Expectation of Geometric Distribution
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Theorem
Let $X$ be a discrete random variable with the geometric distribution with parameter $p$.
Then the expectation of $X$ is given by:
- $E \left({X}\right) = \dfrac p {1-p}$
Proof 1
From the definition of expectation:
- $\displaystyle \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
By definition of geometric distribution:
- $\displaystyle \expect X = \sum_{k \mathop \in \Omega_X} k p^k \paren {1 - p}$
Let $q = 1 - p$:
\(\ds \expect X\) | \(=\) | \(\ds q \sum_{k \mathop \ge 0} k p^k\) | as $\Omega_X = \N$ | |||||||||||
\(\ds \) | \(=\) | \(\ds q \sum_{k \mathop \ge 1} k p^k\) | as the $k = 0$ term is zero | |||||||||||
\(\ds \) | \(=\) | \(\ds q p \sum_{k \mathop \ge 1} k p^{k - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds q p \frac 1 {\paren {1 - p}^2}\) | Derivative of Geometric Sequence | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac p {1 - p}\) | as $q = 1 - p$ |
$\blacksquare$
Proof 2
From the Probability Generating Function of Geometric Distribution, we have:
- $\Pi_X \left({s}\right) = \dfrac q {1 - ps}$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF, we have:
- $E \left({X}\right) = \Pi'_X \left({1}\right)$
We have:
\(\ds \Pi'_X \left({s}\right)\) | \(=\) | \(\ds \frac {\mathrm d} {\mathrm d s} \left({\frac q {1 - p s} }\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {q p} {\left({1 - p s}\right)^2}\) | Derivatives of PGF of Geometric Distribution |
Plugging in $s = 1$:
\(\ds \Pi'_X \left({1}\right)\) | \(=\) | \(\ds \frac {q p} {\left({1 - p}\right)^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac p {1 - p}\) | as $q = 1 - p$ |
Hence the result.
$\blacksquare$