Expectation of Shifted Geometric Distribution/Proof 2

Theorem

Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.

Then the expectation of $X$ is given by:

$\expect X = \dfrac 1 p$

Proof

From the Probability Generating Function of Shifted Geometric Distribution, we have:

$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$

where $q = 1 - p$.

From Expectation of Discrete Random Variable from PGF, we have:

$\expect X = \map {\Pi'_X} 1$

We have:

\(\ds \map {\Pi'_X} s\)

\(=\)

\(\ds \map {\frac \d {\d s} } {\frac {p s} {1 - q s} }\)

\(\ds \)

\(=\)

\(\ds \frac p {\paren {1 - q s}^2}\)

Derivatives of PGF of Shifted Geometric Distribution

Plugging in $s = 1$:

\(\ds \map {\Pi'_X} 1\)

\(=\)

\(\ds \frac p {\paren {1 - q}^2}\)

\(\ds \)

\(=\)

\(\ds \frac p {p^2}\)

as $q = 1 - p$

\(\ds \)

\(=\)

\(\ds \frac 1 p\)

Hence the result.

$\blacksquare$