Derivatives of Moment Generating Function of Logistic Distribution

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Theorem

Let $X$ be a continuous random variable which satisfies the logistic distribution:

$X \sim \map {\operatorname {Logistic} } {\mu, s}$

for some $\mu \in \R, s \in \R_{> 0}$.

Let $\size t < \dfrac 1 s$.

Let $M_X$ denote the moment generating function of $X$.


The $n$th derivative of $M_X$ is given by:

$\ds {M_X}^{\paren n} = \map \exp {\mu t} \sum_{k \mathop = 0}^n \paren {-1}^k \dbinom n k \mu^{n - k} s^k \int_{\to 0}^{\to 1} \map {\ln^k} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u$


Proof

The proof proceeds by induction on $n$.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$\ds {M_X}^{\paren n} = \map \exp {\mu t} \sum_{k \mathop = 0}^n \paren {-1}^k \dbinom n k \mu^{n - k} s^k \int_{\to 0}^{\to 1} \map {\ln^k} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u$


Basis for the Induction

$\map P 0$ is the case:

\(\ds {M_X}^{\paren 0}\) \(=\) \(\ds M_X\) Definition of Higher Derivative
\(\ds \) \(=\) \(\ds \map \exp {\mu t} \int_{\to 0}^{\to 1} \paren {\dfrac {1 - u} u}^{-s t} \rd u\) Moment Generating Function of Logistic Distribution
\(\ds \) \(=\) \(\ds \map \exp {\mu t} \sum_{k \mathop = 0}^0 \paren {-1}^k \dbinom n k \mu^{n - k} s^k \int_{\to 0}^{\to 1} \map {\ln^k} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u\)


Thus $\map P 0$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\ds {M_X}^{\paren k} = \map \exp {\mu t} \sum_{r \mathop = 0}^k \paren {-1}^r \dbinom k r \mu^{k - r} s^r \int_{\to 0}^{\to 1} \map {\ln^r} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u$


from which it is to be shown that:

$\ds {M_X}^{\paren {k + 1} } = \map \exp {\mu t} \sum_{r \mathop = 0}^{k + 1} \paren {-1}^r \dbinom {k + 1} r \mu^{k + 1 - r} s^r \int_{\to 0}^{\to 1} \map {\ln^r} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u$


Induction Step

This is the induction step:

\(\ds {M_X}^{\paren {k + 1} }\) \(=\) \(\ds \dfrac \d {\d t} {M_X}^{\paren k}\) Definition of Higher Derivative
\(\ds \) \(=\) \(\ds \dfrac \d {\d t} \map \exp {\mu t} \sum_{r \mathop = 0}^k \paren {-1}^r \dbinom k r \mu^{k - r} s^r \int_{\to 0}^{\to 1} \map {\ln^r} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \map \exp {\mu t} \paren {\mu \sum_{r \mathop = 0}^k \paren {-1}^r \dbinom k r \mu^{k - r} s^r \int_{\to 0}^{\to 1} \map {\ln^r} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u + \paren {-s \map \ln {\dfrac {1 - u} u} } \sum_{r \mathop = 0}^k \paren {-1}^r \dbinom k r \mu^{k - r} s^r \int_{\to 0}^{\to 1} \map {\ln^r} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u }\) Product Rule, Chain Rule for Derivatives Derivative of Power of Constant and Derivative of Exponential Function
\(\ds \) \(=\) \(\ds \map \exp {\mu t} \paren {\sum_{r \mathop = 0}^k \paren {-1}^r \dbinom k r \mu^{k + 1 - r} s^r \int_{\to 0}^{\to 1} \map {\ln^r} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u + \sum_{r \mathop = 0}^k \paren {-1}^{r + 1} \dbinom k r \mu^{k - r} s^{r + 1} \int_{\to 0}^{\to 1} \map {\ln^{r + 1} } {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u }\)
\(\ds \) \(=\) \(\ds \map \exp {\mu t} \paren {\sum_{r \mathop = 0}^k \paren {-1}^r \dbinom k r \mu^{k + 1 - r} s^r \int_{\to 0}^{\to 1} \map {\ln^r} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u + \sum_{r \mathop = 1}^{k + 1} \paren {-1}^r \dbinom k {r - 1} \mu^{k + 1 - r} s^r \int_{\to 0}^{\to 1} \map {\ln^r} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u }\) re-index the second sum: $r = r + 1$
\(\ds \) \(=\) \(\ds \map \exp {\mu t} \sum_{r \mathop = 0}^{k + 1} \paren {-1}^r \dbinom {k + 1} r \mu^{k + 1 - r} s^r \int_{\to 0}^{\to 1} \map {\ln^r} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u\) Pascal's Rule: $\dbinom n {k - 1} + \dbinom n k = \dbinom {n + 1} k$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\ds \forall n \in \Z_{\ge 0}: {M_X}^{\paren n} = \map \exp {\mu t} \sum_{k \mathop = 0}^n \paren {-1}^k \dbinom n k \mu^{n - k} s^k \int_{\to 0}^{\to 1} \map {\ln^k} {\dfrac {1 - u} u} \paren {\dfrac {1 - u} u}^{-s t} \rd u$

$\blacksquare$