$$p(x)=e^{-2 |x|}$$ with x in [-inf, +inf]. I've calculated the characteristic function as $E[e^{ikx}]=\frac{1}{ik+2}-\frac{1}{ik-2}=\frac{4}{k^2+4}$. Now i'd like the moments.. so I suppose I should manipulate this expression in order to reduce it to a sum of infinite moments, but I can't! Could you help me?

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    $\begingroup$ Note that you have called the characteristic function the moment generating function in your body text. The mgf is $M_X(t)=E(e^{tX})$ not $E(e^{itX})$; they're not identical. Secondly, you should simplify your characteristic function before playing with it further, and then see if you can write it as a power series. $\endgroup$ – Glen_b Jan 12 '17 at 11:18
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    $\begingroup$ Please add the self-study tag and read its tag wiki $\endgroup$ – Glen_b Jan 12 '17 at 11:20
  • $\begingroup$ @Glen_b Ok, fixed question. But I still can't write it as power series. How should I proceed in these cases? $\endgroup$ – Surfer on the fall Jan 12 '17 at 11:33
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    $\begingroup$ The mechanical way uses Taylor's theorem. It's frequently the case that the Binomial theorem applies, as it does here: $$\frac{4}{k^2+4}=(1+(k^2/4))^{-1}=\sum_{i=0}^\infty \binom{-1}{i}\left(\frac{k^2}{4}\right)^i$$ provided $|k^2/4| \lt 1$. $\endgroup$ – whuber Jan 12 '17 at 15:24
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    $\begingroup$ By definition, $$\binom{n}{k} = \frac{n(n-1)\cdots(n-k+1)}{k!}.$$ That's all you need to obtain the moments, but if you want to see the actual coefficients, plug $-1$ in for $n$ to get $$\binom{-1}{k} = \frac{(-1)(-2)\cdots(-k)}{k!}=(-1)^k.$$ $\endgroup$ – whuber Jan 12 '17 at 20:12

One of the nice things about characteristic and moment-generating functions is that you can (usually) read the moments right off of them. You just have to expand them as MacLaurin series (Taylor series around zero). In this answer I use the techniques described at How to compute moments from an MGF explicitly and therefore will refer interested readers to that thread for more information.

This one is handled easily with the Binomial Theorem:

$$\frac{4}{k^2+4}=(1+(k^2/4))^{-1}=\sum_{j=0}^\infty \binom{-1}{j}\left(\frac{k^2}{4}\right)^j = \sum_{i=0}^\infty \frac{(-1)^j(2j)!}{2^{2j}}\frac{k^{2j}}{(2j)!}.$$

To achieve the expression at the right I used $$\binom{-1}{j} = \frac{(-1)(-2)\cdots(-j)}{j!}=(-1)^j$$ and I also inserted a factor of $$\frac{(2j)!}{(2j)!}=1$$ so that the general term in the sum would be in the form

$$(i)^n\mu_n \frac{k^n}{n!}$$

where, evidently, $n=2j$ must be an even number and $i=\sqrt{-1}$. We see, without any further calculation, that

$$\mu_n = \frac{n!}{2^n}.$$

This is the $n^\text{th}$ raw moment. All odd moments are zero.

It's always a good idea to check one's work if possible. To this end, note that some values of the even moments are

$$\matrix{ \text{n}&\mu_n \\ 0 & \frac{0!}{2^0} = 1 \\ 2 & \frac{2!}{2^2} = \frac{1}{2} \\ 4 & \frac{4!}{2^4} = \frac{3}{2} }$$

(I am in the habit of computing the zeroth moment--which is the total probability, always equal to $1$--as a check of the formula.)

These agree with the values for $b=\sqrt{4}=2$ listed in the Wikipedia entry for the Laplace distribution. In particular, the "excess kurtosis" is given by

$$\kappa_4 = \frac{\mu_4}{\mu_2^2}-3 = \frac{3/2}{(1/2)^2} - 3 = 6-3 = 3,$$

exactly as claimed.

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  • $\begingroup$ Unexpected idea: isn't the geometric expansion enough (instead of binomial)? $\endgroup$ – Surfer on the fall Jan 18 '17 at 12:23
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    $\begingroup$ @Surfer Yes, that's correct (and elegant). I like to cite the Binomial Theorem because for many people that is the unexpected result--and it is far more general. $\endgroup$ – whuber Jan 18 '17 at 14:17

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