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:

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

$\blacksquare$


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:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \Pi'_X \left({s}\right)\) \(=\) \(\displaystyle \) \(\displaystyle \frac {\mathrm d} {\mathrm ds} e^{-\lambda \left({1-s}\right)}\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \lambda e^{- \lambda \left({1-s}\right)}\) \(\displaystyle \) \(\displaystyle \)          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:

$e^0 = 1$

$\blacksquare$


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