Expectation of Poisson Distribution/Proof 2
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Theorem
Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the expectation of $X$ is given by:
- $\expect X = \lambda$
Proof
From Probability Generating Function of Poisson Distribution:
- $\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$
From Expectation of Discrete Random Variable from PGF:
- $\expect X = \map {\Pi'_X} 1$
We have:
\(\ds \map {\Pi'_X} s\) | \(=\) | \(\ds \frac \d {\d s} e^{-\lambda \paren {1 - s} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lambda e^{- \lambda \paren {1 - s} }\) | Derivatives of PGF of Poisson Distribution |
Plugging in $s = 1$:
- $\map {\Pi'_X} 1 = \lambda e^{- \lambda \paren {1 - 1} } = \lambda e^0$
Hence the result from Exponential of Zero:
- $e^0 = 1$
$\blacksquare$
Sources
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 4.3$: Moments: Exercise $5$