Moment Generating Function of Geometric Distribution/Formulation 2/Examples/Third Moment
Jump to navigation
Jump to search
Examples of Use of Moment Generating Function of Geometric Distribution/Formulation 2
Let $X \sim \Geometric p$ for some $0 < p < 1$, where $\Geometric p$ is the Geometric distribution.
- $\map X \Omega = \set {0, 1, 2, \ldots} = \N$
- $\map \Pr {X = k} = p \paren {1 - p}^k$
The third moment generating function of $X$ is given by:
- $\map { {M_X}} t = p \paren {1 - p} e^t \paren {\dfrac {1 + 4 \paren {1 - p} e^t + \paren {1 - p}^2 e^{2t} } {\paren {1 - \paren {1 - p} e^t}^4 } }$
Proof
We have:
\(\ds \map { {M_X}} t\) | \(=\) | \(\ds \frac \d {\d t} \map { {M_X}} t\) | Definition of Moment Generating Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac \d {\d t} p \paren {1 - p} e^t \paren {\dfrac {1 + \paren {1 - p} e^t } {\paren {1 - \paren {1 - p} e^t}^3 } }\) | Moment Generating Function of Geometric Distribution: Second Moment | |||||||||||
\(\ds \) | \(=\) | \(\ds p \paren {1 - p} \paren {\frac \d {\d t} \paren {\paren { e^t + \paren {1 - p} e^{2t} } \paren {1 - \paren {1 - p} e^t}^{-3} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds p \paren {1 - p} \paren {\paren {e^t + 2 \paren {1 - p} e^{2t} } \paren {1 - \paren {1 - p} e^t}^{-3} - 3 \paren {e^t + \paren {1 - p} e^{2t} } \paren {1 - \paren {1 - p} e^t}^{-4} \paren {-\paren {1 - p} e^t} }\) | Product Rule for Derivatives, Chain Rule for Derivatives, Derivative of Power, Derivative of Exponential Function | |||||||||||
\(\ds \) | \(=\) | \(\ds p \paren {1 - p} \paren {\dfrac {e^t + 2 \paren {1 - p} e^{2t} } {\paren {1 - \paren {1 - p} e^t}^3 } + \dfrac {3 \paren {1 - p} e^{2t} + 3 \paren {1 - p}^2 e^{3t} } {\paren {1 - \paren {1 - p} e^t}^4 } }\) | gathering terms | |||||||||||
\(\ds \) | \(=\) | \(\ds p \paren {1 - p} \paren {\dfrac {e^t + 2 \paren {1 - p} e^{2t} } {\paren {1 - \paren {1 - p} e^t}^3 } \dfrac {\paren {1 - \paren {1 - p} e^t} } {\paren {1 - \paren {1 - p} e^t} } + \dfrac {3 \paren {1 - p} e^{2t} + 3 \paren {1 - p}^2 e^{3t} } {\paren {1 - \paren {1 - p} e^t}^4 } }\) | multiplying by $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds p \paren {1 - p} e^t \paren {\dfrac {1 + 2 \paren {1 - p} e^t - \paren {1 - p} e^t - 2 \paren {1 - p}^2 e^{2t} + 3 \paren {1 - p} e^t + 3 \paren {1 - p}^2 e^{2t} } {\paren {1 - \paren {1 - p} e^t}^4 } }\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds p \paren {1 - p} e^t \paren {\dfrac {1 + 4 \paren {1 - p} e^t + \paren {1 - p}^2 e^{2t} } {\paren {1 - \paren {1 - p} e^t}^4 } }\) | simplifying |
$\blacksquare$