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:

$E \left({X}\right) = \lambda$


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:

$e^0 = 1$

$\blacksquare$


Also see


Sources

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