Expectation of Shifted Geometric Distribution/Proof 2
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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.
$\blacksquare$