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Comment on “Cumulant mapping as the basis of multi-dimensional spectrometry” by Leszek J. Frasinski, Phys. Chem. Chem. Phys., 2022, 24, 20776–20787

Åke Andersson
Department of Physics, University of Gothenburg, 412 96 Gothenburg, Sweden. E-mail: ake.andersson@physics.gu.se

Received 31st May 2023 , Accepted 2nd November 2023

First published on 22nd November 2023


Abstract

I state a general formula for the n-variate joint cumulant of the first order and prove that it satisfies the desired properties listed in Section 3.3 of Phys. Chem. Chem. Phys., 2022, 24, 20776–20787.


Motivation

A recent article by Frasinski1 develops a theory of cumulant mapping, which extends covariance mapping2 to any number of fragments. The central object of this theory is the n-variate joint cumulant of the first order, abbreviated nth cumulant. In Section 3.3, Frasinski lists what properties the nth cumulant should satisfy, and then gives explicit expressions for up to the 6th cumulant. How these can be found in practice is not elaborated upon.

The purpose of this comment is to show how we can find the nth cumulant—in theory and practice. I will do this by providing a general formula and describe how to evaluate it. Additionally, I will use the general formula to prove that cumulants fulfill some useful properties.

The general formula

Let X1,…,Xn be random variables. Their multivariate cumulant-generating function is3
 
image file: d3cp02525j-t1.tif(1)
From this function the nth cumulant is defined as3
 
image file: d3cp02525j-t2.tif(2)
This definition is simple and useful for proving properties, but difficult to evaluate. Later in this comment I will derive the more explicit expression
 
image file: d3cp02525j-t3.tif(3)
where image file: d3cp02525j-t4.tif means all partitions of {1,…,n} into k sets.

How to evaluate it

Let us say we want to find the 7th cumulant. The main thing we should do is to find the partitions of 7 into k numbers that each are 2 or greater. For k = 1 there is the trivial 7; for k = 2 there is 5 + 2 and 4 + 3; and for k = 3 there is 3 + 2 + 2. From each nontrivial partition we then create a sum over all congruent products of covariances. Hopefully the rule becomes apparent by looking at the result
 
image file: d3cp02525j-t5.tif(4)
where I have used i as a shorthand for Xi − 〈Xi〉 inside 〈[thin space (1/6-em)]〉. The prefactor of each sum is simply (−1)k+1(k − 1)!. The number of products in a sum can be calculated as
 
image file: d3cp02525j-t6.tif(5)
where #m is the number of parts of size m. Matlab and Python code implementing the nth cumulant is available as ESI.

Useful properties

The four desired properties listed in Section 3.3 of the original article1 are

χn(…) ≠ 0 only if all arguments are collectively correlated;

χn(…) has units of the product of all arguments;

χn(…) is linear in the arguments;

χn(…) is invariant under interchange of any two arguments.

I will now prove that the cumulant has these desired properties, starting with the interchange of arguments.

Property 1 (symmetric). The nth cumulant is invariant under permutation of its arguments:

 
χn(Xπ(1),…,Xπ(n)) = χn(X1,…,Xn).(6)

Proof. Commutativity of addition, and of differentiation. □

The desired properties about linearity and units are combined into one, because the former implies the latter.

Property 2 (multilinear). The nth cumulant is linear in each of its arguments:

 
χn(aX + bY,Z2,…) = n(X,Z2,…) + n(Y,Z2,…).(7)

Proof. Because of symmetry we only have to prove linearity in the first argument. By expanding the expression

 
KaX+bY,…(t1,…) − aKX,…(t1,…) − bKY,…(t1,…)(8)
in its first argument t1, we find that the first-order terms cancel out. Differentiating with respect to t1 (among others) and evaluating at the origin will therefore give
 
image file: d3cp02525j-t7.tif(9)
Next, I phrase the desired property about correlation conversely.

