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.

$\blacksquare$