Law of total cumulance

Only in case n = either 2 or 3 is the nth cumulant the same as the nth central moment. The case n = 2 is well-known (see law of total variance). Below is the case n = 3. The notation μ₃ means the third central moment.

${\displaystyle \mu _{3}(X)=\operatorname {E} (\mu _{3}(X\mid Y))+\mu _{3}(\operatorname {E} (X\mid Y))+3\operatorname {cov} (\operatorname {E} (X\mid Y),\operatorname {var} (X\mid Y)).}$

For general 4th-order cumulants, the rule gives a sum of 15 terms, as follows:

${\displaystyle {\begin{aligned}&\kappa (X_{1},X_{2},X_{3},X_{4})\\[5pt]={}&\kappa (\kappa (X_{1},X_{2},X_{3},X_{4}\mid Y))\\[5pt]&\left.{\begin{matrix}&{}+\kappa (\kappa (X_{1},X_{2},X_{3}\mid Y),\kappa (X_{4}\mid Y))\\[5pt]&{}+\kappa (\kappa (X_{1},X_{2},X_{4}\mid Y),\kappa (X_{3}\mid Y))\\[5pt]&{}+\kappa (\kappa (X_{1},X_{3},X_{4}\mid Y),\kappa (X_{2}\mid Y))\\[5pt]&{}+\kappa (\kappa (X_{2},X_{3},X_{4}\mid Y),\kappa (X_{1}\mid Y))\end{matrix}}\right\}({\text{partitions of the }}3+1{\text{ form}})\\[5pt]&\left.{\begin{matrix}&{}+\kappa (\kappa (X_{1},X_{2}\mid Y),\kappa (X_{3},X_{4}\mid Y))\\[5pt]&{}+\kappa (\kappa (X_{1},X_{3}\mid Y),\kappa (X_{2},X_{4}\mid Y))\\[5pt]&{}+\kappa (\kappa (X_{1},X_{4}\mid Y),\kappa (X_{2},X_{3}\mid Y))\end{matrix}}\right\}({\text{partitions of the }}2+2{\text{ form}})\\[5pt]&\left.{\begin{matrix}&{}+\kappa (\kappa (X_{1},X_{2}\mid Y),\kappa (X_{3}\mid Y),\kappa (X_{4}\mid Y))\\[5pt]&{}+\kappa (\kappa (X_{1},X_{3}\mid Y),\kappa (X_{2}\mid Y),\kappa (X_{4}\mid Y))\\[5pt]&{}+\kappa (\kappa (X_{1},X_{4}\mid Y),\kappa (X_{2}\mid Y),\kappa (X_{3}\mid Y))\\[5pt]&{}+\kappa (\kappa (X_{2},X_{3}\mid Y),\kappa (X_{1}\mid Y),\kappa (X_{4}\mid Y))\\[5pt]&{}+\kappa (\kappa (X_{2},X_{4}\mid Y),\kappa (X_{1}\mid Y),\kappa (X_{3}\mid Y))\\[5pt]&{}+\kappa (\kappa (X_{3},X_{4}\mid Y),\kappa (X_{1}\mid Y),\kappa (X_{2}\mid Y))\end{matrix}}\right\}({\text{partitions of the }}2+1+1{\text{ form}})\\[5pt]&{\begin{matrix}{}+\kappa (\kappa (X_{1}\mid Y),\kappa (X_{2}\mid Y),\kappa (X_{3}\mid Y),\kappa (X_{4}\mid Y)).\end{matrix}}\end{aligned}}}$

Suppose Y has a Poisson distribution with expected value λ, and X is the sum of Y copies of W that are independent of each other and of Y.

${\displaystyle X=\sum _{y=1}^{Y}W_{y}.}$

All of the cumulants of the Poisson distribution are equal to each other, and so in this case are equal to λ. Also recall that if random variables W₁, ..., W_m are independent, then the nth cumulant is additive:

${\displaystyle \kappa _{n}(W_{1}+\cdots +W_{m})=\kappa _{n}(W_{1})+\cdots +\kappa _{n}(W_{m}).}$

We will find the 4th cumulant of X. We have:

${\displaystyle {\begin{aligned}\kappa _{4}(X)={}&\kappa (X,X,X,X)\\[8pt]={}&\kappa _{1}(\kappa _{4}(X\mid Y))+4\kappa (\kappa _{3}(X\mid Y),\kappa _{1}(X\mid Y))+3\kappa _{2}(\kappa _{2}(X\mid Y))\\&{}+6\kappa (\kappa _{2}(X\mid Y),\kappa _{1}(X\mid Y),\kappa _{1}(X\mid Y))+\kappa _{4}(\kappa _{1}(X\mid Y))\\[8pt]={}&\kappa _{1}(Y\kappa _{4}(W))+4\kappa (Y\kappa _{3}(W),Y\kappa _{1}(W))+3\kappa _{2}(Y\kappa _{2}(W))\\&{}+6\kappa (Y\kappa _{2}(W),Y\kappa _{1}(W),Y\kappa _{1}(W))+\kappa _{4}(Y\kappa _{1}(W))\\[8pt]={}&\kappa _{4}(W)\kappa _{1}(Y)+4\kappa _{3}(W)\kappa _{1}(W)\kappa _{2}(Y)+3\kappa _{2}(W)^{2}\kappa _{2}(Y)\\&{}+6\kappa _{2}(W)\kappa _{1}(W)^{2}\kappa _{3}(Y)+\kappa _{1}(W)^{4}\kappa _{4}(Y)\\[8pt]={}&\kappa _{4}(W)\lambda +4\kappa _{3}(W)\kappa _{1}(W)\lambda +3\kappa _{2}(W)^{2}+6\kappa _{2}(W)\kappa _{1}(W)^{2}\lambda +\kappa _{1}(W)^{4}\lambda \\[8pt]={}&\lambda \operatorname {E} (W^{4})\qquad {\text{(the punch line -- see the explanation below).}}\end{aligned}}}$

We recognize the last sum as the sum over all partitions of the set { 1, 2, 3, 4 }, of the product over all blocks of the partition, of cumulants of W of order equal to the size of the block. That is precisely the 4th raw moment of W (see cumulant for a more leisurely discussion of this fact). Hence the cumulants of X are the moments of W multiplied by λ.

In this way we see that every moment sequence is also a cumulant sequence (the converse cannot be true, since cumulants of even order ≥ 4 are in some cases negative, and also because the cumulant sequence of the normal distribution is not a moment sequence of any probability distribution).

Suppose Y = 1 with probability p and Y = 0 with probability q = 1 − p. Suppose the conditional probability distribution of X given Y is F if Y = 1 and G if Y = 0. Then we have

${\displaystyle \kappa _{n}(X)=p\kappa _{n}(F)+q\kappa _{n}(G)+\sum _{\pi <{\widehat {1}}}\kappa _{\left|\pi \right|}(Y)\prod _{B\in \pi }(\kappa _{\left|B\right|}(F)-\kappa _{\left|B\right|}(G))}$

where ${\displaystyle \pi <{\widehat {1}}}$ means π is a partition of the set { 1, ..., n } that is finer than the coarsest partition – the sum is over all partitions except that one. For example, if n = 3, then we have

${\displaystyle \kappa _{3}(X)=p\kappa _{3}(F)+q\kappa _{3}(G)+3pq(\kappa _{2}(F)-\kappa _{2}(G))(\kappa _{1}(F)-\kappa _{1}(G))+pq(q-p)(\kappa _{1}(F)-\kappa _{1}(G))^{3}.}$