Mean and Variance of the Poisson Distribution
Formulas
$X \sim \text{Poi}(\lambda)$ Surface $$ E(X) = \lambda \\ \operatorname{Var}(X) = \lambda $$
Derivation
Strategy: Directly deduce from the definition of the Poisson distribution. The trick of splitting factorials and series is important.
Definition of Poisson Distribution: For $\lambda > 0$, an discrete probability distribution that has the following probability mass function $\text{Poi} ( \lambda )$ is called a Poisson distribution. $$ p(x) = {{ e^{-\lambda} \lambda^{x} } \over { x! }} \qquad , x = 0 , 1 , 2, \cdots $$
Mean
$$ \begin{align*} E(X) =& \sum _{ x=0 }^{ \infty }{ x\frac { { \lambda ^ x }{ e ^ { - \lambda } } }{ x! } } \\ =& { e ^ { - \lambda } }\sum _{ x=0 }^{ \infty }{ x\frac { { \lambda ^ x } }{ x! } } \\ =& { e ^ { - \lambda } }\sum _{ x=1 }^{ \infty }{ \frac { { \lambda \cdot \lambda } ^{ x-1 } }{ (x-1)! } } \\ =& { \lambda e } ^{ -\lambda } \sum _{ x=1 }^{ \infty }{ \frac { { \lambda } ^{ x-1 } }{ (x-1)! } } \\ =& \lambda { e ^ { - \lambda } }{ e^ \lambda } \\ =& \lambda \end{align*} $$
■
Variance
$$ \begin{align*} E({ X }^{ 2 }) =& \sum _{ x=0 }^{ \infty }{ { x } ^{ 2 } \frac { { \lambda ^ x }{ e ^ { - \lambda } } }{ x! } } \\ =& { e ^ { - \lambda } }\sum _{ x=1 }^{ \infty }{ x\frac { { \lambda ^ x } }{ (x-1)! } } \\ =& { e ^ { - \lambda } }\sum _{ x=1 }^{ \infty }{ \frac { { (x-1+1)\lambda } ^{ x } }{ (x-1)! } } \\ =& { e ^ { - \lambda } }\sum _{ x=1 }^{ \infty }{ \frac { { (x-1)\lambda } ^{ x }+{ \lambda ^ x } }{ (x-1)! } } \\ =& { e ^ { - \lambda } }\sum _{ x=1 }^{ \infty }{ \left\{ \frac { { (x-1)\lambda } ^{ x } }{ (x-1)! }+\frac { { \lambda ^ x } }{ (x-1)! } \right\} } \\ =& { e ^ { - \lambda } }\left\{ \sum _{ x=2 }^{ \infty }{ \frac { { \lambda ^ x } }{ (x-2)! } }+\sum _{ x=1 }^{ \infty }{ \frac { { \lambda ^ x } }{ (x-1)! } } \right\} \\ =& { e ^ { - \lambda } }\left\{ \sum _{ x=2 }^{ \infty }{ \frac { { { \lambda ^ 2 }\cdot \lambda } ^{ x-2 } }{ (x-2)! } }+\sum _{ x=1 }^{ \infty }{ \frac { { \lambda \cdot \lambda } ^{ x-1 } }{ (x-1)! } } \right\} \\ =& { e ^ { - \lambda } }({ \lambda ^ 2 }{ e^ \lambda }+\lambda { e^ \lambda }) \\ =& { \lambda ^ 2 }+\lambda \end{align*} $$ Therefore $$ \operatorname{Var} (X)=E({ X }^{ 2 })-{ \left\{ E(X) \right\} } ^{ 2 } ={ (\lambda }^{ 2 }+\lambda )-{ \lambda }^{ 2 }=\lambda $$
■