2
$\begingroup$

I would like some help with the following statistical problem.

We have a coin with probability $\theta$ for heads, with prior for $\theta$ being a Beta(a,a) distribution (a is a known parameter). After tossing the coin n times, we get n-1 times heads.

Assuming that we toss the coin again, we would like to find the minimum n for which heads is at least two times more likely than tails.

I tried to address the problem in the following way:

We have that $\pi(\theta)=Beta(a,a)$ and $L(\theta;x)=f(x;\theta)=Bin(n,x)$
Therefore we get: $\pi(\theta|x) \propto Beta(a,a) \times L(\theta;x) \implies \pi(\theta|x) =Beta(a+x,a+n-x)$

Obviously, the posterior of $\theta$ is a function of n and x

But for x=n-1 we get $\pi_n(\theta|x) \propto Beta(a+n-1,a+1)$, that is the posterior of $\theta$ depends only on n.

The n+1 toss is a Bernoulli trial with parameter $\theta$ following the beta distribution described by $\pi_n(\theta|x)$, i.e. $Beta(a+n-1,a+1)$.

In order for heads to be at least two times more likely than tails, we must have $\theta \geq 2/3$.

I am not quite sure, how to continue with the problem after this point.

After skimming some Bayesian Statistics textbooks, I came to the conclusion that the standard Bayesian approach would be to take :$E[\pi_n(\theta)] \geq 2/3$ Still, I am not sure about this.

I have, also, thought of $\frac{\int_{2/3}^{1}\pi_n(\theta)}{\int_{0}^{2/3}\pi_n(\theta)} > 1$

So, summing up, I am not sure how to continue after having calculated the posterior distribution $\pi_n(\theta|x)$ of $\theta$.

Any help would be greatly appreciated.

$\endgroup$
3
  • 1
    $\begingroup$ The probability of getting heads given a Beta(a, b) distribution is $(a+1)/(a+b+2)$. Should be possible solve for this being at least 2/3. $\endgroup$
    – fblundun
    Commented Dec 29, 2020 at 16:40
  • $\begingroup$ I assume that you arrived at this by taking the mean value of Beta, in the manner of the Bayesian approach I described. Could you elaborate on why this is the correct approach? $\endgroup$
    – Filipg
    Commented Dec 29, 2020 at 16:48
  • $\begingroup$ Oops, I meant $a/(a + b)$. For example, if you start with a Beta(1, 1) prior and then see 3 heads and 1 tails, you should assign probability 4/6 to getting heads on the next flip. $\endgroup$
    – fblundun
    Commented Dec 29, 2020 at 17:19

1 Answer 1

0
$\begingroup$

After skimming some Bayesian Statistics textbooks, I came to the conclusion that the standard Bayesian approach would be to take :$$𝐸[𝜋_𝑛(𝜃)]≥2/3$$ Still, I am not sure about this.

Correct. We want the probability of the next toss being heads greater than or equal to $2/3$. This means $$P(H|\mathcal D)=\int P(H|\theta)\pi(\theta|\mathcal D)d\theta=\int \theta \pi(\theta|\mathcal D)d\theta = E[\theta|\mathcal D]$$ which is the posterior expectation. $\mathcal D$ represents the data.

$\endgroup$

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.