Expectation of Shifted Geometric Distribution/Proof 1
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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 definition of expectation:
- $\expect X = \ds \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
By definition of shifted geometric distribution:
- $\expect X = \ds \sum_{k \mathop \in \Omega_X} k p \paren {1 - p}^{k - 1}$
Let $q = 1 - p$:
\(\ds \expect X\) | \(=\) | \(\ds p \sum_{k \mathop \ge 0} k q^{k - 1}\) | as $\Omega_X = \N$ | |||||||||||
\(\ds \) | \(=\) | \(\ds p \sum_{k \mathop \ge 1} k q^{k - 1}\) | The term in $k = 0$ vanishes | |||||||||||
\(\ds \) | \(=\) | \(\ds p \frac 1 {\paren {1 - q}^2}\) | Derivative of Geometric Sequence | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac p {p^2}\) | as $q = 1 - p$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 p\) |
$\blacksquare$
Sources
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 2.4$: Expectation: Example $24$