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I have a PDF defined by: $$Pr(X=k)=p \left({n\choose k}a^k(1-a)^{n-k}\right)+(1-p) \left({n\choose k}b^k(1-b)^{n-k}\right)$$ I am not given any information about the parameters $n, p, a, b$. I am given a data-set $x=(x_1,\ldots,x_N)$, generated from this distribution with unknown parameters, and my aim is to find estimators for these parameters. My main issue is with the parameter for $n$.

So far I have used $\hat{n}=\max{x_i}$. I also tried using the MLE, along with the log-likelihood, but could not find a way to simplify the formula to be able to work with it.

Is using MLE a viable option in this case? If yes, how should I go about it? Is there any other way (not involving MLE) as a good estimator for n?

I recently asked a similar question, but for a Binomial Distribution.

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    $\begingroup$ Show us the likelihood expression you came up with. You should be able to put that in a global mixed integer nonlinear optimize, such as BARON, and let it crank out a numerical answer. You have one integer variable and 3 continuous variables, which is a rather low dimensional problem. You have nice lower and upper bounds for the continuous variables; max $x_i$ serves as lower bound for n, and if you use BARON, you will need to create an upper bound for n. You could bypass a mixed integer capability by solving separate optimization problems for each candidate value of n, but not recommended. $\endgroup$
    – Mark L. Stone
    Feb 9, 2018 at 19:36
  • $\begingroup$ will give this a shot $\endgroup$
    – Dylan Zammit
    Feb 9, 2018 at 19:55

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Not a full answer but hopeful a few hints:

What you have is called a mixture of two binomials. Mixtures models are usually fitted with MLE using Expectation Maximization algorithm. The log-likelihood is usually not concave.

Here is a tutorial where $p$ and $n$ are known: http://ai.stanford.edu/~chuongdo/papers/em_tutorial.pdf and a question on CV here: EM algorithm for a binomial distribution.

Maybe you can find a way to extend to method to unknown $p$ and $n$.

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  • $\begingroup$ perfect! I'd rather be pointed in a direction than having the full answer so I can learn new techniques. Thanks for the help $\endgroup$
    – Dylan Zammit
    Feb 9, 2018 at 19:52

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