mean_and_variance_of_poisson_distribution
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mean_and_variance_of_poisson_distribution [2020/11/21 01:50] – [Mean] hkimscil | mean_and_variance_of_poisson_distribution [2024/10/28 07:51] (current) – hkimscil | ||
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+ | ====== Mean and Variance of Poisson Distribution ====== | ||
====== Mean ====== | ====== Mean ====== | ||
Mean Poisson distribution = $\lambda$ | Mean Poisson distribution = $\lambda$ | ||
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\end{eqnarray*} | \end{eqnarray*} | ||
- | 우선 Taylor series을 이용하면 | + | 우선 |
\begin{eqnarray*} | \begin{eqnarray*} | ||
e^{a} = \sum_{y=0}^{\infty} \frac{a^y}{y!} \\ | e^{a} = \sum_{y=0}^{\infty} \frac{a^y}{y!} \\ | ||
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& = & \lambda \cdot e^{-\lambda} \cdot \sum_{y=0}^{\infty} \frac{\lambda^{y}}{(y)!} \\ | & = & \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)!}} \\ | & = & \lambda \cdot e^{-\lambda} \cdot \underline{\sum_{y=0}^{\infty} \frac{\lambda^{y}}{(y)!}} \\ | ||
- | \text{from Taylor series} & \\ | + | \text{recall: } \; \\ |
+ | e^{a} = \sum_{y=0}^{\infty} \frac{a^y}{y!} \\ | ||
& = & \lambda \cdot e^{-\lambda} \cdot e^{\lambda} \\ | & = & \lambda \cdot e^{-\lambda} \cdot e^{\lambda} \\ | ||
& = & \lambda \\ | & = & \lambda \\ | ||
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- | ====== Variance ====== | + | ====== Variance ====== |
+ | Variance는 위의 binomial 케이스처럼 좀 복잡하다. | ||
Variance of Poisson distribution = $\lambda$ | 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.1605891028.txt.gz · Last modified: 2020/11/21 01:50 by hkimscil