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mean_and_variance_of_poisson_distribution

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*}

mean_and_variance_of_poisson_distribution.txt · Last modified: 2020/11/21 02:22 by hkimscil