Hello, fellow Wilmotters. I was wondering if any of you guys are familiar with Bayesian Decision Theory. The book we are using in class is Berger's Statistical Decision Theory and Bayesian Analysis, which in some examples does not describe all steps. In page 12 (example 4), it says the following:Assuming we desire to estimate Theta under squared error loss (theta - a)^2, suppose that Then, for the decision rule ,TextMy question is: If pi(theta) is normal with mean 0, why is the expected value of Theta^2 = Tau^2?Shouldn't it be zero? Thanks a lot for any help you can give me on this.

I'm assuming here that pi(theta) is understood to mean the distribution of the random variable theta. If that is so, then your question is not about decision theory, but just basic probability theory: if theta is a normal random variable with mean 0 and variance tau^2, then the expected value of theta^2 is tau^2.At the very least, it should be obvious why the expected value of the square of a normal random variable can't be 0: For any X, if E(X^2) = 0, then with probability 1, X = 0! The proof that E(theta^2) = tau^2 is an elementary integration exercise, which you should be able to find in any introductory probability theory textbook.

It was pure and basic stats, as you said. Var X: Ex^2 - (Ex)^2Ex^2 cannot be determined directly. Thanks for your input.