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:

$\expect X = \lambda$


Proof 1

From the definition of expectation:

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

By definition of Poisson distribution:

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

Then:

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

$\blacksquare$


Proof 2

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:

$e^0 = 1$

$\blacksquare$


Proof 3

From Moment Generating Function of Poisson Distribution, the moment generating function of $X$, $M_X$, is given by:

$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$

By Moment in terms of Moment Generating Function:

$\expect X = \map {M_X'} 0$

We have:

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

Setting $t = 0$ gives:

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

$\blacksquare$


Also see


Sources