Variance of Poisson Distribution/Proof 1

Theorem

Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.

Then the variance of $X$ is given by:

$\var X = \lambda$

Proof

From the definition of Variance as Expectation of Square minus Square of Expectation:

$\var X = \expect {X^2} - \paren {\expect X}^2$

From Expectation of Function of Discrete Random Variable:

$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \, \map \Pr {X = x}$

So:

\(\ds \expect {X^2}\)

\(=\)

\(\ds \sum_{k \mathop \ge 0} {k^2 \dfrac 1 {k!} \lambda^k e^{-\lambda} }\)

Definition of Poisson Distribution

\(\ds \)

\(=\)

\(\ds \lambda e^{-\lambda} \sum_{k \mathop \ge 1} {k \dfrac 1 {\paren {k - 1}!} \lambda^{k - 1} }\)

Note change of limit: term is zero when $k=0$

\(\ds \)

\(=\)

\(\ds \lambda e^{-\lambda} \paren {\sum_{k \mathop \ge 1} {\paren {k - 1} \dfrac 1 {\paren {k - 1}!} \lambda^{k - 1} } + \sum_{k \mathop \ge 1} {\frac 1 {\paren {k - 1}!} \lambda^{k - 1} } }\)

straightforward algebra

\(\ds \)

\(=\)

\(\ds \lambda e^{-\lambda} \paren {\lambda \sum_{k \mathop \ge 2} {\dfrac 1 {\paren {k - 2}!} \lambda^{k - 2} } + \sum_{k \mathop \ge 1} {\dfrac 1 {\paren {k - 1}!} \lambda^{k - 1} } }\)

Again, note change of limit: term is zero when $k-1=0$

\(\ds \)

\(=\)

\(\ds \lambda e^{-\lambda} \paren {\lambda \sum_{i \mathop \ge 0} {\dfrac 1 {i!} \lambda^i} + \sum_{j \mathop \ge 0} {\dfrac 1 {j!} \lambda^j} }\)

putting $i = k - 2, j = k - 1$

\(\ds \)

\(=\)

\(\ds \lambda e^{-\lambda} \paren {\lambda e^\lambda + e^\lambda}\)

Taylor Series Expansion for Exponential Function

\(\ds \)

\(=\)

\(\ds \lambda \paren {\lambda + 1}\)

\(\ds \)

\(=\)

\(\ds \lambda^2 + \lambda\)

Then:

\(\ds \var X\)

\(=\)

\(\ds \expect {X^2} - \paren {\expect X}^2\)

\(\ds \)

\(=\)

\(\ds \lambda^2 + \lambda - \lambda^2\)

Expectation of Poisson Distribution: $\expect X = \lambda$

\(\ds \)

\(=\)

\(\ds \lambda\)

$\blacksquare$