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i've looked everywhere I could for an answer to this question but no luck ! If I have $X_1 .... X_n$ random variables that are independent and identically distributed such as ∀ $1 <$ $i$ $<n$, $X_i$ ~ B(n,$\theta$) (binomial distribution)

I know that the likelihood is :

$P_n$($\theta$,x)=$\prod\limits_{i}$ $\binom{n}{x_i}p^{x_i}(1-p)^{n-x_i}$

but then it seems kind of hard to calculate as product, I tried to calculate log($p_n$) but then the $x_i$! causes me problems

Can you enlighten me ? what is the max-likelihood of a binomial distribution ?

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Take the log-likelihood function, i.e. $$L(p) = \log \prod_i \binom{n}{x_i} p^{x_i}(1-p)^{n-x_i}$$ which becomes $$L(p) = \sum_i \log\binom{n}{x_i} p^{x_i}(1-p)^{n-x_i}$$ even more $$L(p) = \sum_i \log\binom{n}{x_i} + \sum_i x_i\log p + \sum_i(n-x_i)\log(1-p)$$ Since you're interested in the ML estimate of $p$. let's derive with respect to $p$ and set it to zero. $$\frac{d}{dp}L(p) = \frac{1}{p}\sum_i x_i - \frac{1}{1-p}\sum_i(n-x_i) = 0$$ which gives you $$\frac{1}{1-p}\sum_i(n-x_i) = \frac{1}{p}\sum_i x_i$$ or $$\frac{\sum_i(n-x_i)}{\sum_i x_i} = \frac{1-p}{p} = \frac{1}{p} - 1$$ Now find $p$. $$\frac{1}{p} = \frac{\sum_i(n-x_i) + \sum_i x_i}{\sum_i x_i} = \frac{\sum_i n}{\sum_i x_i} $$

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  • $\begingroup$ thank you, it was just a matter of concentration and calculating step by step $\endgroup$ – Blueberry Feb 1 '19 at 16:03

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