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.

  • 1
    $\begingroup$ 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. $\endgroup$ Feb 9, 2018 at 19:36
  • $\begingroup$ will give this a shot $\endgroup$ Feb 9, 2018 at 19:55

1 Answer 1


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 $n$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.