Expectation of Negative Binomial Distribution/Second Form
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Theorem
Let $X$ be a discrete random variable with the negative binomial distribution (second form) with parameters $n$ and $p$.
Then the expectation of $X$ is given by:
- $\expect X = \dfrac n p$
Proof
From Probability Generating Function of Negative Binomial Distribution (Second Form), we have:
- $\map {\Pi_X} s = \paren {\dfrac {p s} {1 - q s} }^n$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF, we have:
- $\expect X = \map {\Pi'_X} 1$
We have:
\(\ds \map {\Pi'_X} s\) | \(=\) | \(\ds \map {\frac \d {\d s} } {\frac {p s} {1 - q s} }^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n p \paren {\frac {\paren {p s}^{n - 1} } {\paren {1 - q s}^{n + 1} } }\) | First Derivative of PGF of Negative Binomial Distribution/Second Form |
Plugging in $s = 1$:
\(\ds \map {\Pi'_X} 1\) | \(=\) | \(\ds n p \paren {\frac {p^{n - 1} } {\paren {1 - q}^{n + 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n p \paren {\frac {p^{n - 1} } {p^{n + 1} } }\) | as $p = 1 - q$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \expect X\) | \(=\) | \(\ds \frac n p\) | simplifying |
$\blacksquare$
Sources
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 4.3$: Moments: Exercise $6$