Åke
Andersson
Department of Physics, University of Gothenburg, 412 96 Gothenburg, Sweden. E-mail: ake.andersson@physics.gu.se
First published on 22nd November 2023
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.
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.
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• 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,…) = aχn(X,Z2,…) + bχn(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) |
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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 A∪B vanishes:
χn(A1,…,Am, Bm+1,…,Bn) = 0. | (10) |
Proof. Because and are independent, we can separate the generating function like
KA1,…,Am,Bm+1,…,Bn(…) = KA1,…,Am(…) + KBm+1,…,Bn(…). | (11) |
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. □
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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 . Its coefficient will then be .
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 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 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.
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Footnote |
† Electronic supplementary information (ESI) available. See DOI: https://doi.org/10.1039/d3cp02525j |
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