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If a random variable $x$ has a characteristic function $\phi(\omega)$, then the $n^{\mathrm{th}}$ moment of the distribution of $x$, $\mu_n$ can be calculated as:

$$\mu_n = \imath^{-n}\left[\frac{d^n}{d\omega^n}\phi(\omega)\right]_{\omega=0}$$

Is there a similar formula to compute the $n^{\mathrm{th}}$ cumulant of the distribution, $\kappa_n$ in terms of the characteristic function? Wolfram MathWorld lists a formula as:

$$\ln[\phi(\omega)]=\sum_{n=0}^\infty\kappa_n \frac{(\imath \omega)^n}{n!}$$

However, this is not helpful in obtaining the $n^\mathrm{tm}$ cumulant (or if it is, I don't know how).

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2 Answers 2

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Yes, that expansion is all you need. Let $\ln[\phi(\omega)]=c(\omega)$, then $$\kappa_n = i^{-n}c^{(n)}(\omega)|_{\omega=0}$$ where $c^{(n)}$ is the $n$-th derivative of the cumulant generating function with respect to $\omega$. Also, it is possible to get the cumulants from the moments using the recursions in this link

http://en.wikipedia.org/wiki/Cumulant#Cumulants_and_moments

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$$\kappa_n = \imath^{-n}\left[\frac{\mathrm d^n}{\mathrm d\omega^n}\log\phi(\omega)\right]_{\omega=0}$$

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