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I'm confusing myself with this question.

Suppose there is a Normal r.v $X \sim \mathcal{N}(\mu, \sigma^2)$. We known the variance $\sigma^2$ however don't know the mean $\mu$, and choose to use Bayesian inference to update our belief.

Suppose further we have a prior distribution of $\mu$, also Normal, i.e $\mu \sim \mathcal{N}(\mu_0, \sigma_0^2)$, the observation is $x_0$.

Then what is the conditional expectation $\mathbb{E}(X|x_0)$ ? Would you please show me the derivation steps?

I know how to get the posterior distribution on $\mu$ when assuming $\sigma_0^2$ is fixed, i.e

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

Many thanks!

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  • $\begingroup$ I'm not entirely sure what $x_0$ is supposed to be. Is it a realization of some $X' \sim \mathcal N(\mu, \sigma^2)$ and $X$ is some unobserved random variable? If that's the case, just take $E(X | x_0) = E(E(X_0 | x_0, \mu)|x_0) = E(\mu | x_0) = \mu'_0$. $\endgroup$ – guy Mar 2 '12 at 23:47
  • $\begingroup$ I don't understand your last sentence. The posterior distribution of $\mu$ given $X$ would generally have a smaller variance than the marginal distribution of $\mu$. Is your phrase "when $\sigma_0$ is fixed" supposed to imply something to the contrary? $\endgroup$ – Michael Hardy Mar 3 '12 at 0:05
  • $\begingroup$ If, by $E(X\mid x_0)$ you mean $E(X \mid X=x_0)$, then that expectation is $x_0$. But to speak of things like $E(X \mid X=6)$ is to speak of trivialities. I can't help suspect that you meant something else. $\endgroup$ – Michael Hardy Mar 3 '12 at 3:27
  • $\begingroup$ @guy I believe that's what I was looking for. Thanks. $\endgroup$ – ShuaiYuan Mar 3 '12 at 9:19

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