Variance of Poisson Distribution/Proof 1
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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: $\expect X = \lambda$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \lambda\) |
$\blacksquare$