Expectation of Poisson Distribution
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:
- $\displaystyle E \left({X}\right) = \sum_{x \mathop \in \operatorname{Im} \left({X}\right)} x \Pr \left({X = x}\right)$
By definition of Poisson distribution:
- $\displaystyle E \left({X}\right) = \sum_{k \mathop \ge 0} k \frac 1 {k!} \lambda^k e^{-\lambda}$
Then:
| \(\displaystyle E \left({X}\right)\) | \(=\) | \(\displaystyle \lambda e^{-\lambda} \sum_{k \mathop \ge 1} \frac 1 {\left({k-1}\right)!} \lambda^{k-1}\) | $\quad$ as the $k = 0$ term vanishes | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \lambda e^{-\lambda} \sum_{j \mathop \ge 0} \frac {\lambda^j} {j!}\) | $\quad$ putting $j = k - 1$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \lambda e^{-\lambda} e^{\lambda}\) | $\quad$ Taylor Series Expansion for Exponential Function | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \lambda\) | $\quad$ | $\quad$ |
$\blacksquare$
Proof 2
From the Probability Generating Function of Poisson Distribution, we have:
- $\Pi_X \left({s}\right) = e^{-\lambda \left({1-s}\right)}$
From Expectation of Discrete Random Variable from PGF, we have:
- $E \left({X}\right) = \Pi'_X \left({1}\right)$
We have:
| \(\displaystyle \Pi'_X \left({s}\right)\) | \(=\) | \(\displaystyle \frac {\mathrm d} {\mathrm ds} e^{-\lambda \left({1-s}\right)}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \lambda e^{- \lambda \left({1-s}\right)}\) | $\quad$ Derivatives of PGF of Poisson Distribution | $\quad$ |
Plugging in $s = 1$:
- $\Pi'_X \left({1}\right) = \lambda e^{- \lambda \left({1-1}\right)} = \lambda e^0$
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:
- $\displaystyle \map {M_X} t = e^{\lambda \paren {e^t - 1} }$
By Moment in terms of Moment Generating Function:
- $\displaystyle \expect X = \map {M_X'} 0$
We have:
| \(\displaystyle \map {M_X'} t\) | \(=\) | \(\displaystyle \frac \d {\d t} \paren {e^{\lambda \paren {e^t - 1} } }\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac \d {\d t} \paren {\lambda \paren {e^t - 1} } \frac \d {\d \paren {\lambda \paren {e^t - 1} } } \paren {e^{\lambda \paren {e^t - 1} } }\) | $\quad$ Chain Rule | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \lambda e^t e^{\lambda \paren {e^t - 1} }\) | $\quad$ Derivative of Exponential Function | $\quad$ |
Setting $t = 0$ gives:
| \(\displaystyle \expect X\) | \(=\) | \(\displaystyle \lambda e^0 e^{\lambda \paren {e^0 - 1} }\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \lambda\) | $\quad$ Exponential of Zero | $\quad$ |
$\blacksquare$
Also see
Sources
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 2.4$: Expectation: Exercise $10$
