Let $X_1, X_2, . . . , X_n$ be a random sample from a Bernoulli distribution with parameter $θ$. Compute the maximum likelihood estimator for $θ$.

In my opinion this is the correct way to solve it:
$l'(\theta)=\frac{n}{\theta}-\frac{n}{1-\theta}=0$, $\theta=\frac{1}{2}$
For the official solution it should be $\frac{\sum_{i=1}^nX_i}{n}$, why MLE is incorrect, where should I put the $x$ that in my solution is not present?

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    $\begingroup$ bernoulli has no pdf, calculating max with derivatives is here not possible $\endgroup$ – user620174 Jan 25 '19 at 14:52
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    $\begingroup$ Your likelihood formula is wrong - your formula doesn't depend on the data that you see, only $n$. When you use the correct formula you can differentiate the log likelihood to get the MLE. $\endgroup$ – Alex Jan 25 '19 at 15:28
  • $\begingroup$ @Alex what do you mean? $\endgroup$ – Luke Marci Jan 25 '19 at 15:31
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    $\begingroup$ The likelihood should depend on the data you see. E.g. suppose you just had $n = 1$ and observed $X_1 = x_1$. If that was a success ($x_1 = 1$) then the likelihood would be $\theta$, and if it was a failure ($x_1 = 0$) then the likelihood would be $(1 - \theta)$. In the first case the MLE would be $\hat{\theta} = 1$ and in the second case the MLE would be $\hat{\theta} = 0$. $\endgroup$ – Alex Jan 25 '19 at 15:36

The likelihood function for Bernoulli can be written as follows: $$ L(\theta)=\begin{cases}\theta, & X_1=1\cr 1-\theta, & X_1=0\end{cases}\times \begin{cases}\theta, & X_2=1\cr 1-\theta, & X_2=0\end{cases}\times \cdots\times\begin{cases}\theta, & X_n=1\cr 1-\theta, & X_n=0\end{cases} $$ Note that $\sum_{i=1}^n X_i=n\overline X$ calculates the number of units in a sample, and $n-n\overline X$ calculates the number of zeros. In the above product, $\theta$ is multiplied as many times as there are units in the sample. And $1-\theta$ is multiplied as many times as there are zeros in the sample. So $$ L(\theta) = \theta^{n\overline X}\cdot (1-\theta)^{n-n\overline X}. $$ Starting from here you can calculate MLE with derivatives.


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