# Bayesian inference using Dirichlet: muddled outcome case

In relation to my previous question (Bayesian inference for Beta distribution after an uncertain outcome),

Suppose that $$(x_1,x_2,x_3)\sim Dirichlet(a_1,a_2,a_3)$$

and an associated Mutinoulli trial has probability vector $$(x_1,x_2,x_3)$$. We want to (partially) identify these probabilities.

The thing I'm curious about is what happens when we observe only a binary outcome from the multinoulli trial: when $$(1,0,0)$$ or $$(0,1,0)$$, we observe ''success'' and when $$(0,0,1)$$, we observe ''failure''.

If this is the case, what would be the Bayesian inference after observing a ''success''?

My initial intuition was the posterior is $$(x_1,x_2,x_3)\sim Dirichlet(a_1+\frac{a_1}{a_1+a_2},a_2+\frac{a_2}{a_1+a_2},a_3)$$

as $$(1,0,0)$$ happens with probability $$x_1$$, which has an expected value of $$\frac{a_1}{a_1+a_2}$$ and $$(0,1,0)$$ happens with probability $$x_2$$, which has an expected value of $$\frac{a_2}{a_1+a_2}$$, but it doesn't seem to be right.

To calculate the posterior, \begin{align*} \pi(x_1,x_2,x_3|success)=&\gamma~\mathbb P[success|x_1,x_2,x_3]\pi(x_1,x_2,x_3)\\ =&\gamma~(x_1+x_2)\frac{\Gamma(a_1+a_2+a_3)}{\Gamma(a_1)\Gamma(a_2)\Gamma(a_3)}x_1^{a_1-1}x_2^{a_2-1}x_3^{a_3-1}\\ =&\tilde \gamma (x_1^{a_1}x_2^{a_2-1}x_3^{a_3-1}+x_1^{a_1-1}x_2^{a_2}x_3^{a_3-1}). \end{align*}

Here, the last term clearly looks different from the pdf of $$Dirichlet(a_1+\frac{a_1}{a_1+a_2},a_2+\frac{a_2}{a_1+a_2},a_3)$$..

Any idea or help?

If you observe $$y\sim\mathcal Bin(n,x_1+x_2)$$and if$$(x_1,x_2,x_3)\sim \mathcal Dir(a_1,a_2,a_3)$$then $$(x_1,x_2,x_3,y)\sim p(x_1,x_2,x_3,y)\propto (x_1+x_2)^y x_3^{n-y} x_1^{a_1}x_2^{a_2-1}x_3^{a_3-1}$$ that is, $$\pi(x_1,x_2,x_3\mid y)\propto \sum_{i=0}^{n-y} {n-y \choose i}x_1^{a_1+i}x_2^{a_2-1+n-y-i}x_3^{a_3-1+n-y}$$ The posterior is therefore a mixture of $$n-y+1$$ Dirichlet distributions

• Thank you very much for your comment. So basically, the posterior is a convex combination of Dir distributions using binomial weights, right? Just to make it clear for myself, if $n=1$ and one success, the posterior then is a Dirichlet distribution? (mixture of 1-1+1=1 Dir dist.) – Andeanlll Jul 26 '19 at 8:00
• I inverted the meaning for success and failure: it is when there is one failure out of one trial that the posterior remains a Dirichlet. – Xi'an Jul 26 '19 at 16:04
• Oh, I got it. Thanks! – Andeanlll Jul 27 '19 at 16:05