I just got an unexpected result, and wanted to check if I made any mistakes in my calculations, or if you can point me to posts/literature on the subject I describe below.

Suppose you sample $n$ random individuals from a large population (so you can assume replacement although you don't actually replace), and test them for a given feature that can be present or absent.
You find that the feature is present in $s$ individuals out of the $n$ tested.

Normally I would conclude that the expected fraction ($p_0$) of individuals with the feature in the overall population is $\frac s n$.

Concerning the distribution of $p_0$, I used the binomial distribution for the probability density:

$P(s \ out \ of \ n | p_0 = p) = \binom n s p^s (1-p)^{n-s}$

and Bayes' theorem with a uniform prior:

$P(p_0 = p | s \ out \ of \ n) = \frac {P(s \ out \ of \ n | p_0 = p)}{P(s \ out \ of \ n)} = \frac {P(s \ out \ of \ n | p_0 = p)}{\int_0^1 P(s \ out \ of \ n | p_0 = p) \cdot dp} $

Given that:

$\int_0^1 P(s \ out \ of \ n | p_0 = p) dp = \int_0^1 \binom n s p^s (1-p)^{n-s} \cdot dp = \frac 1 {n+1}$

it follows that:

$P(p_0 = p | s \ out \ of \ n) = (n+1) \binom n s p^s (1-p)^{n-s} $

which is the distribution I was looking for.

I tried plotting it for $p \in [0,1]$ with some numerical values of $n, s$, and the curve seemed OK. It was narrower when $n$ was larger, and it looked more symmetrical when $\frac s n$ was far from the extremes $0,1$.
Differentiating it w.r.t. $p$ and solving for $p$ gave $p = \frac s n$, which I took as an indication that the 'most frequent' value of $p_0$ was indeed the assumed one.

However, when I applied the 'expected value' method:

$E[p_0] = \int_0^1 P(p_0 = p | s \ out \ of \ n) \cdot p \cdot dp = \int_0^1 (n+1) \binom n s p^s (1-p)^{n-s} \cdot p \cdot dp = \frac {s+1}{n+2}$

This is the same as $\frac s n$ when $s = \frac n 2$, whereas in all other cases it gives a 'skewed' result.
E.g. for $s=2, n=6$, it gives $\frac 3 8$ instead of the 'assumed' $\frac 2 6 = \frac 1 3$.

Do you think this makes sense, or is it instead a consequence of my (wrong?) calculations or Bayesian prior?

EDIT: prior resulting in an expected value equal to the observed % ($\frac s n$)

Starting from the concept that a uniform Bayesian prior might not be appropriate, as suggested by user Jarle Tufto in the comments, I examined a different prior.

In particular, suppose that percentages (here the fraction $p_0$) are not uniformly distributed, whereas their logit ($u$) is:

$u=ln(\frac p {1-p})$

This implies that:

$P(p_0=p) = \frac 1 {p (1-p)}$

(the way I obtained this is quite nebulous even to me, and I will need to post a new question for that).

Reapplying Bayes' theorem with this new prior gives:

$P(p_0 = p | s \ out \ of \ n) = \frac {P(s \ out \ of \ n | p_0 = p) P(p_0=p)}{\int_0^1 P(s \ out \ of \ n | p_0 = p) P(p_0=p) \cdot dp} $

The denominator integrates to:

$\int_0^1 P(s \ out \ of \ n | p_0 = p) P(p_0=p) \cdot dp = \int_0^1 \binom n s p^s (1-p)^{n-s} \frac 1 {p (1-p)} \cdot dp = \frac n {s (n-s)} $


$P(p_0 = p | s \ out \ of \ n) = \frac {\binom n s p^s (1-p)^{n-s} \frac 1 {p (1-p)}}{\frac n {s (n-s)}} = \binom {n-1} {s-1} (n-s) p^{s-1} (1-p)^{n-s-1}$

And the expected value is:

