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Define:

$y= \theta + \varepsilon + a,$

where $a$ is a choice variable in a behavioral economic model, with equilibrium solution $a^e$, and $\theta$ and $\varepsilon$ are independently distributed random variables with distributions:

$\theta \sim \mathcal{N}(\bar{\theta},\sigma_{\theta}^2)$,

and

$\varepsilon \sim \mathcal{N}(0,\sigma_{\varepsilon}^2)$.

A paper I am reading, on page 173 at bottom states:

$E[E[\theta|y]]=\bar{\theta} + \phi E[\theta + \varepsilon + a - a^e - \bar{\theta}]$,

where

$\phi=\sigma_{\theta}^2/(\sigma_{\theta}^2+\sigma_{\varepsilon}^2)$.

It refers to this result as a "well known" signal extraction result.

Googling the latter I found that for $y= a + b$, with $a$ and $b$ i.i.d standard normal (i.e. mean zero) then:

$E[a|y] = \frac{\sigma_{a}^2}{\sigma_{y}^2}y$.

This differs from the case above in that the latter is not standard normal with mean zero and the definition of $y$ includes a constant.

Hence I am having difficulty deriving the result in the paper. Grateful if someone could explain the steps. Thanks!

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    The hour is late and I'll look at this tomorrow (by which time 20 answers may have appeared already.....) but may I comment that it seems a bit odd to use the same letter, $\theta$, to refer both to a random variable and to its expected value.2011-08-20
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    @Michael Hardy All expected values have a bar, as in $\bar{\theta}$, except for the definition of the mean in $\mathcal{N}(\theta,\sigma_{\theta}^2)$ where I corrected typo.2011-08-20

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