Moment Generating Function of Geometric Distribution/Formulation 2

Theorem

Let $X$ be a discrete random variable with a geometric distribution with parameter $p$ for some $0 < p < 1$.

$\map X \Omega = \set {0, 1, 2, \ldots} = \N$

$\map \Pr {X = k} = p \paren {1 - p}^k$

Then the moment generating function $M_X$ of $X$ is given by:

$\map {M_X} t = \dfrac p {1 - \paren {1 - p} e^t}$

for $t < -\map \ln {1 - p}$, and is undefined otherwise.

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Proof

From the definition of the geometric distribution, $X$ has probability mass function:

$\map \Pr {X = k} = p \paren {1 - p}^k$

From the definition of a moment generating function:

$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{k \mathop = 0}^\infty \map \Pr {X = k} e^{k t}$

So:

\(\ds \sum_{k \mathop = 0}^\infty \map \Pr {X = k} e^{k t}\)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty p \paren {1 - p}^k e^{k t}\)

\(\ds \)

\(=\)

\(\ds p \sum_{k \mathop = 0}^\infty \paren {\paren {1 - p } e^t}^k\)

By Sum of Infinite Geometric Sequence, for this sum to be convergent we must have:

$\size {\paren {1 - p} e^t} < 1$

In the case $p = 1$, this demand is satisfied immediately regardless of $t$.

Otherwise, as both $e^t$ and $p$ are positive:

$e^t < \dfrac 1 {1 - p}$

So, by Logarithm of Power:

$t < -\map \ln {1 - p}$

is the range of $t$ for which the expectation is well-defined.

Now applying Sum of Infinite Geometric Sequence, we have for this range of $t$:

$\ds \map {M_X} t = p \sum_{k \mathop = 0}^\infty \paren {\paren {1 - p} e^t}^k = \frac p {1 - \paren {1 - p } e^t}$

$\blacksquare$

First Moment

The first moment generating function of $X$ is given by:

$\map { {M_X}'} t = \dfrac {p \paren {1 - p} e^t } {\paren {1 - \paren {1 - p} e^t}^2 }$

Second Moment

The second moment generating function of $X$ is given by:

$\map { {M_X}} t = p \paren {1 - p} e^t \paren {\dfrac {1 + \paren {1 - p} e^t } {\paren {1 - \paren {1 - p} e^t}^3 } }$

Third Moment

The third moment generating function of $X$ is given by:

$\map { {M_X}} t = p \paren {1 - p} e^t \paren {\dfrac {1 + 4 \paren {1 - p} e^t + \paren {1 - p}^2 e^{2t} } {\paren {1 - \paren {1 - p} e^t}^4 } }$

Fourth Moment

The fourth moment generating function of $X$ is given by:

$\map { {M_X}^{\paren 4} } t = p \paren {1 - p } e^t \paren {\dfrac {1 + 11 \paren {1 - p} e^t + 11 \paren {1 - p}^2 e^{2t} + \paren {1 - p}^3 e^{3t} } {\paren {1 - \paren {1 - p} e^t}^5 } }$