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There are $n$ bins and there is a guy who throws a ball to a randomly chosen bin. Initially we believe that each bin has equal probability to receive a ball, $p_i = \frac{1}{n}$.

My question is, how should we update our beliefs about $p_1,p_2, ... p_n$ if we observed $N$ independent throws and there $a_1$ balls in bin $1$, $a_2$ balls in bin $2$ and $a_n$ balls in bin $n$? $\sum_{i}a_i = N$

I dont think maximum likelihood estimation will work in this case. For example if $N = 1$, then $p_j = 1$, $p_k = 0$, for $k \neq j$. So I need to use Bayes to incorporate my prior belief about $p_i$.

But i am not sure how to apply Bayes here. I guess that each $p_i$ is $\mathbb{U}(0,1)$ but with constraint that $\sum_{i}p_i = 1$, and I don't know how to proceed from here

Any help is very much appreciated

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The standard approach is to use Dirichlet distribution as a prior distribution of initial probabilities $P\in H = \{p\in[0,1]^n: \sum p_j = 1\}$, which has density $$\pi(p_1, \dots, p_n; \alpha_1, \dots, \alpha_n) = \frac{\Gamma(\alpha_1+\dots+\alpha_n)}{\Gamma(\alpha_1)\dots\Gamma(\alpha_n)}p_1^{\alpha_1-1}\dots p_n^{\alpha_n-1} \sim Dir(\alpha_1, \dots, \alpha_n)$$ w.r.t. surface measure of $H$ with the parameters $\alpha_j > 0$. Let $M=(M_1, \dots, M_n)$ be numbers of balls in each bin after fixed amount of tosses. When you calculate the posterior conjugacy emerges, i.e.

$$f(p|m) \propto f(m|p)\pi(p) \propto p_1^{m_1}\dots p_n^{m_n} \ p_1^{\alpha_1-1}\dots p_n^{\alpha_n-1} \sim Dir(\alpha_1 + m_1, \dots, \alpha_n+m_n).$$

Setting all $\alpha_j=1$ gives $P\sim U(H)$. Note that you can't have $P_j \sim U(0,1)$, because after summing and taking expectation you would obtain $$\mathbb{E}\sum_{j=1}^n P_j = \frac n2.$$ More on Dirichlet distribution can be found here: https://en.wikipedia.org/wiki/Dirichlet_distribution

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  • $\begingroup$ But why choose the Dirichlet ? The problem states that the prior is (discrete) uniform on n. You need some further assumptions though. Joint behaviour of the p_i's. $\endgroup$
    – theoGR
    Sep 9, 2019 at 1:22
  • $\begingroup$ Prior distribution reflects your beliefs about possible values of $p_i$'s. In this approach we take it to be uniform over all possible values. Distribution of the first chosen bin is indeed discrete $U{1, \dots, n}$, which I believe you mistake as prior distribution. Why choose Dirichlet? Because of conjugacy, which is a standard notion in bayesian statistics. It does not only helps to avoid unsolvable integrals but is also somewhat natural. More on conjugacy can be found here: en.wikipedia.org/wiki/Conjugate_prior $\endgroup$
    – mbartczak
    Sep 10, 2019 at 11:02

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