Double exponential function Expected value Hi I'm having trouble calculating high moment of a double exponential function. $$f(x\mid\mu,\sigma)=\frac{1}{2\sigma}e^{-\left\lvert\frac{x-\mu}{\sigma}\right\rvert}$$
How do I calculate $E(X^{2009})$
I tried to calculate the moment generating function MGF but it does not work for this expectation since I have to take the derivative 2009 times!
Any suggestions? Thanks!
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\begin{align}
{\rm E}\pars{x^{n}}
&=\int_{-\infty}^{\infty}{1 \over 2\sigma}\,
\exp\pars{-\,{\verts{x - \mu} \over \sigma}}x^{n}\,\dd x
\\[3mm]&=
{1 \over 2\sigma}\bracks{%
\int_{-\infty}^{\mu}\exp\pars{x - \mu \over \sigma}x^{n}\,\dd x
+
\int_{\mu}^{\infty}\exp\pars{\mu - x\over \sigma}x^{n}\,\dd x}
\\[3mm]&=
{1 \over 2\sigma}\bracks{%
\expo{-\mu/\sigma}\sigma^{n + 1}
\int_{-\infty}^{\mu/\sigma}\expo{x}x^{n}\,\dd x
+
\expo{\mu/\sigma}\sigma^{n + 1}
\int_{\mu/\sigma}^{\infty}\expo{-x}x^{n}\,\dd x}
\\[3mm]&=
\half\,\sigma^{n}\bracks{%
\expo{-\mu/\sigma}\pars{-1}^{n + 1}
\int_{\infty}^{-\mu/\sigma}\expo{-x}x^{n}\,\dd x
+
\expo{\mu/\sigma}
\int_{\mu/\sigma}^{\infty}\expo{-x}x^{n}\,\dd x}
\\[3mm]&=
\half\,\sigma^{n}
\expo{-\mu/\sigma}\pars{-1}^{n + 1}
\bracks{-\Gamma\pars{n + 1} + \gamma\pars{n + 1,-\,{\mu \over \sigma}}}
\\[3mm]&\phantom{=}+
\\[3mm]&\phantom{=}
\half\,\sigma^{n}\expo{\mu/\sigma}
\bracks{-\gamma\pars{n + 1,{\mu \over \sigma}} + \Gamma\pars{n + 1}}
\\[3mm]&=
\half\sigma^{n}\bracks{\pars{-1}^{n}\expo{-\mu/\sigma} + \expo{\mu/\sigma}}
\Gamma\pars{n + 1}
\\[3mm]&\phantom{=}+
\\[3mm]&\phantom{=}
\half\sigma^{n}\bracks{%
\expo{-\mu/\sigma}\pars{-1}^{n + 1}\gamma\pars{n + 1,-\,{\mu \over \sigma}}
-
\expo{\mu/\sigma}\gamma\pars{n + 1,{\mu \over \sigma}}}
\end{align}
where $\Gamma\pars{z}$ is the Gamma function and $\gamma\pars{\alpha,z}$ is an incomplete gamma function. 

$\Gamma\pars{2009 + 1} = 2009!$. The $\gamma$'s are approximated by
  $
\gamma\pars{\alpha,x} \approx {x^{\alpha} \over \alpha} 
$ when $\alpha \gg 1$. Then
  $$
\gamma\pars{2009 + 1,\pm\,{\mu \over \sigma}} \approx {\pars{\pm\,\mu/\sigma}^{2010} \over 2010}
$$

A: Write the integral that is used to evaluate $E(X^{2009})$ then use reduction formulae
A: The generating function of $X$ is $E[\mathrm e^{tX}]=\mathrm e^{t\mu}/(1-\sigma^2t^2)$ hence
$$
\sum_{n\geqslant0}E[X^n]t^n/n!=\sum_{i\geqslant0}\mu^it^i/i!\cdot\sum_{j\geqslant0}\sigma^{2j}t^{2j}.
$$
Equating the coefficients of $t^n$, one gets
$$
E[X^n]=n!\sum_{i+2j=n}\mu^i\sigma^{2j}/i!,
$$
hence
$$
E[X^{2009}]=2009!\sum_{j=0}^{1004}\mu^{2009-2j}\frac{\sigma^{2j}}{(2009-2j)!}=2009!\sum_{k=0}^{1004}\mu^{2k+1}\frac{\sigma^{2008-2k}}{(2k+1)!}.
$$
If $1004\cdot\sigma/\mu$ is large, this can be approximated by the sum of the full series, that is,
$$
E[X^{2009}]\approx2009!\sigma^{2009}\sinh(\mu/\sigma).
$$
A: Let $W=(X-\mu)/\sigma$.  Then
$$
f_W(x) = \frac 1 2 e^{-|x|}.
$$
\begin{align}
E\left(X^{2009}\right) & = E\left((\sigma W+\mu)^{2009}\right) \\[10pt]
& = \sum_{k=0}^{2009} \binom{2009}{k} \sigma^k E(W^k)\mu^{2009-k} \\[10pt]
& = \sum_{k=0}^{2009} \binom{2009}{k} \sigma^k \mu^{2009-k} \frac 1 2 \int_{-\infty}^\infty x^k e^{-|x|} \,dx \\[10pt]
& = \sum_{k=0}^{2009} \binom{2009}{k} \sigma^k \mu^{2009-k} \int_0^\infty x^k e^{-x}\,dx \\[10pt]
& = \sum_{k=0}^{2009} \binom{2009}{k} \sigma^k \mu^{2009-k} k! \\[10pt]
& = \sum_{k=0}^{2009} \frac{2009!}{(2009-k)!} \sigma^k\mu^{2009-k}
\end{align}
I don't know how much, if anything, can be done beyond that.
