Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

I want to ask,

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.

Thanks for your help!

share|cite|improve this question
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
up vote 0 down vote accepted

Starting from your last equation for $U$, one gets $$ U = \frac{\sigma_0^2 \mu + \sigma^2 \mu_0}{\sigma_0^2 + \sigma^2}. $$

share|cite|improve this answer
Do you mind giving more details on this? – shuaiyuancn Mar 3 '12 at 11:10
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
Trying to put it in another way. Could you please give some hints on the derivation of your r.h.s? – shuaiyuancn Mar 3 '12 at 11:48
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
I get it. Thanks for the help! – shuaiyuancn Mar 3 '12 at 16:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.