Derivatives of Moment Generating Function of Logistic Distribution

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 for Derivatives, 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:

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

$\blacksquare$