# Expectation of normal random variable conditionally on one observation

I got stuck with this problem,

Suppose there is a Normal r.v $X \sim \mathcal{N}(\mu, \sigma^2)$, where $\sigma^2$ is known and $\mu$ is unknown and will be updated using Bayesian inference.

We give a prior distribution of $\mu \sim \mathcal{N}(\mu_0, \sigma_0^2)$ and update it after observation $x_0$ to

$$\mu \sim \mathcal{N}(\mu'_0, \sigma_0^2), \;\;\; \mu'_0 = \frac{\sigma_0^2 x_0 + \sigma^2 \mu_0}{\sigma_0^2 + \sigma^2}$$

The problem is, actually we don't have any observations but want to see the result over all possible situations,

$$U = \int p(x_0) \mathbb{E}(X|x_0) dx_0$$

which is used in the Dynamic programming formulation for the forward planning.

1- does this $U$ equation even make sense?

2- what's the final analytical form of $U$, if it makes sense? I can expand it as,

$$U = \int \frac{1}{\sqrt{2 \pi \sigma^2}} \exp \{ - \frac{(x_0 - \mu)^2}{2 \sigma^2} \} \frac{\sigma_0^2 x_0 + \sigma^2 \mu_0}{\sigma_0^2 + \sigma^2} dx_0$$

but don't have a clue to how to proceed.

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I don't know the answer to the question, so please forgive me for commenting here, as I am just too--- happy to see you here. SHIXIONG, WO YE SHI BEIHANG DE, WO GEN HEYING YIGE DAOSHI. – ShinyaSakai Mar 3 '12 at 10:42
@ShinyaSakai: Why not use OP's email, given on the webpage the OP's profile links to? – Did Mar 3 '12 at 10:58

Starting from your last equation for $U$, one gets $$U = \frac{\sigma_0^2 \mu + \sigma^2 \mu_0}{\sigma_0^2 + \sigma^2}.$$
The integral in the RHS of the last equation in your post depends on $\mu$, $\mu_0$, $\sigma^2$ and $\sigma_0^2$ and this integral equals the RHS of my answer, whether this suits you or not. If you have a different question in mind, please post it. – Did Mar 3 '12 at 11:17
Yes: the RHS is by definition the expectation of $(\sigma^2_0\xi+\sigma^2\mu_0)/(\sigma^2_0+\sigma^2)$ for any $\xi$ of distribution $\mathcal N(\mu,\sigma^2)$, and the expectation of any such $\xi$ is $\mu$. – Did Mar 3 '12 at 13:42