Expectation of Shifted Geometric Distribution

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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 1

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 $q = 1 - p$
\(\ds \) \(=\) \(\ds \frac 1 p\)

$\blacksquare$


Proof 2

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.

$\blacksquare$


Sources

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