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--- mean_and_variance_of_poisson_distribution [2020/11/21 01:27] – [Variance] hkimscil
+++ mean_and_variance_of_poisson_distribution [2020/11/21 02:22] (current) – hkimscil
@@ Line 1: / Line 1: @@
+====== Mean and Variance of Poisson Distribution ======
 ====== Mean ======
 Mean Poisson distribution = $\lambda$
-====== Variance ======
+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*}