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 $\theta^{\sum_{i=1}^n x_i} (1-\theta)^{n-\sum_{i=1}^n x_i}$,

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


1 Answer 1


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?

  • $\begingroup$ It is trivial for $T=n$, which is 1, but still not clear the case of otherwise. $\endgroup$ Feb 2, 2023 at 9:28
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.