# Quick Bayesian Probability Question

• September 15th 2010, 07:22 AM
Guy
Quick Bayesian Probability Question
Hey all,

I need to check my understanding of this. Let $\theta$ have some prior on it and observe $X_1, ..., X_n, X_{n + 1}$, which are independent conditional on $\theta$, such that $EX_i|\theta = \theta$. I'm asked to find the posterior mean of $X_{n + 1}$ given $X_1, ..., X_n$. I'll spare the extra details.

My reasoning is

$EX_{n + 1} | X_1, ..., X_n = E\left(EX_{n + 1} | \theta, X_1, ..., X_n \right) | X_1, ..., X_n$
$= E\left(EX_{n + 1} | \theta \right) | X_1, ..., X_n$ (from conditional independnce)
$= E\theta | X_1, ..., X_n$ (since EX|theta = theta)

So the posterior mean of $X_{n + 1}$ is the posterior mean of $\theta$. This seems a little off to me for some reason I can't explain. Does this look okay? It seems like this is working out to an expectation over the joint distribution of $\theta|X_1, ..., X_n$ and $X_{n + 1} | \theta$ but I guess I'm not sure if that's what is meant by asking for the posterior mean.

More broadly, if I'm asked for anything posterior, should I get something that is free of theta?
• September 15th 2010, 04:37 PM
Guy
Nothing? After thinking about it I think this is fine, but I'd like to get some confirmation that I'm not making some fundamental mistake.