How can I found 3rd and 4th central moments of gumbel distribution with characteristic function?

Question

I tried many ways(using digamma function, partial integration etc.) to find moments using by characteristic function but I couldn't. Is there any trick or suggestion? (I couldn't calculate integral of 2nd, 3nd and 4th derivatives of gamma function.)

Thanks in advance..

Answer 1

The characteristic function for the Gumbel distribution with location parameter $\alpha$ and scale parameter $\beta$ is $$ \varphi_X(t) = \mathrm{e}^{\mathrm{i} t \alpha} \Gamma(1+\mathrm{i}t \beta) \text{.} $$ The $n^\text{th}$ (noncentral) moment can be found from the characteristic function using $$ E[X^n] = \mathrm{i}^{-n} \left. \frac{\mathrm{d}^n}{\mathrm{d}t^n} \varphi_X(t) \right|_{t = 0} \text{.} $$

So, \begin{align*} E[X^0] &= 1 \cdot \mathrm{e}^0 \Gamma(1) = 1 \\ E[X^1] &= -\mathrm{i} \cdot \left( \mathrm{e}^{\mathrm{i} \cdot 0 \cdot \alpha} \Gamma'(1 + 0)(\mathrm{i}\beta) + \mathrm{i}\alpha \mathrm{e}^{\mathrm{i}\cdot 0 \cdot \alpha} \Gamma(1+0)\right) \\ &= - \gamma \beta + \alpha \\ E[X^2] &= \cdots \\ &= \alpha ^2-2 \gamma \alpha \beta +\frac{1}{6} \left(6 \gamma ^2+\pi ^2\right) \beta ^2 \\ E[X^3] &= \cdots \\ &= \alpha ^3-3 \gamma \alpha ^2 \beta +\frac{1}{2} \left(6 \gamma ^2+\pi ^2\right) \alpha \beta ^2-\frac{1}{2} \beta ^3 \left(2 \gamma ^3+\gamma \pi ^2-2 \psi^{(2)}(1)\right) \\ E[X^4] &= \cdots \\ &= \alpha ^4-4 \gamma \alpha ^3 \beta +\left(6 \gamma ^2+\pi ^2\right) \alpha ^2 \beta ^2-2 \alpha \beta ^3 \left(2 \gamma ^3+\gamma \pi ^2-2 \psi^{(2)}(1)\right)+\frac{1}{20} \beta ^4 \left(20 \gamma ^4+20 \gamma^2 \pi^2 + 3 \pi^4 - 80 \gamma \psi^{(2)}(1)\right) \text{,} \end{align*} where

• $\gamma$ is the Euler-Mascheroni constant and
• $\psi^{(2)}$ is the polygamma function of order $2$.

Now we compute the central moments from the noncentral moments. The $n^\text{th}$ central moment, $\mu_n$ is $$ \mu_n = \begin{cases} 1 ,& n = 0 \\ E[X^1] ,& n = 1 \\ \sum_{j=0}^n \binom{n}{j} (-1)^{n-j} E[X^j] E[X^1]^{n-j} ,& n \geq 2 \end{cases} \text{.} $$ So, \begin{align*} \mu_1 &= E[X^1] \\ &= -\gamma \beta + \alpha \\ \mu_2 &= \alpha ^2-2 \gamma \alpha \beta +(i \alpha -i \gamma \beta )^2+\frac{\pi^2 \beta ^2}{6}+\gamma ^2 \beta ^2 \\ &= \frac{\pi^2 \beta^2}{6} \\ \mu_3 &= 3 i \left(\alpha ^2-2 \gamma \alpha \beta +\frac{\pi ^2 \beta ^2}{6}+\gamma ^2 \beta ^2\right) (i \alpha -i \gamma \beta )+i \left(-i \alpha ^3+3 i \gamma \alpha ^2 \beta +3 i \alpha \left(-\frac{1}{6} \pi ^2 \beta ^2-\gamma ^2 \beta ^2\right)+\frac{1}{2} i \gamma \pi ^2 \beta ^3+i \gamma ^3 \beta ^3-i \beta ^3 \psi ^{(2)}(1)\right)+2 i (i \alpha -i \gamma \beta )^3 \\ &= \beta ^3 \psi ^{(2)}(1) \end{align*} and you should see how to compute $\mu_4$.