Talk:Cumulant

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The formula

E\left(e^{tX}\right)=\exp\left(\sum_{n=1}^\infty\kappa_n t^n/n!\right)\,

might be written

\log E\left(e^{tX}\right)=\sum_{n=1}^\infty\kappa_n t^n/n!\,

The constant term is found by setting t = 0:

\log E\left(e^0\right)=0

Zero is not a cumulant, and so the function

\frac{d}{dt}\log E\left(e^{tX}\right)=\sum_{n=0}^\infty\kappa_{n+1} t^n/n!=\mu+\sigma^2t+\cdots

better deserves the name 'cumulant generation function'.

Bo Jacoby 12:55, 4 January 2006 (UTC)

I agree that there's no zeroth-order cumulant. But I don't think that's a reason to change the convention to what you've given here. In that version, the coefficient of tn/n! is not the nth cumulant, and that is potentially confusing. Besides, to speak of a zeroth cumulant and say that it's zero regardless of the probability distribution seems harmless at worst. Michael Hardy 00:03, 9 January 2006 (UTC)

I understand your reservations against changing conventions. Note, however, the tempting simplification obtained by differentiation.

The (new cumulant generation function of the) degenerate distribution is 0;
The (..) normal distribution is t.
The (..) bernoulli distribution is (1+(p−1−1)e−t)−1
The (..) binomial distribution is n(1+(p−1−1)e−t)−1
The (..) geometric distribution is (−1+(1−p)−1e−t)−1
The (..) negative binomial distribution is n(−1+(1−p)−1e−t)−1
The (..) poisson distribution is λet

Bo Jacoby 12:21, 31 January 2006 (UTC)

The last point kills your proposal: one wants to be able to speak not only of compositions of cumulant-generating functions, but of compositional inverses in cases where the expected value is not 0. So one wants the graph of the function to pass through (0, 0), with nonzero slope when the expected value is not 0. Michael Hardy 22:54, 31 January 2006 (UTC)

What do you mean? Please explain first and conclude later. The two definitions allow the same operations. The new definition just does not contain a superfluous zero constant term. The graph of the new cumulant-generating function passes through (0, μ) having the slope σ2. Curvature shows departure from normality: μ+σ2t. Bo Jacoby 09:14, 1 February 2006 (UTC)

I don't like that kind of definition for the cumulant generating function. Imagine you have a random variable X and a constant a. If K(t) is the cumulant generating function for X, then what is the cumulant generating function for aX? Using the standard definition it's K(at) whereas using your definition it would be aK(at) which is more complicated. Ossi 22:41, 4 April 2006 (UTC)

I'll comment further on Bo Jacoby's comments some day. But for now, let's note that what is in the article has been the standard convention in books and articles for more than half a century, and Wikipedia is not the place to introduce novel ideas. Michael Hardy 22:55, 4 April 2006 (UTC)

Contents

[edit] cumulant (encyclopedia)

I came randomly to see the article : no explanation about a cumulant were in view. A TOC was followed by formulas.

We math people love what we do. Let us try to do more : explain what we do (for this, I need help).

  • What class of math object is that
  • Who uses it and for what
  • Are there plain related concepts to invoke, &c. ? Thanks --DLL 18:47, 9 June 2006 (UTC)

P.S. Wolfram, for example, gives links to : Characteristic Function, Cumulant-Generating Function, Fourier Transform, k-Statistic, Kurtosis, Mean, Moment, Sheppard's Correction, Skewness, Unbiased Estimator, Variance. [Pages Linking Here]. Though I cannot tell if it is pertinent here, maybe a little check might be done ? Thanks again. --DLL 18:56, 9 June 2006 (UTC)

The very first sentence in the article says what cumulants are. Michael Hardy 21:20, 9 June 2006 (UTC)

[edit] joint cumulant

(please forgive my english)

I think that in the formula

\kappa(X_1,\dots,X_n) =\sum_\pi\prod_{B\in\pi}(|B|-1)!(-1)^{|B|-1}E\left(\prod_{i\in B}X_i\right)

the number |B| of elements in B should be replaced by the number | π | of blocks in π.


