underline{\sum_{y=0}^{\infty} \frac{\lambda^{y}}{(y)!}} \\
\text{recall: } \; \\ 
e^{a} = \sum_{y=0}^{\infty} \frac{a^y}{y!} \\
& = & \lambda \cdot e^{-\lambda} \cdot e^{\lambda} \\
& = & \lambda \\
\end{eqnarray*}





 ====== Variance ======
Variance는 위의 binomial 케이스처럼 좀 복잡하다. 
Variance of Poisson distribution = $\lambda$

\begin{eqnarray*}
Var(X) & = & E \left[(X-\mu)^2 \right] \\
& = & \sum_{x=1}^{n}(x-\mu)^2 \cdot p(x) \\
\end{eqnarray*}

또한 
\begin{eqnarray*}
E \left[(X-\mu)^2 \right] 
& = & E(X^2) - \left[E(X) \right]^2 \\
\end{eqnarray*}

이 중에서 우선 $E(X^2)$을 우선 다루면
\begin{eqnarray*}
E(X^2) & = & \sum_{x=1}^{n}x^2 \cdot p(x) \\
& = & \sum_{x=1}^{n}x^2 \cdot \frac{e^{-\lambda} \cdot \lambda^{x}}{x!}\\
& = & e^{-\lambda} \cdot \sum_{x=1}^{n}x^2 \cdot \frac{\lambda^{x}}{x!}\\
\end{eqnarray*}

이상태로는 $X^2$를 없앨 수는 없으므로 계산을 우회하기로 하면

\begin{eqnarray*}
E[X(X-1)] & = & \sum_{x=0}^{\infty} x(x-1) \cdot p(x) \\
& = & \sum_{x=0}^{\infty} x(x-1) \cdot \frac{e^{-\lambda} \cdot \lambda^{x}} {x!} \\
& = & e^{-\lambda} \cdot \sum_{x=2}^{n} x(x-1) \cdot \frac{\lambda^{x}}{x(x-1) \cdot (x-2)!} \\
& = & e^{-\lambda} \cdot \sum_{x=2}^{n} \frac{\lambda^{x}}{(x-2)!} \\
& = & e^{-\lambda} \cdot \sum_{x=2}^{n} \frac{\lambda^2 \cdot \lambda^{x-2}}{(x-2)!} \\
& = & \lambda^2 \cdot e^{-\lambda} \cdot \sum_{x=2}^{n} \frac{\lambda^{x-2}}{(x-2)!} \\
\text{let } \; y = x-2 \\
& = & \lambda^2 \cdot e^{-\lambda} \cdot \underline{\sum_{y=0}^{n} \frac{\lambda^{y}}{y!}} \\
& = & \lambda^2 \cdot e^{-\lambda} \cdot \underline{\sum_{y=0}^{n} \frac{\lambda^{y}}{y!}} \\
\text{underlined part } = e^{\lambda} \\
& = & \lambda^2 \cdot e^{-\lambda} \cdot e^{\lambda} \\
& = & \lambda^2 
\end{eqnarray*}

따라서 
\begin{eqnarray*}
E[X(X-1)] & = & E[X^2-X] = E(X^2) - E(X) \\
& = & \lambda^2  \\
E(X^2) & = & \lambda^2 + \lambda \\
\end{eqnarray*}

다시 원래대로 돌아가서
\begin{eqnarray*}
Var(X) & = & E \left[(X-\mu)^2 \right]  \\
& = & E(X^2) - \left[E(X) \right]^2 \\
& = & \lambda^2 + \lambda - \lambda^2 \\
& = & \lambda 
\end{eqnarray*}