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I am working with cumulants of a distribution. I have an example of how the second cumulant may be simplified from:

$k_2 = p\alpha^2\left\{\left(\sum a_i\right)^2 - 2\sum_{i<j}a_ia_j\left(1-\rho^{j-i}\right)\right\}$

to:

$k_2 = p\alpha^2\left\{\sum a_i^2 +2\sum_{i<j}a_ia_j\rho^{j-i}\right\}$

Following this simplification, the terms $a_j$ represent a function of another variable, $b$ such that $a_j = \left(1-b\right)b^j$. When this is substituted into the expression for $k_2$, the following expression is found:

$k_2 = p\alpha^2\frac{\left(1-b\right)^2}{1-b^2}\left[1+\frac{2b\rho}{1-b\rho}\right]$

and how the third cumulant may be simplified from:

$k_3 = 2p\alpha^3 \left\{\left(\sum a_i\right)^3 -3\sum a_i \sum_{i<j}a_ia_j\left(1-\rho^{j-i}\right)+3\sum_{i<j<k}a_ia_ja_k\left(1-\rho^{j-i}\right)\left(1-\rho^{k-j}\right)\right\}$

to:

$k_3 = 2p\alpha^3\left\{\sum a_i^3+3\sum_{i<j} a_i a_j^2 \rho^{j-i} +3\sum_{i<j} a_i^2a_j\rho^{j-i}+6\sum_{i<j<k} a_ia_ja_k\rho^{k-i}\right\}$

Using a similar substitution for $a_j$ as above, $k_3$ can be expressed as:

$k_3 = 2p\alpha^3\frac{\left(1-b\right)^3}{1-b^3}\left[1+\frac{3b\rho}{1-b\rho}+\frac{3b^2\rho}{1-b^2\rho}+\frac{6b^3\rho^2}{\left(1-b\rho\right)\left(1-b^2\rho\right)}\right]$

These results are derived in a paper by Warren in the Journal of Hydrology 1986 http://dx.doi.org/10.1016/0022-1694(86)90080-6. I am currently able to use these results for $k_2$ and $k_3$, however, I need to derive the result for $k_4$. Helpfully, Warren details an interim step for this in his later paper http://link.springer.com/article/10.1007%2FBF01581449 in Stochastic Hydrology and Hydraulics in 1992. He provides the following form for $k_4$:

$k_4 = 6p\alpha^4\left\{\left(\sum a_i\right)^4 -4\left(\sum a_i\right)^2\sum_{i<j}a_ia_j\left(1-\rho^{j-i}\right)+4\sum a_i \sum_{i<j<k} a_i a_j a_k \left(1-\rho^{j-i}\right)\left(1-\rho^{k-j}\right) + 2\left[\sum_{i<j}a_ia_j\left(1-\rho^{j-i}\right)\right]^2 - 4\sum_{i<j<k<l} a_i a_j a_k a_l \left(1-\rho^{j-i}\right)\left(1-\rho^{k-j}\right)\left(1-\rho^{l-k}\right)\right\}$

My question is: How are $k_2$ and $k_3$ simplified as shown above, and how can the same technique be applied to simplify $k_4$ so it may be similarly expressed? I realise that the resulting expression for $k_4$ may be very complex!

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    $\begingroup$ you write that $a_j= (1-b)b^j$ --- this implies $a_j$ and $a_k$ are correlated (given $a_j$ I can calculate $a_k$); who says that is the case? $\endgroup$
    – Carlo Beenakker
    Commented May 5, 2017 at 17:38
  • $\begingroup$ There is a vote to close but this looks interesting and related to the linked cluster theorem in statistical mechanics $\endgroup$
    – Abdelmalek Abdesselam
    Commented May 5, 2017 at 21:08
  • $\begingroup$ @Carlo: I think the $a_j$'s are variables but not random ones. $\endgroup$
    – Abdelmalek Abdesselam
    Commented May 5, 2017 at 21:52
  • $\begingroup$ @CarloBeenakker After a careful look into p.132 of Warren's paper, I think what the OP wanted to ask for is the cumulants for the stationary Markov (autoregressive) chain defined by $P$, where $a_i$'s are rate instead of random quantities. $\endgroup$
    – Henry.L
    Commented May 5, 2017 at 21:53
  • $\begingroup$ @AbdelmalekAbdesselam I do not see how it is related to cluster theorem and thought this is no more than a calculation, could you explain how we can view it in that way? However, the calculation itself is kind of interesting so I upvoted it. thanks! $\endgroup$
    – Henry.L
    Commented May 5, 2017 at 21:54

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