Signal extraction from multivariate normal

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. – Michael Hardy Aug 20 '11 at 5:20
@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. – Fred Aug 20 '11 at 5:32

I'm not surprised you're having difficulty deriving the result; I think it's rather badly explained in the paper. Here's how I would describe the derivation:

In the paper, $a^e$ is introduced not as an equilibrium solution, but as "the public's perception of $a$" (p. 172, above (4)). In (3) and (10), the expectation values $E(\theta|x)$ and $E(\theta|y)$, respectively, refer to calculations performed by the public, whereas the expectation operator $\mathsf E$ (you didn't reproduce the typographical distinction made in the paper) refers to calculations performed by the bureaucrat (who can use the correct value $a$). The public decides about the career prospects of the bureaucrat in accordance with its estimation of the bureaucrat's talents, and it can only do so based on its perception $a^e$ of the bureaucrat's effort $a$ and on the policy outcome $y$. The paper does not specify how the public models the difference between its perception $a^e$ and the actual value $a$, but it seems from (4) that the public simply calculates as if the actual effort where known to be $a^e$. If we transfer that to the present setting including the unobservable noise $\varepsilon$, the public will use the model

$$y=\theta^e+\varepsilon+a^e\;,$$

where $\theta^e$ is the public's idea of the talent $\theta$, which it models according to the given normal distribution. Since $y=\theta+\varepsilon+a$, we have $\theta^e=\theta+a-a^e$. Now the public will apply your result, which I'll rewrite for $c$, $d$ and $x$ since the version with $a$, $b$ and $y$ is rather confusing (with $a$ referring to $\theta$ rather than $a$ in the present case): For $x=c+d$, with $c$ and $d$ i.i.d. normally distributed random variables with zero mean, $E[c|x]=(\sigma_c^2/\sigma_x^2)x$. Rewritten in terms of zero-mean variables, the public's model is

$$y-\bar\theta-a^e=(\theta^e-\bar\theta)+\varepsilon\;,$$

so we can apply the general result with $x=y-\bar\theta-a^e$, $c=\theta^e-\bar\theta$ and $d=\varepsilon$. Thus, the public's calculation of the expected talent of the bureaucrat given its perception $a^e$ of the effort and the observed policy outcome $y$ is

$$\begin{eqnarray} E[\theta^e|y] &=&\bar\theta+E[c|x] \\ &=& \bar\theta+(\sigma_c^2/\sigma_x^2)x \\ &=& \bar\theta+(\sigma_\theta^2/\sigma_y^2)(y-\bar\theta-a^e) \\ &=& \bar\theta+(\sigma_\theta^2/(\sigma_\theta^2+\sigma_\varepsilon^2))(\theta^e-\bar\theta+\varepsilon) \\ &=& \bar\theta+(\sigma_\theta^2/(\sigma_\theta^2+\sigma_\varepsilon^2))(\theta+a-a^e-\bar\theta+\varepsilon)\;, \end{eqnarray}$$

in agreement with the result in the paper.

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Many thanks, this is very useful. I was thinking precisely along those line i.e. centering the variables around zero. What I am not so clear is on the jump from $E[\theta^e|y]$ to $E[\theta^e|y]=\bar{\theta} + E[c|x]$. What is the general rule that allows you to do this? How was the $\bar{\theta}$ factored out? I understand the intuition, but not the formal derivation. BTW after the third equal sign in the last derivation just substitute $y=\theta^e+\varepsilon+a^e$ for $y$ and you are done. – Fred Aug 20 '11 at 14:43
I meant substitute $y=\theta+\varepsilon+a$ for $y$ and you are done. – Fred Aug 20 '11 at 14:49
to be clear, you have implicitly done the following: $E[\theta|y]=E[\theta] + E[\theta-\bar{\theta}|y-\bar{y}] = E[\theta] + E[\theta-\bar{\theta}|y-\bar{\theta}-a]$ and I am unclear what axioms of expected probability where involved in the process. Then, in eq. $a=a^e$ – Fred Aug 20 '11 at 15:08