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Let's say I have a model that's like,

$$ Y \;|\; \theta_1 \sim P(Y \;|\; \theta_1) $$ $$\theta_1 \;|\; \theta_2 \sim P(\theta_1 \;|\; \theta_2) $$ $$ \theta_2 \;|\; \theta_3 \sim P(\theta_2 \;|\; \theta_3) $$

where $Y$ is data and $\theta_2$ is the mean of $\theta_1 \;|\; \theta_2$.

Is it necessarily true that $\theta_2 \;|\; Y$ is the mean of $\theta_1 \;|\; Y$?

If it's not, how do I interpret $\theta_2 \;|\; Y$?

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2 Answers 2

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No. $\theta_2 \mid Y$ is a random variable, while the mean of $\theta_1 \mid Y$ is simply a number.

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    $\begingroup$ In Bayesian analysis isn't the mean a random variable? I'm going to edit my answer to put a prior on $\theta_2$ and make this clear. $\endgroup$
    – TrynnaDoStat
    Commented Jul 23, 2018 at 15:40
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    $\begingroup$ It might help to write an equation with exactly what you mean by "mean." $\endgroup$
    – Danica
    Commented Jul 23, 2018 at 15:40
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Is it necessarily true that $\theta_2 | Y$ is the mean of $\theta_1 |Y$?

No, they just have the same "centers." The mean of $\theta_1 | Y$ is $$ E[\theta_1 \mid Y] = E[ E(\theta_1 \mid \theta_2) \mid Y] = E[ \theta_2 \mid Y]. $$

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  • $\begingroup$ This is very good! How would I then interpret a wide posterior for $\theta_2 \mid Y$? Surely this tells me something about my uncertainty in $E[\theta_1 \mid Y]$? $\endgroup$
    – TrynnaDoStat
    Commented Jul 23, 2018 at 17:56
  • $\begingroup$ What do you mean by "uncertainty?" $\endgroup$
    – Taylor
    Commented Jul 23, 2018 at 18:15
  • $\begingroup$ What I mean is that the posterior for $\theta_1$ can be quite wide but the posterior for it's mean, $\theta_2$ can be quite narrow. In this case our posterior belief in the value of the mean of $\theta_1$ is quite "certain" but we're not "certain" what a particular draw of $\theta_1$ will look like. How do I quantify my certainty in the posterior mean of $\theta_1$? If $E[\theta_1 | Y]$ was a distribution the uncertainty is quantified in the width of the posterior (as pointed out - it's clearly not a distribution). $\endgroup$
    – TrynnaDoStat
    Commented Jul 23, 2018 at 19:03
  • $\begingroup$ @TrynnaDoStat " but the posterior for it's mean, $\theta_1$ can be quite narrow" <- $E[\theta_1 |Y]$ is a point and doesn't have a posterior. The (joint) posterior $p(\theta_1, \theta_2 | y)$ is a surface over a subset of $\mathbb{R}^2$. You can take the variance of either component, or you can take the conditional variance of either component, conditioning on specific values of the other component. You also might be interested in the law of total variance, which, in this case, ties together the marginal variance of one component with the marginal variance (among other things) of the other. $\endgroup$
    – Taylor
    Commented Jul 23, 2018 at 19:09
  • $\begingroup$ What does a wide posterior of $\theta_2$ tell me then? Given that in the model spec it's the prior mean of $\theta_1$? $\endgroup$
    – TrynnaDoStat
    Commented Jul 23, 2018 at 19:19

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