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?


1 Answer 1


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

  • $\begingroup$ 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.) $\endgroup$
    – Andeanlll
    Jul 26, 2019 at 8:00
  • $\begingroup$ 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. $\endgroup$
    – Xi'an
    Jul 26, 2019 at 16:04
  • $\begingroup$ Oh, I got it. Thanks! $\endgroup$
    – Andeanlll
    Jul 27, 2019 at 16:05

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.