Re: Is n a parameter of a binomial distribution?

On Oct 18, 1:23 pm, Vinayak Rao <vinayak....@xxxxxxxxx> wrote:
Hi,
Is the number of experiments 'n' a parameter of the binomial
distribution

p(k|n,p) = nCk p^k (1-p)^(n-k) ?

The answer would seem to be 'yes' almost by definition, but if I
express the binomial distribution in the "exponential family form", it
has only 1 natural parameter log(p/(1-p)). If n were a parameter,
would this mean that the binomial belongs to the exponential family
only w.r.to parameter p (and not n)?

Vinayak

Vinayak,
n is not usually considered a parameter in the binomial
distribution. It is a fixed number by the design of the experiment. We
usually estimate p in a given number of n trials where there are k
successes, and k is a random variable ranging between zero and n.

Your expression of the density as a member of the exponential family
is another reason why n is not usually considered a "parameter".

Mark

.

Relevant Pages

  • Re: Is n a parameter of a binomial distribution?
    ... express the binomial distribution in the "exponential family form", ... and k is a random variable ranging between zero and n. ... Dept. of Mathematics and Statistics ...
    (sci.stat.math)
  • Is n a parameter of a binomial distribution?
    ... express the binomial distribution in the "exponential family form", ... has only 1 natural parameter log). ... would this mean that the binomial belongs to the exponential family ...
    (sci.stat.math)
  • Re: Binomial Distribution
    ... n and p (success probability). ... The conditional distribution of X_1 given T is hypergeometric, ... from a population of n_1+n_2 items containing t successes. ...
    (sci.math)
  • Re: Testing Hypotheses
    ... Significance level = 5%, observed value X=58 ... successes in 100 trials at alpha=5% significance ... value to which the distribution function is precisely ... between the Distribution Function and the Critical ...
    (sci.stat.math)
  • Re: expected maximum of a sequence of binomial variables
    ... Suppose that $X_1$, ..., $X_t$ are an independent sequence of ... each $X_i$ represents the number of successes of $n$ ... I can think of how one might come up with a relatively messy formula ... I think you will ultimately need the joint distribution of Y and Z. ...
    (sci.math)