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Consider the following setup with known standard deviation $\sigma$: \begin{align} X|\theta &\sim N(\theta,\sigma^2)\\ \theta &\sim N(\mu_0,\sigma_0^2) \end{align}

and $(\mu_n,\sigma_n^2)$ is the belief after $n$ samples.

Can we say that the posterior mean $\mu_n$ is always finite?

The updating equation seems to suggest that it can be infinite only if the average of samples taken is $\infty$, which has the probability of zero under normal distribution. So, it should be finite. Does it sound right?

Or should I say $\mu_n$ is finite almost surely (since as $x$ goes to infinity, the probability goes to zero, but we can not say the probability of infinity is zero)?

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    $\begingroup$ For any finite $n$, the posterior mean is finite with probability 1. $\endgroup$ – Alex R. May 19 '17 at 21:39
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Since $$\mathbb{E}[\theta]=\mu=\mathbb{E}^{X_{1:N}}\left\{\mathbb{E}[\theta|X_{1:N}]\right\}$$ the quantity $$\mathbb{E}[\theta|X_{1:N}]$$ is almost surely finite. Which means that for almost any realisation of the random vector $X_{1:N}$ the conditional expectation is finite. Or, equivalently, that$$\mathbb{P}(|\mathbb{E}[\theta|X_{1:N}]|=+\infty)=0$$

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