Expectation of Poisson Distribution/Proof 2

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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:

$\expect X = \lambda$


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:

$e^0 = 1$