Consider a sequence of independent bernoulli random variables $X_1 X_2 ... X_n$ with parameter $\theta$ and $0<\theta<1$,where $P(X_i=1)=1-P(X_i=0)=\theta$. Assume the prior of $\theta$ follows Beta(3,2). Given that the first 9 observations are such that $\sum_{X_i = 1}^9 X_i = 7 $. what would be the (Bayesian) probability that $X_{10} =1$?

So I find out the posterior probability...

$P( \theta |X) \propto \theta^9(1-\theta)^3 $

And question ask me that $P(X_{10} = 1| H=7, T=2)$

But I don't have any idea for solving this question...


1 Answer 1



  • Your posterior distribution for $\theta$ with density proportional to $\theta^9(1-\theta)^3$ is another Beta distribution. You should try to find its parameters

  • The posterior probability that the next value will be $1$ is then equal to the posterior expected value of $\theta$

  • One approach (not the quickest if you are familiar with Beta distributions) might be to find $$\dfrac{\int\limits_0^1 \theta \cdot \theta^9(1-\theta)^3\, d \theta}{\int\limits_0^1 \,\,\,\,\,\,\, \theta^9(1-\theta)^3\, d \theta}$$

  • $\begingroup$ How about the next value will be 0? Is it still the expected value of theta? $\endgroup$ Oct 21, 2018 at 0:06
  • $\begingroup$ No - the probability that the next value will be $0$ is equal to the expected value of $1-\theta$ $\endgroup$
    – Henry
    Oct 21, 2018 at 0:08

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