Expectation of Poisson Distribution

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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:

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

By definition of Poisson distribution:

$\displaystyle \expect X = \sum_{k \mathop \ge 0} k \frac 1 {k!} \lambda^k e^{-\lambda}$


\(\displaystyle \expect X\) \(=\) \(\displaystyle \lambda e^{-\lambda} \sum_{k \mathop \ge 1} \frac 1 {\paren {k - 1}!} \lambda^{k-1}\) as the $k = 0$ term vanishes
\(\displaystyle \) \(=\) \(\displaystyle \lambda e^{-\lambda} \sum_{j \mathop \ge 0} \frac {\lambda^j} {j!}\) putting $j = k - 1$
\(\displaystyle \) \(=\) \(\displaystyle \lambda e^{-\lambda} e^{\lambda}\) Taylor Series Expansion for Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle \lambda\)


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:

\(\displaystyle \Pi'_X \left({s}\right)\) \(=\) \(\displaystyle \frac {\mathrm d} {\mathrm ds} e^{-\lambda \left({1-s}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda e^{- \lambda \left({1-s}\right)}\) Derivatives of PGF of Poisson Distribution

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:

$e^0 = 1$


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:

\(\displaystyle \map {M_X'} t\) \(=\) \(\displaystyle \frac \d {\d t} \paren {e^{\lambda \paren {e^t - 1} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac \d {\d t} \paren {\lambda \paren {e^t - 1} } \frac \d {\d \paren {\lambda \paren {e^t - 1} } } \paren {e^{\lambda \paren {e^t - 1} } }\) Chain Rule
\(\displaystyle \) \(=\) \(\displaystyle \lambda e^t e^{\lambda \paren {e^t - 1} }\) Derivative of Exponential Function

Setting $t = 0$ gives:

\(\displaystyle \expect X\) \(=\) \(\displaystyle \lambda e^0 e^{\lambda \paren {e^0 - 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \lambda\) Exponential of Zero


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