I have a (biased) coin that has an unknown Head probability $p\in(0,1)$. To point estimate $p$, say that I'm going to use two approaches.

Approach 1. I can use the Bayesian inference technique. Starting from a Beta prior $p^0\sim Beta(a^0_H,a^0_T)$, I calculate the posterior from the observations. When I observed $n_H$ Heads and $n_T$ Tails in round $n_H+n_T$, The posterior will follow $Beta(a^0_H+n_H,a^0_T+n_T)$. As a point estimate for $p$, I can take the posterior mean, which is $\hat p_B=\frac{a^0_H+n_H}{a^0_H+n_H+a^0_T+n_T}$

Approach 2. I can use MLE. The MLE in this case is given by $\hat p_F=\frac{n_H}{n_H+n_T}$.

My question is what is the relationship between $\hat p_B$ and $\hat p_F$? Especially,

  1. It looks like when I have enough observations, the two estimates coincide. In this case, can I say $\hat p_B$ and $\hat p_T$ are asymptotically equal? or is there any other terminology that I can formally describe the asymptotic relationship between the two?
  2. When I only have a handful of observations, what can I say about the relationship? Can I say the two differ only by constant terms? or, again, is there any other formal description of the two under the small sample situation?
  • $\begingroup$ I guess that you mean $p^0\sim Beta(a_H^0,a_T^0)$. It looks like you are choosing the Beta parameters so that they sum up to 1. It would be a very strange and strong prior... $\endgroup$
    – Sergio
    Commented Sep 8, 2020 at 14:39
  • $\begingroup$ @Sergio a Beta prior whose coefficients sum to unity would be a very weak prior. However, where do you get the impression the OP is assuming any such constraint? I haven't been able to find such an implication. $\endgroup$
    – whuber
    Commented Sep 8, 2020 at 14:47
  • $\begingroup$ @whuber (a) if $p\sim Beta(0.5,0.5)$, then $p=0$ and $p=1$ have the maximum density. What am I missing? (b) Why $(a_H,a_T)$ instead of $(\alpha,\beta)$? $\endgroup$
    – Sergio
    Commented Sep 8, 2020 at 16:11
  • 1
    $\begingroup$ @Sergio (b) is a matter of taste. (a) To see why Beta$(1/2,1/2)$ is a weak prior compared to a Beta distribution with larger parameters $\alpha$ and $\beta,$ compare the two posterior distributions obtained upon observing $H$ heads and $T$ tails. One of them is Beta$(H+1/2,T+1/2)$ and the other is Beta$(H+\alpha,T+\beta).$ As $\alpha$ and $\beta$ increase, the latter becomes more and more concentrated around the prior: that's what a strong prior does. $\endgroup$
    – whuber
    Commented Sep 8, 2020 at 16:59
  • $\begingroup$ @whuber Thanks for your reply. $\endgroup$
    – Sergio
    Commented Sep 8, 2020 at 17:57

1 Answer 1


The posterior point estimate is a weighted combination of the prior point estimate, $\frac{a^0_H}{a^0_H+a^0_T}$, and the maximum likelihood estimate $\frac{n_H}{n_H+n_T}$.

The weights are simply $\omega_{\text{Prior}} = \frac{n_{\text{Prior}}}{n_{\text{Prior}} + n_{\text{Lik}}}$ and $\omega_{\text{Lik}} = \frac{n_{\text{Lik}}}{n_{\text{Prior}} + n_{\text{Lik}}}$, where $n_{\text{Prior}} = a^0_H+a^0_T$ and $n_{\text{Lik}} = n_H+n_T$.

You can confirm this by substitution:

$$ \begin{align} \text{Posterior Mean} &= \frac{a^0_H}{a^0_H+a^0_T} \omega_{\text{Prior}} + \frac{n_H}{n_H+n_T} \omega_{\text{Lik}}\\ &= \frac{a^0_H}{a^0_H+a^0_T} \times \frac{a^0_H + a^0_T}{a^0_H+a^0_T+n_H+n_T} + \frac{n_H}{n_H+n_T} \times \frac{n_H+n_T}{a^0_H+a^0_T+n_H+n_T}\\ &= \frac{a^0_H}{a^0_H+a^0_T+n_H+n_T} + \frac{n_H}{a^0_H+a^0_T+n_H+n_T}\\ &= \frac{a^0_H+n_H}{a^0_H+n_H+a^0_T+n_T} \end{align} $$

  1. With lots of observations, $\omega_{\text{Lik}} \gg \omega_{\text{Prior}}$, and so the posterior mean is almost the same as the maximum likelihood estimate.
  2. With few observations, $\omega_{\text{Lik}} \ll \omega_{\text{Prior}}$, and so the posterior mean is almost the same as the prior mean.

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