Property 3 (discerning). Let (Ai)mi=1 and (Bj)nj=m+1 be nonempty tuples of random variables such that Ai and Bj are independent. Then the nth cumulant of AB vanishes:

 
χn(A1,…,Am, Bm+1,…,Bn) = 0.(10)

Proof. Because image file: d3cp02525j-t8.tif and image file: d3cp02525j-t9.tif are independent, we can separate the generating function like

 
KA1,…,Am,Bm+1,…,Bn(…) = KA1,…,Am(…) + KBm+1,…,Bn(…).(11)
Differenting with respect to tn and t1 (among others) will annihilate both terms. □

Finally, I note an important property that follows from the last two. It tells us that independent signals simply add their contributions to a cumulant.

Property 4 (additive). Let (Ai)ni=1 and (Bj)nj=1 be equal-length tuples of random variables such that Ai and Bj are independent. Then the nth cumulant distributes over the addition of these tuples:

 
χn(A1 + B1,…,An + Bn) = χn(A1,…,An) + χn(B1,…,Bn).(12)

Proof. By repeatedly using linearity, we can expand the left hand side into 2n terms. The mixed terms containing both some Ai and some Bj vanish because of the discerning property. □

Explicit expression

Our strategy will be to approximate the generating function with Taylor series, starting from the inside. The exponential can be truncated by removing terms containing any ti2
 
image file: d3cp02525j-t10.tif(13)
The last product turns into 2n terms, one for each subset of I = {1,…,n}. The term corresponding to S contains ti if and only if S contains i. Explicitly,
 
image file: d3cp02525j-t11.tif(14)
Applying the expectation value is straightforward linearity
 
image file: d3cp02525j-t12.tif(15)
In anticipation of taking the logarithm, we extract a 1 from our expectation value taking out the term where S is the empty set
 
image file: d3cp02525j-t13.tif(16)
Now, we plug this x into the Taylor series of the logarithm
 
image file: d3cp02525j-t14.tif(17)
Recall that the cumulant is obtained by differentiating the above expression with respect to each variable and evaluating at the origin. This can be thought of as extracting the coefficient of the image file: d3cp02525j-t15.tif term. Hence we will focus on contributions to it.

The first term, x, contains one subterm for each nonempty subset S of I. Only the subterm corresponding to S = I will be proportional to image file: d3cp02525j-t16.tif. Its coefficient will then be image file: d3cp02525j-t17.tif.

The second term, −x2/2, contains when expanded one subterm for every ordered pair of nonempty subsets (S1,S2) of I. In order to get a term proportional to image file: d3cp02525j-t18.tif each index i must be an element in exactly one of S1 and S2. In other words, {S1,S2} must be a partition of I. For every partition of I there are 2! matching ordered pairs, each contributing image file: d3cp02525j-t19.tif to the coefficient.

By now the rule for the kth term is clear. It will contain subterms corresponding to each partition of I into k nonempty sets. Each subterm will be a product of the prefactor (−1)k+1(k − 1)! and k expectation values.

 
image file: d3cp02525j-t20.tif(18)
We can make single-variable expectations vanish by replacing Xi with Xi − 〈Xi〉 everywhere. Using the additive property with Ai = Xi and Bi = −〈Xi〉, we see that this change preserves the cumulant.

Conflicts of interest

There are no conflicts to declare.

References

  1. L. J. Frasinski, Phys. Chem. Chem. Phys., 2022, 24, 20776–20787 RSC.
  2. V. Zhaunerchyk, L. Frasinski, J. H. Eland and R. Feifel, Phys. Rev. A: At., Mol., Opt. Phys., 2014, 89, 053418 CrossRef.
  3. A. Stuart and K. Ord, Kendall's advanced theory of statistics, distribution theory, John Wiley & Sons, 2010, vol. 1 Search PubMed.

Footnote

Electronic supplementary information (ESI) available. See DOI: https://doi.org/10.1039/d3cp02525j

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