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$ \widehat p = \overline x$ is an unbiased estimator? I think the most likelihood estimator to p is $ \widehat p=\frac{\overline x}{n}$, but what shall I use? A unbiased or a Most Likelihood estimator?

My teacher said to always use unbiased estimators, but she also said that the estimator of p was $ \widehat p=\frac{\overline x}{n}$ , however, I saw on internet that proportion is $ \widehat p=\overline x$ .. I'm confused.

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  • $\begingroup$ It depends of what $\bar{x}$ means. Is it the mean of a number of samples from a binomial distribution? Or is it based on a single observation from the binomial distribution, giving the proportion of successes? $\endgroup$ – Henry May 24 '17 at 22:16
  • $\begingroup$ @Henry thanks , you're right $\endgroup$ – Vitor Aguiar May 24 '17 at 22:18
  • $\begingroup$ If $X$ is the number of Successes in $n$ trials, then $\hat p = X/n,$ the proportion of Successes, is the usual estimator. It is the Maximum Likelihood Estimator and it is unbiased. Specifically, if I get $X = 53$ Heads in $n = 100$ tosses of a coin then $\hat p = 53/100 = 0.53$ is the estimated probability of Heads. $\endgroup$ – BruceET May 24 '17 at 23:38
  • $\begingroup$ @BruceET but, isn't this estimator to a bernoulli distribution? $\endgroup$ – Vitor Aguiar May 24 '17 at 23:52
  • $\begingroup$ If $X = V_1 + V_2 + \cdots + V_n,$ where $V_i$ are $n$ independent Bernoulli's, each with Success probability $p,$ then $X \sim \mathsf{Binom}(n, p).$ So you can speak of estimating Bernoulli $p$ or Binomial $p,$ which is the same. $\endgroup$ – BruceET May 25 '17 at 0:53

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