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From the definition of KGF (cumulant generating function) we can write: \begin{align} K_x(t) &= \log_eM_x(t) \\ &= \log_e\left[1+\sum_{r=1}^{\infty}\frac{t^r}{r!}\mu_r^{'}\right] \\ &= k_1t+k_2\frac{t^2}{2!}+\cdots+k_r\frac{t^r}{r!}+\cdots \\ &= \log_e\left[ 1+t\mu_1^{'}+\frac{t^2}{2!}\mu_2^{'}+\cdots+\frac{t^r}{r!}\mu_r^{'}+\cdots \right] \end{align} But now I am confused how to get $r$th cumulant expressed in raw moments. My book shows that: $$ k_1=\mu_1^{'},\quad k_2=\mu_2^{'}-(\mu_1^{'})^2,\quad k_3=\mu_3^{'}-3\mu_2^{'}\mu_1^{'}+2(\mu_1^{'})^3,\quad \ldots $$ How do they get cumulants in term of moments?

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  • $\begingroup$ Why does the second equation hold? From $\log_e[...]$ to $k_1t+k_2\frac{t^2}{2!}...$. $\endgroup$
    – WCMC
    Aug 18 '20 at 21:37
  • $\begingroup$ Did you get the solution? I couldn’t expand the series.. @emonhossain $\endgroup$ Feb 5 at 8:58
  • $\begingroup$ Yes I manage to get that solution @MalihaSiddika. I use Multinomial theorem to collect the coefficient from the $\ln{(1+\star)}$. If you need detail answer let me know. By the way, I am also from math dept at same University 😃 $\endgroup$ Feb 25 at 19:43
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We can use the series expansion of $\ln\left(1 + x\right)$ to write $$\ln\left(1 + t\mu_{1}' + \dfrac{t^{2}}{2!}\mu_{2}' + \dots\right) = \sum_{j = 1}^{\infty}\left(-1\right)^{j - 1}\dfrac{\left(\frac{\mu_{1}'t}{1!} + \frac{\mu_{2}'t^{2}}{2!} + \dots\right)^{j}}{j} $$ The general technique is to then collect for powers of $t$ in $$k_{1}t + k_{2}\dfrac{t^{2}}{2!} + \dots = \sum_{j = 1}^{\infty}\left(-1\right)^{j - 1}\dfrac{\left(\frac{\mu_{1}'t}{1!} + \frac{\mu_{2}'t^{2}}{2!} + \dots\right)^{j}}{j}$$ and equate on both sides to solve for the cumulants in terms of the moments. It's relatively straightforward to do for the first few cumulants but becomes more tedious to do by hand for higher cumulants.

A good reference is 2.2.5 of Expansions and Asymptotics for Statistics by Small.

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  • $\begingroup$ I know that.But how could I expand $ln(1+x)$ where it's a large term in $x$.Can you provide links or something else where I can get the information @rzch Sir.Thanks sir for your great answer :) $\endgroup$ Dec 7 '18 at 13:48
  • $\begingroup$ Did this link from the original answer help? It doesn't matter how many terms are in $x$, you can still plug it in the series. $\endgroup$
    – rzch
    Dec 7 '18 at 15:50

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