Expectation of Poisson Distribution
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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 1
From the definition of expectation:
- $\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$
By definition of Poisson distribution:
- $\ds \expect X = \sum_{k \mathop \ge 0} k \frac 1 {k!} \lambda^k e^{-\lambda}$
Then:
| \(\ds \expect X\) | \(=\) | \(\ds \lambda e^{-\lambda} \sum_{k \mathop \ge 1} \frac 1 {\paren {k - 1}!} \lambda^{k-1}\) | as the $k = 0$ term vanishes | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lambda e^{-\lambda} \sum_{j \mathop \ge 0} \frac {\lambda^j} {j!}\) | putting $j = k - 1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lambda e^{-\lambda} e^{\lambda}\) | Taylor Series Expansion for Exponential Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lambda\) |
$\blacksquare$
Proof 2
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:
- $e^0 = 1$
$\blacksquare$
Proof 3
From Moment Generating Function of Poisson Distribution, the moment generating function of $X$, $M_X$, is given by:
- $\map {M_X} t = e^{\lambda \paren {e^t - 1} }$
By Moment in terms of Moment Generating Function:
- $\expect X = \map {M_X'} 0$
We have:
| \(\ds \map {M_X'} t\) | \(=\) | \(\ds \map {\frac \d {\d t} } {e^{\lambda \paren {e^t - 1} } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map {\frac \d {\d t} } {\lambda \paren {e^t - 1} } \frac \d {\map \d {\lambda \paren {e^t - 1} } } \paren {e^{\lambda \paren {e^t - 1} } }\) | Chain Rule for Derivatives | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lambda e^t e^{\lambda \paren {e^t - 1} }\) | Derivative of Exponential Function |
Setting $t = 0$ gives:
| \(\ds \expect X\) | \(=\) | \(\ds \lambda e^0 e^{\lambda \paren {e^0 - 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lambda\) | Exponential of Zero |
$\blacksquare$
Also see
Sources
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 2.4$: Expectation: Exercise $10$