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I have $X_1,X_2,\dots,X_n$ as random samples from a binomial distribution, with probability function:

$$p_X(x) = Pr(X=x) = {m \choose{n}}\alpha^x(1-\alpha)^{m-x},x=0,1,2,\dots,m$$ where $m$ is given and $\alpha \in (0,1)$ is an unknown parameter.

I want to show that the maximum likelihood estimator of $\alpha$ is given by the sample mean over $m$.

Now I imagine the first thing I must do is, find the maximum likelihood estimator.

I believe this is found by finding the joint distribution for my data, and equating the first derivative to be $0$, is this the right direction?

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  • $\begingroup$ Yes, that and factoring everything that does not depend on $\alpha$ in the joint likelihood. $\endgroup$ – Did Aug 21 '14 at 11:02
  • $\begingroup$ @Did Could you verify the answer from Nicolas? I would just like a second opinion. If his first assumption is correct, I can see that all of the following method is correct. $\endgroup$ – Tony Aug 21 '14 at 12:22
  • $\begingroup$ Indeed the answer had some misprints, corrected now. $\endgroup$ – Did Aug 21 '14 at 13:23
  • $\begingroup$ @Did Thank you for that. $\endgroup$ – Tony Aug 23 '14 at 11:10
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The likelihood function, assuming the samples are independent, is $$L(\alpha) = \prod_{i=1}^N p_X(x_i)= \prod_{i=1}^N {m \choose {x_i} } \alpha^{x_i} (1-\alpha)^{m-x_i}.$$ The next step is to take the logarithm of the likelihood function $$ \log L(\alpha) = \sum_{i=1}^N\log{m \choose {x_i} }+\sum_{i=1}^Nx_i \log(\alpha) +\sum_{i=1}^N(m-x_i) \log(1-\alpha), $$ and to equate its derivative with respect to $\alpha $ to zero, that is, $$ \frac{\partial \log L(\alpha)}{\partial \alpha}=\frac{1}{\alpha}\sum_{i=1}^N x_i -\frac{1}{1-\alpha}\sum_{i=1}^N (m-x_i)=0. $$ From this we deduce that the MLE is $$\hat \alpha=\frac1N\sum_{i=1}^N \frac{x_i}m.$$

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  • $\begingroup$ Why have we assumed $P_X = L(\alpha)$? I don't doubt the result, but I thought that we had the joint pdf, normally I work with IID and just taking $\Pi P_X$ $\endgroup$ – Tony Aug 21 '14 at 11:54
  • $\begingroup$ Sorry, you are right. I edited this answer above. $\endgroup$ – nico Aug 21 '14 at 12:40
  • $\begingroup$ If you really want to be precise, you should also verify that the found $\hat \alpha$ really is a maximum by calculating the second derivative. $\endgroup$ – Aahz Aug 21 '14 at 13:33
  • $\begingroup$ @NicolasThorstensen I forgot to thank you, this was very helpful! $\endgroup$ – Tony Aug 26 '14 at 13:55

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