Talk:Degenerate distribution

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[edit] PMF vs PDF

I think talking about the Dirac Delta function as a "generalized PDF" is mixing apples and oranges. The degenerate distribution is a discrete distribution, notions of continuity don't apply here. The PMF is zero everywhere except at, say, zero, where it is unity. Thats not a delta function. It not that I think the present statement is flawless, its just that the new statment is more obscure. I wouldn't mind if we could come up with a more specific statement. Miguel mentioned "generalized PDF". Can we clarify what that means? PAR 20:36, 8 May 2005 (UTC)

Yes, the Dirac Delta function is not a function, at best a "generalized function" in physics parlance (and a synonim with distribution in the sense of Schwartz). You wrote the discrete distribution is the discrete counterpart of the Dirac delta function, which IMHO makes no sense as it is exactly the same thing. For instance, if X is degenerate at x0, then the expected value of f(X) is f(x0). Precisely:
E(f(X))=f(x_0)=\int f(x)\delta_{x_0}(x)dx=\int f(x)d\theta(x-x_0)
which means that if you wish you can interpret the dirac delta as the pdf of the degenerate distribution using the theory of distributions of Laurent Schwartz. You can also use the Heaviside step function as a Stieltjes integrator, like any cdf, and I would advocate that if Stieltjes integrals didn't scare people off. So, the Dirac delta in not just "a counterpart" in any sense. However, since I agree talking of "generalized pdf" may be confusing, I feel like reverting the article to before your previous edits on this basis. I will restore the infobox, though. — Miguel 11:26, 2005 May 10 (UTC)

I changed the beginning to emphasize the fact that the degenerate distribution is a discrete rather than continuous distribution. In other words, its domain is countable, and can be assigned integer values.

Thats where my problem with the Dirac delta function comes in. We can't be using continuum ideas here, because they don't apply. There is no such thing as a value of 3.22 when rolling a die. It makes no sense to integrate over the value of k when k is restricted to being integers. A function fk which is only defined for integer values of k, and is zero for all values of k except it is unity at k=0 is not the same as the Dirac delta function. I also disagree with the idea that the cumulative distribution function is the Heaviside step function. The cumulative distribution function is likewise only defined for integer values of the argument:

F_k = \sum_{j=-\infty}^k f_j

and is zero for k<0 and unity for k ≥ 0. (In other words, I still have problems with the article, but I don't want to use the article as a forum for our disagreements) PAR 15:20, 10 May 2005 (UTC)

[edit] Continuous v. discrete

I am not convinced by the argument above, and I think the article needs more discussion about degenerate distributions and point probabilities from a discrete and continuous perspective. I could easily imagine somebody saying that a die (dice) with a probability mass function

Prob(X=x)={1 \over 6} \text{ for }x=1,2,3,4,5,6

could be written using Kronecker deltas as

Prob(X=n)={\delta_{x1} + \delta_{x2} + \delta_{x3} + \delta_{x4} + \delta_{x5} + \delta_{x6} \over 6}

so in a sense might be considered as if it had a density

p(x)={\delta(x-1)+ \delta(x-2)+ \delta(x-3)+ \delta(x-4)+ \delta(x-5)+ \delta(x-6) \over 6}

to the extent that any Dirac delta densities are meaningful. They could then extend this to the case where the die was fixed to produce a single result with probability 1. --Rumping (talk) 17:32, 3 March 2008 (UTC)

Yes, I believe that is entirely correct within the context of the theory of distributions. 207.62.177.230 (talk) 19:14, 7 April 2008 (UTC)