$E[p_0] = \int_0^1 P(p_0 = p | s \ out \ of \ n) \cdot p \cdot dp = \int_0^1 \binom {n-1} {s-1} (n-s) p^{s-1} (1-p)^{n-s-1} \cdot p \cdot dp = \frac s n$

However now it's the maximum value that does not seem to make much sense: by differentiating and solving for $p$, I get:

$p = \frac {s-1}{n-2}$


  • 1
    $\begingroup$ Just to mention that your calculations are right. The posterior that you have obtained is beta distribution with $\alpha = s+1, \beta=n-s+1$. $\endgroup$
    – Dayne
    Nov 23, 2020 at 2:42
  • $\begingroup$ @Dayne: do you want to post your comment(s) as an answer? Better to have a short answer than no answer at all. Anyone who has a better answer can post it. $\endgroup$ Nov 23, 2020 at 16:22
  • $\begingroup$ @kjetilbhalvorsen: actually I don't know the answer this question. In fact, I found this question quite interesting so following this. As to why, MLE and expected value of the posterior is different (though there is no reason for them to be same either but still) $\endgroup$
    – Dayne
    Nov 23, 2020 at 16:52
  • $\begingroup$ @Dayne : indeed, I don't get why the expected value is different from the most frequent value, not mathematically but conceptually. Until I did this calculation, I had no doubt that the best estimate of the % of successes in the population was the % of successes in the sample. Instead, this result seems to say that if I find 1 success in a sample of size 1000, the expected value of the population % is closer to 0.2% than to 0.1%, which I would say is not insignificant. I thought that maybe integrating rather than summing was wrong, but in a large population p is essentially continuous, so... $\endgroup$ Nov 23, 2020 at 19:10
  • 1
    $\begingroup$ The posterior mean is not only influenced by the observed data but also by the prior pulling the posterior mean towards the prior mean of 1/2. $\endgroup$ Nov 23, 2020 at 19:21

1 Answer 1


You start using an Uniform prior, equivalent to a $Be(1,1)$. This two 1's can be understood as pseudo observations of "previous" experiments, let's say, having obtained 1 Yes, 1 No previously.

Now let's make an experiment where we obtain 2 Yes and 4 No, as in your example. The posterior is now a $Be(1+2,1+4)=Be(3,5)$.

The mode of the posterior is equivalent to the MLE in the frequentist method, this is, 2/6. Because you used an Uniform prior.

E.g. for $s=2, n=6$, it gives $\frac 3 8$ instead of the 'assumed' $\frac 2 6 = \frac 1 3$.

The expected value of the posterior is the MLE if we are taking into account the pseudo observations, this is, (2+1)/(6+1+1)=2/8.

In the second case I would say you are using the Haldane prior, $Be(0,0)$, where there are no "previous" observations. So, the expected value of the posterior is now the frequentist MLE.

  • $\begingroup$ Thanks! I will need to study this 'Be' function, it seems to simplify everything compared to all the integrals here. The part I don't get is how one automatically concludes 'The mode of the posterior is equivalent to the MLE in the frequentist method, this is, 2/6. Because you used an Uniform prior.'. How come the uniform prior has such a profound effect on the expected value but not on the mode? Unless that is another property of the 'Be' function. $\endgroup$ Nov 25, 2020 at 7:52
  • $\begingroup$ @user6376297: mode is not necessarily equal to expected value. That happens for symmetric distributions. The posterior happens to be non-symmetric, is all. The interesting part was why (conceptually) the MLE from frequentist approach was not equal to the expected value of posterior despite a flat prior. This answer seems to give quite a good reasoning for this. $\endgroup$
    – Dayne
    Nov 25, 2020 at 10:32
  • $\begingroup$ @Dayne : sure, the reasoning is good, but not obvious at all, at least for me. I wonder if people usually think of a uniform prior as influencing results this way. And I still do not get why it should be so obvious that the mode instead is not influenced by the uniform prior. $\endgroup$ Nov 25, 2020 at 10:44

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