Question: Show that the discrete Poisson-distributed random variable (X) with discrete density [operatorname{Pr}{X=k}=frac{e^{-lambda} lambda^{k}}{k !}, k=0,1,2, ldots] has an expectation (E{X}=lambda).
Show that the discrete Poisson-distributed random variable \(X\) with discrete density
\[\operatorname{Pr}\{X=k\}=\frac{e^{-\lambda} \lambda^{k}}{k !}, k=0,1,2, \ldots\]
has an expectation \(E\{X\}=\lambda\).
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