The Strongly Intensive Cumulants Made Simple: A Simplifjed Explanation Geared
T
- wards Intuitive Understanding
The Strongly Intensive Cumulants Made Simple: A Simplifjed - - PowerPoint PPT Presentation
The Strongly Intensive Cumulants Made Simple: A Simplifjed Explanation Geared T owards Intuitive Understanding Evan Sangaline I was going to reuse a talk, but... Wojtek bets on a bottle of a good wine that it is impossible to get strongly
Wojtek bets on a bottle of a good wine that it is impossible to get strongly intensive measures of the order higher then three. He could not get through your
(or right) in a simpler way? ~Viktor Begun
convolved variables The Taylor series coeffjcients/cumulants simply add.
T ake a tiny little box with a tiny little cumulant generating function ~dV ~V And add them up to make a big box with a big cumulant generating function
difgerent volumes
moments are mixed
There is an explicit dependency on the distribution
Reality Check: It’s OK if you don’t understand what this function “means.” It’s just a mathematical tool that makes this volume dependence explicit in a simple way. What you must understand is that:
aylor series coeffjcients of each grouped ξ term correspond to the moments that you measure in an experiment.
in the volume.
So, we know we can measure the coeffjcients of And we know that these depend on bad volume fmuctuations. Let’s come up with a new function that doesn’t.
We defjne a new generating function Ψ* using this quantity which we just showed was explicitly “strongly intensive” And the strongly intensive cumulants, κ*, are then defjned as the T aylor series coeffjcients of Ψ*. This is in direct analogy to how the cumulants are quite literally defjned as the coeffjcients of their generating function Ψ.
Use it to come up with recursion relationships that you can use to calculate arbitrary strongly intensive cumulants in terms of moments. I’ve already done this for you in the strongly intensive cumulants paper (https://arxiv.org/pdf/1505.00261.pdf). This is the same thing that you have to do for cumulants and, no, it’s not particularly easy. Can you calculate κ4,1 without looking up the equation? If the answer is no then you shouldn’t expect to be able to do it for κ*4,1 either. They’re equally diffjcult, and intimately related, problems.
The recursion relation simplifjes dramatically for κ*n,0 which leads to simple, ready to use expressions
Challenge: Can you get the Mrowczynski measure, which in this case is ********************************************************************* Q_1 kappa_3 (A/<A> − A/<A>) = kappa_3 (a/<a> − b/<b>) from your technique? ******************** ~Wojtek Broniowski
From https://arxiv.org/abs/1704.01532
Symmetric combinations of strongly intensive cumulants
From an email to Wojtek...
a[r1_, r2_] := Sum[Sum[If[r1 == 0 && r2 == 0, 1/Moment[{0, 1}], If[i1 == r1 && i2 == r2, 0,
Moment[{r1 - i1, r2 + 1 - i2}]/Moment[{0, 1}] ]], {i1, 0, r1}], {i2, 0, r2}] k[r1_, r2_] := Sum[Sum[ Binomial[r1 - 1, i1]*Binomial[r2, i2]*a[i1, i2]* Moment[{r1 - i1, r2 - i2}] , {i1, 0, r1 - 1}], {i2, 0, r2}]
These equations are trivial to verify explicitly if you work
This is literally just plugging and chugging.
This is literally just plugging and chugging.
This is literally just plugging and chugging.