3
$\begingroup$

If one does not define cumulants via the cumulant generating function (cgf), e.g. because the cgf does not exist, then an alternative way is to use the recusion \begin{align*} \kappa_n=\mu'_n-\sum_{m=1}^{n-1}{n-1 \choose m-1}\kappa_m \mu_{n-m}', \end{align*} where $\mu_i'$ denotes the $i$th uncentered moment.

For this definition, what is the best way to show that the cumulants are additive under an independence assumption? More precisely, how do we show that \begin{align*} \kappa_n(X+Y) = \kappa_n(X) + \kappa_n(Y), \end{align*} if $X$ and $Y$ are two independent random variables?

$\endgroup$

1 Answer 1

7
$\begingroup$

I assume all your moments are finite. In that case, a simple proof is to approximate your random variable by truncation, use the additivity for the approximated cummulants, and then pass to the limit using dominated convergence.

$\endgroup$
2
  • $\begingroup$ would an induction actually work? $\endgroup$
    – user45183
    Mar 24, 2015 at 15:58
  • $\begingroup$ @seno44 I see no reason why the binomial theorem would not give you what you want by induction. $\endgroup$
    – Igor Rivin
    Mar 24, 2015 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.