Negative binomail distribution and its variance

AI Thread Summary
The discussion focuses on estimating the variance of the negative binomial distribution using maximum likelihood estimation and Fisher's information. The original poster seeks feedback on their solution but expresses difficulty in deriving the variance despite understanding its theoretical basis. A suggestion is made to reformulate the problem in the context of the exponential family to facilitate the estimation process. Additionally, there is a critique of the maximum likelihood estimator (MLE) provided, specifically questioning the calculation of the log-likelihood function. The conversation emphasizes the need for clarity in the derivation process and the importance of accurate calculations in statistical estimation.
_joey
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Hi,

I am trying to estimate variance for negative Binomial distribution using
maximum likelihood estimation and Expected (Fisher's) information to determine its variance. I know what variance is for this distrubution but I cannot derive it.


Here is my solution. Any comments and suggestions are much appreciated

Thanks!

Solution: http://img825.imageshack.us/img825/7545/ndist.png
 
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I think I know what the problem is. I've been solving problems for many hours and I need a break. :)
 
Have you tried putting it into the form of the exponential family first?

Your MLE = x bar is incorrect. What did you get for your log-likelihood function?
 
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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