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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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4 Answers 4

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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 $$ \gamma\pars{2009 + 1,\pm\,{\mu \over \sigma}} \approx {\pars{\pm\,\mu/\sigma}^{2010} \over 2010} $$

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Write the integral that is used to evaluate $E(X^{2009})$ then use reduction formulae

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Not sure I understand your suggestion. –  Did Sep 4 '13 at 11:11

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, $$ E[X^{2009}]\approx2009!\sigma^{2009}\sinh(\mu/\sigma). $$

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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.

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