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?
Solved – Relations between moments and cumulants
cumulantsmoment-generating-functionmoments
Related Solutions
Question asks: "is the $n$th cumulant equivalent to the $n$th central moment (i.e. about the mean)?"
Answer is: only for $n = 1, 2$ or $3$.
Here, for example, are the first 9 cumulants of the population in terms of central moments $\mu_i$ of the population:
using mathStatica's CumulantToCentral function.
More generally
In a multivariate world, the product cumulant will only be identical to the product central moments if 1 < (sum of the indexes) $\le$ 3. For example, $\kappa_{i,j,k}$ will be equal to $\mu_{i,j,k}$ provided $1 < i+j+k \le 3$. Here are some bivariate product cumulants expressed in terms of product central moments of the population:
The equation given by Wikipedia connects cumulants to moments (generally).
A proof of a formula connecting cumulants to central moments is found in A Recursive Formulation of the Old Problem of Obtaining Moments from Cumulants and Vice Versa
Letting $K(t)$ be the cumulant-generating function, and $M(t)$ the moment-generating function. The relationship between the two is \begin{equation} M(t)=\exp{\left[K(t)\right]} \end{equation} The proof follows by differentiation of this expression and noting that the $n$th derivative can be written as \begin{equation} D^n[M(t)]=\sum_{i=0}^{n-1}\binom{n-1}{i}D^{n-i}[K(t)]D^i[M(t)] \end{equation} Where $D^k$ denotes the $k$th derivative. Now setting $t=0$: \begin{equation} \theta_n=\sum_{i=0}^{n-1}\binom{n-1}{i}\kappa_{n-i}\theta_i\\ \theta_n=\kappa_n+\sum_{i=1}^{n-1}\binom{n-1}{i}\kappa_{n-i}\theta_i\\ \end{equation} Rewriting yields: \begin{equation} \kappa_n = \theta_n-\sum_{i=1}^{n-1}\binom{n-1}{i}\kappa_{n-i}\theta_i \end{equation} In terms of the central moments and cumulants.
Best Answer
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.