<p>An improper prior means that you are working with a prior &quot;measure&quot; rather than a prior &quot;probability measure&quot;. For the (improper) uniform prior over the real line, you just take the density <span class="math-container">$f(\theta)$</span> that is constantly equal to <span class="math-container">$1$</span> over <span class="math-container">$\mathbb{R}$</span> (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 <a href="https://en.wikipedia.org/wiki/Prior_probability#Improper_priors" rel="noreferrer">Wikipedia</a>.</p>
<p>If we let <span class="math-container">$\phi$</span> denote the density of a standard normal, we have for the signal pdfs <span class="math-container">$$\phi\left(\frac{y-\theta}{\sigma_\eta}\right) \quad \text{and} \quad  \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right).$$</span>
If we apply Bayes' rule, the posterior density
<span class="math-container">$$
f(\theta|y,x_i)=\frac{1\cdot\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)}{\int_{-\infty}^{\infty}1\cdot\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)~d\theta}
$$</span>
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
<span class="math-container">$$
\int_{-\infty}^{\infty} \theta\cdot\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)~d\theta=\frac{\alpha y+\beta x_i}{\alpha+\beta}.
$$</span>
See <a href="https://ccrma.stanford.edu/%7Ejos/sasp/Product_Two_Gaussian_PDFs.html" rel="noreferrer">here</a> 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.</p>

CC BY-SA 4.0
2022-01-31T02:18:00.957Z
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 $\int_{-\infty}^{\infty}\theta f(\theta|y,x_{i})d\theta=\frac{\int_{-\infty}^{\infty}\theta\cdot\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)d\theta}{\int_{-\infty}^{\infty}1\cdot\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)~d\theta}$?.
John
CC BY-SA 4.0
2022-01-31T10:32:55.670Z
Your formula for the conditional mean is correct. I just already used there that the denominator equals 1.
axelniemeyer
CC BY-SA 4.0
2022-02-01T01:30:42.843Z
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.
John
CC BY-SA 4.0
2022-02-01T14:15:31.280Z
$\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)$ 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.
axelniemeyer
CC BY-SA 4.0
2022-01-31T00:32:37.920Z
2022-01-31T00:32:37.920Z
axelniemeyer
<p>I am reading <em>Social Value of Public Information by Morris and Shin(2002)</em> and I have a question about calculating the conditional expectation after observing both public and private signals.</p>
<blockquote>
<p>In their model, the state <span class="math-container">$\theta$</span> is drawn from an improper uniform prior over the real line and the agent <span class="math-container">$i$</span> observes a public signal <span class="math-container">$$y=\theta+\eta$$</span> Where <span class="math-container">$\eta \sim N(0,\sigma_{\eta}^{2})$</span> and <span class="math-container">$\eta$</span> is independent of <span class="math-container">$\theta$</span>. Additionally, the agent will also receive a private signal <span class="math-container">$$x_{i}=\theta+\epsilon_{i}$$</span> Where <span class="math-container">$\epsilon_{i}\sim N(0,\sigma_{\epsilon}^{2})$</span>.</p>
<p>Denote the precision of the public signal as <span class="math-container">$\alpha$</span> and that of private signal as <span class="math-container">$\beta$</span>. In particular, <span class="math-container">$\alpha=\frac{1}{\sigma_{\eta}^{2}}$</span> and <span class="math-container">$\beta=\frac{1}{\sigma_{\epsilon}^{2}}$</span>.</p>
</blockquote>
<p>They claim that conditional on observing <span class="math-container">$y$</span> and <span class="math-container">$x_{i}$</span>, <span class="math-container">$$\mathbb E[\theta|y,x_{i}]=\frac{\alpha y+\beta x_{i}}{\alpha+\beta}$$</span>.</p>
<p>My question is that, how do we derive the above expression for conditional expectation?</p>
<p>In particular, the main difficulty for me is to understand what an improper uniform prior means. For example, if <span class="math-container">$\theta$</span> is also drawn from a normal distribution, then we know that <span class="math-container">$\theta$</span>, <span class="math-container">$x_{i}$</span> and <span class="math-container">$y$</span> are jointly normal, then we can use the formula for conditional distribution of multivariate normal distribution to derive the desired conditional expectation. However, since <span class="math-container">$\theta$</span> 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.</p>
<p>Thanks in advance!</p>

<game-theory>
Conditional Expectation of state after observing public and private signals
John
# Conditional Expectation of state after observing public and private signals

(Asked by: username_0 on 2022-01-30)

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!

**Tags:** game-theory


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## 1 answers

### Answer by username_1 on 2022-01-31. Score: 5 | Accepted answer

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](https://en.wikipedia.org/wiki/Prior_probability#Improper_priors).

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](https://ccrma.stanford.edu/%7Ejos/sasp/Product_Two_Gaussian_PDFs.html) 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.

> **Comments:**
> **username_0**: 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 $\int_{-\infty}^{\infty}\theta f(\theta|y,x_{i})d\theta=\frac{\int_{-\infty}^{\infty}\theta\cdot\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)d\theta}{\int_{-\infty}^{\infty}1\cdot\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)~d\theta}$?.  
> **username_1**: Your formula for the conditional mean is correct. I just already used there that the denominator equals 1.  
> **username_0**: 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.  
> **username_1**: $\phi\left(\frac{y-\theta}{\sigma_\eta}\right)\cdot \phi\left(\frac{x_i-\theta}{\sigma_\epsilon}\right)$ 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.  

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