I am reading Social Value of Public Information by Morris and Shin(2002) and I have a question about calculating the conditional expectation after observing both public and private signals.

In their model, the state $\theta$ is drawn from an improper uniform prior over the real line and the agent $i$ observes a public signal $$y=\theta+\eta$$ Where $\eta \sim N(0,\sigma_{\eta}^{2})$ and $\eta$ is independent of $\theta$. Additionally, the agent will also receive a private signal $$x_{i}=\theta+\epsilon_{i}$$ Where $\epsilon_{i}\sim N(0,\sigma_{\epsilon}^{2})$.

Denote the precision of the public signal as $\alpha$ and that of private signal as $\beta$. In particular, $\alpha=\frac{1}{\sigma_{\eta}^{2}}$ and $\beta=\frac{1}{\sigma_{\epsilon}^{2}}$.

They claim that conditional on observing $y$ and $x_{i}$, $$\mathbb E[\theta|y,x_{i}]=\frac{\alpha y+\beta x_{i}}{\alpha+\beta}$$.

My question is that, how do we derive the above expression for conditional expectation?

In particular, the main difficulty for me is to understand what an improper uniform prior means. For example, if $\theta$ is also drawn from a normal distribution, then we know that $\theta$, $x_{i}$ and $y$ are jointly normal, then we can use the formula for conditional distribution of multivariate normal distribution to derive the desired conditional expectation. However, since $\theta$ is improperly uniformly distributed, I don't know how to derive the formula above, and I would really appreciate it if someone can give me some help on this.

Thanks in advance!


1 Answer 1


An improper prior means that you are working with a prior "measure" rather than a prior "probability measure". For the (improper) uniform prior over the real line, you just take the density $f(\theta)$ that is constantly equal to $1$ over $\mathbb{R}$ (i.e., the density of Lebesgue measure). You can do this as long as all the posterior densities are well-defined probability densities (i.e., integrate to 1). See also Wikipedia.

If we let $\phi$ denote the density of a standard normal, we have for the signal pdfs $$\phi\left(\frac{y-\theta}{\sigma_\eta}\right) \quad \text{and} \quad \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right).$$ If we apply Bayes' rule, the posterior density $$ f(\theta|y,x_i)=\frac{1\cdot\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)}{\int_{-\infty}^{\infty}1\cdot\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)~d\theta} $$ is a well-defined probability density because the product of two normal pdfs is again a normal pdf (so the numerator integrates to 1). In particular, for the mean we have $$ \int_{-\infty}^{\infty} \theta\cdot\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)~d\theta=\frac{\alpha y+\beta x_i}{\alpha+\beta}. $$ See here for a reference, but I guess you could try to verify this for yourself. I suppose the point of using a uniform prior and normal signals is that one gets this very convenient formula.

  • $\begingroup$ For the last step, why the conditional mean is $\int_{-\infty}^{\infty} \theta\cdot\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)~d\theta$ instead of $\int_{-\infty}^{\infty}\theta f(\theta|y,x_{i})d\theta=\frac{\int_{-\infty}^{\infty}\theta\cdot\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)d\theta}{\int_{-\infty}^{\infty}1\cdot\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)~d\theta}$?. $\endgroup$
    – John
    Jan 31, 2022 at 2:18
  • $\begingroup$ Your formula for the conditional mean is correct. I just already used there that the denominator equals 1. $\endgroup$ Jan 31, 2022 at 10:32
  • $\begingroup$ Why the denominator equals $1$? I think the denominator serves as the constant which makes posterior density well-defined, but it needs not to be $1$. Additionally, I think the formula in the reference has taken this constant into account. $\endgroup$
    – John
    Feb 1, 2022 at 1:30
  • $\begingroup$ $\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)$ is again a density of a normal distribution with mean and variance as in the reference, and the density of some probability distribution integrates to 1. $\endgroup$ Feb 1, 2022 at 14:15

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