For example, in the given case n=3

\kappa(X,Y,Z)=E(XYZ)-E(XY)E(Z)-E(XZ)E(Y)-E(YZ)E(X)+2E(X)E(Y)E(Z).\,

, the constant before the term E(XYZ) (which corresponds to π = {{1,2,3}} : only one block of 3 items, i.e. |\pi|=1 and \pi={B} with |B|=3) is 1 = ( | π | − 1)! and not 2 = ( | B | − 1)!.

That is certainly correct in this case, and I think probably more generally. I've changed it; I'll come back and look more closely later. Michael Hardy 17:41, 19 September 2006 (UTC)

[edit] Intro

We need an intro --dudzcom 04:52, 24 December 2006 (UTC)

Seconded. I'm a stats n00b and without a basic intro, the encyclopaedic content lacks a basic context. Jddriessen 14:12, 3 March 2007 (UTC)

[edit] k-statistics

I think we should have a section on k-statistics. Could someone knowledgeable write a section describing them and explaining why they are unbiased estimators for the cumulants. Ossi 18:04, 30 December 2006 (UTC)

[edit] Cumulant "basis"?

It appears that you can reconstruct a function from its cumulants; that is, it seems like the cumulants define a "basis" of sorts the same way the sin and cos functions define a Fourier basis. Of course, a function isn't a linear combination of its cumulants, so it's not a linear basis, but in some sense it still seems like a basis. Comments? 155.212.242.34 (talk) 22:23, 11 December 2007 (UTC)

If two finite multisets of numbers have the same cumulant generating function, they are equal. The concept of a random variable is somewhat more general than just a multiset of numbers, and complications arise. It is worth while to understand multisets before trying to understand random variables. The derivative of the cumulant distribution function of a continuous random variable should be considered the limiting case of a series of derivatives of cumulant distribution functions of finite multisets of numbers. As probability density functions are nonnegative, they do not make vector spaces, and so the concept of basis does not immediately apply. Bo Jacoby (talk) 16:33, 19 March 2008 (UTC).

[edit] Error in formula

Quote: Some writers prefer to define the cumulant generating function, via the characteristic function, as h(t) where

h(t)=\log(E (e^{i t X}))=\sum_{n=1}^\infty\kappa_n \cdot\frac{(it)^n}{n!}=\mu\cdot t - \sigma^2\cdot\frac{ t^2}{2} +\cdots\,.

I suppose the formula should be:

h(t)=\log(E (e^{i t X}))=\sum_{n=1}^\infty\kappa_n \cdot\frac{(it)^n}{n!}=\mu\cdot i\cdot t - \sigma^2\cdot\frac{ t^2}{2} +\cdots\,.

Is there a reference? Bo Jacoby (talk) 00:43, 20 March 2008 (UTC).

You're right; the factor of i was missing. I don't think it should be too hard to find references. I wouldn't be surprised if this is in McCullagh's book. Michael Hardy (talk) 18:06, 21 March 2008 (UTC)
References added to article, for this point at least. The Kendall and Stuart ref would be good for many other of the results quoted (but a later edition might be sought out?). Melcombe (talk) 09:18, 17 April 2008 (UTC)

[edit] Improve intro ?

At the end of the intro, the final sentance says: "This characterization of cumulants is valid even for distributions whose higher moments do not exist." This seems to dangle somewhat...

  • exactly what is refered to by "this characterisation"?
  • it seems to imply there are other characterisations?
  • it seems to imply that cumulants might exist even if higher moments do not exist?

Melcombe (talk) 09:30, 17 April 2008 (UTC)

Probably could be improved; I'll think about it. When higher moments do not exist, then neither do higher cumulants. In that case, the characterization of cumulants that says the cumulant-generating function is the logarithm of the moment-generating function is problematic. That is what is meant. As far as other characterizations go, yes of course there are. Michael Hardy (talk) 21:10, 17 April 2008 (UTC)