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We have $X \sim \text{bernoulli} \left(p \right), p \geq0.50$. Goal is to to estimate the MLE of $p$

With unconstraint case, I can calculate the MLE is the sample mean.

For constraint case, there is some discussion in Estimator when a coin is either fair or has two heads - however I cant fully understand this.

In particular, how can I derive this expression using the log-likelihood $\theta^{\sum_{i=1}^n x_i} (1-\theta)^{n-\sum_{i=1}^n x_i}$,

$\hat{\theta}= \begin{cases} \frac{1}{2} &\text{if} &\sum_{i=1}^n X_i\ne n; \\ 1 &\text{if} &\sum_{i=1}^n X_i=n. \end{cases}$

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Let $T=\sum_{1\leq i\leq n} X_i.$

Log likelihood is

$$\mathcal L= T\ln \theta +(n-T) \ln(1-\theta).\tag 1\label 1$$

Differentiate $\eqref 1$ w.r.t. $\theta$ to get

$$\frac{\mathrm d\mathcal L}{\mathrm d\theta} =\frac{T}{\theta} -\frac{n-T}{1-\theta};$$ therefore $$\frac{\mathrm d\mathcal L}{\mathrm d\theta} = 0\implies \hat{\theta}_{\textrm{MLE}}=\frac{T}{n}.\tag 2\label 2$$ What happens to $\hat{\theta}_{\textrm{MLE}}$ in $\eqref 2$ if $T=n?$ What happens if it isn't? Given is $\theta\in\{1/2, 1\}.$

If $\theta=1, $ then $T=n$ for all realizations would be success. However, the moment one observes $T\ne n, $ one can readily conclude $\theta \ne 1.$ So, if latter is the case and you only can have $\theta$ among $1/2$ and $1, $ what would $\hat{\theta}_{\textrm{MLE}}$ be?

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  • $\begingroup$ It is trivial for $T=n$, which is 1, but still not clear the case of otherwise. $\endgroup$ Feb 2, 2023 at 9:28
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    $\begingroup$ @BrianSmith try drawing the likelihood function and mark in the constraint. $\endgroup$
    – Glen_b
    Feb 2, 2023 at 9:30
  • $\begingroup$ @BrianSmith check the edited post. $\endgroup$ Feb 2, 2023 at 9:48

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