For a Bernoulli population, show that the maximum likelihood estimate of $\theta =p^2$ is $\bar{x}^2$.

I'm just looking for a hint to get started.

Obviously, I can find the MLE of a Bernoulli by maximizing the likelihood function. How can I maximize $\theta = p^2 $ if $p^2$ is not in the original distribution

  • $\begingroup$ Simply replace $p$ with $\sqrt{\theta}$ in the likelihood function and then maximize with respect to $\theta$. $\endgroup$ Aug 2, 2020 at 17:18
  • $\begingroup$ @LeanderTilstedKristensen I facepalmed. Thank you! $\endgroup$
    – M1996rg
    Aug 2, 2020 at 17:27
  • $\begingroup$ Consider the invariance property of MLEs $\endgroup$
    – pedernv
    Aug 2, 2020 at 18:06

1 Answer 1


Note that if $\alpha$ is the MLE of $\beta$ , for any function $f(\alpha)$ is MLE of $f(\beta)$. Just consider $f(x)=x^2$.


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