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:

$\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$

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

• Expectation and Variance of Poisson Distribution equal its Parameter

Sources

• 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 2.4$: Expectation: Exercise $10$
• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Appendix $13$: Probability distributions
• 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Appendix $15$: Probability distributions