# Bayesian probability estimation

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...

• 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 $$\dfrac{\int\limits_0^1 \theta \cdot \theta^9(1-\theta)^3\, d \theta}{\int\limits_0^1 \,\,\,\,\,\,\, \theta^9(1-\theta)^3\, d \theta}$$
• No - the probability that the next value will be $0$ is equal to the expected value of $1-\theta$ Oct 21, 2018 at 0:08