An urn contain 5 balls, $ \theta $ white and $ 5 - \theta $ green. The experiment consists in grab 2 balls from the urn and register the pair $(x_1, x_2)$, where $x_i = 1$ if we observe a white ball and $x_i = 0$ otherwise. What is the bayes estimator $ \theta^* $ for $ \theta$ considering the squared loss function? (i.e $l(\theta,\theta^*) = (\theta - \theta^*)^2 $)
I can't figure out which posterior distribution I should use or even if I need to use one. I calculated my loss function considering all the possible values for $ \theta $ and $ \theta^* $ but I can't calculate the risk function without the posterior function. Can someone help me with it??
I can't find out what to do with the ordered pair, I just calculated the probability of each pair
Start with your prior. Since you have no information about the value of $\theta$ except that it's an integer between $0$ and $5$ a uniform prior is appropriate $P(\theta=0)=\dots=P(\theta=5)=1/6$
Then calculate your likelihoods $P(X_1=x_1,X_2=x_2|\theta=n)$ (which I'll abbreviate to $P(x_1,x_2|n)$).
Finally to use Bayes' rule you need $P(x_1,x_2)$ to go on the denominator. You can calculate this from the things you already have by using: $$P(x_1,x_2)=\sum_{n=0}^5 P(x_1,x_2|n).P(\theta=n)$$ (This is called the Law of Total Probability)
Then you can calculate your posterior with Bayes: $$P(\theta=n|x_1,x_2)=\frac{P(x_1,x_2|n).P(\theta=n)}{P(x_1,x_2)}$$ and from there you can minimise your expected loss (it's a property of the squared loss function that it's minimised by the mean of the posterior).