<p>I have figured out the solution. Here is how for the community if one day someone needs it;</p>
<p>The crucial thing to consider is that we have <span class="math-container">$\mathbb{E}\left(p_{t+1}\right)=\mathbb{E}\left(ap_{t}\right)$</span> since the mean of the error term is <span class="math-container">$v_t$</span> is zero. So, we know that</p>
<p><span class="math-container">$$p\left(a\mid p_{t+1}\right)\propto p\left(a\right)p\left(p_{t+1}\mid a\right)$$</span></p>
<p>Then, we write</p>
<p><span class="math-container">$$p\left(a\right)=\left(2\pi\sigma_{t}^{2}\right)^{-\frac{1}{2}}\text{exp}\left(-\frac{1}{2\sigma_{0}^{2}}\left(a-m_{t}\right)^{2}\right)$$</span></p>
<p>and</p>
<p><span class="math-container">$$p\left(p_{t+1}\mid a\right)=\left(2\pi\sigma_{c}^{2}\right)^{-\frac{1}{2}}\text{exp}\left(-\frac{1}{2\sigma_{c}^{2}}\left(p_{t+1}-ap_{t}\right)^{2}\right)$$</span></p>
<p>We write <span class="math-container">$ap_t$</span> without any expectation term because this is known once we are in <span class="math-container">$t+1$</span>. In other words, it becomes something &quot;known&quot;. Then,</p>
<p><span class="math-container">$$p\left(a\right)p\left(p_{t+1}\mid a\right)\propto\text{exp}\left(-\frac{1}{2\sigma_{0}^{2}}\left(a-m_{t}\right)^{2}-\frac{1}{2\sigma_{c}^{2}}\left(p_{t+1}-ap_{t}\right)^{2}\right)$$</span></p>
<p>Note that we substitute variance by precision; <span class="math-container">$\frac{1}{\sigma_{c}^{2}}=r$</span> and <span class="math-container">$\frac{1}{\sigma_{t}^{2}}=h_{t}$</span>. By reformulating</p>
<p><span class="math-container">$$p\left(a\right)p\left(p_{t+1}\mid a\right)\propto\text{exp}\left(-\frac{h_{t}}{2}\left(a^{2}-2am_{t}+a_{t}^{2}\right)-\frac{r}{2}\left(p_{t+1}^{2}-2p_{t+1}p_{t}a+a^{2}p_{t}^{2}\right)\right)\overset{\text{def}}{=}\text{exp}\left(-\frac{h_{t+1}}{2}\underset{=a^{2}-2am_{t+1}+m_{t+1}^{2}}{\underbrace{\left(a-m_{t+1}\right)^{2}}}\right)$$</span></p>
<p>We reformulate again, by factorizing both RHS and LHS by <span class="math-container">$a^2$</span> and <span class="math-container">$a$</span>;</p>
<p><span class="math-container">$$p\left(a\right)p\left(p_{t+1}\mid a\right)\propto\text{exp}\left(-\frac{a^{2}}{2}\left(h_{t}+r_{t}p_{t}^{2}\right)-\frac{2a}{2}\left(m_{t}h_{t}+rp_{t+1}p_{t}\right)-\left(\frac{m_{t}^{2}h_{t}+rp_{t+1}^{2}}{2}\right)\right)\overset{\text{def}}{=}\text{exp}\left(-\frac{a^{2}}{2}\left(h_{t+1}\right)-\frac{2a}{2}\left(h_{t+1}m_{t+1}\right)-h_{t+1}\frac{m_{t+1}^{2}}{2}\right)$$</span>.</p>
<p>So by mathcing coefficients of both RHS and LHS, we see easily</p>
<p><span class="math-container">$$\begin{cases}
-\frac{a^{2}}{2}\left(h_{t}+r_{t}p_{t}^{2}\right)=-\frac{a^{2}}{2}\left(h_{t+1}\right)\\
-\frac{2a}{2}\left(m_{t}h_{t}+rp_{t+1}p_{t}\right)=-\frac{2a}{2}\left(h_{t+1}m_{t+1}\right)
\end{cases}$$</span></p>
<p>From here, it is easy to write</p>
<p><span class="math-container">$$\begin{cases}
h_{t+1}=h_{t}+r_{t}p_{t}^{2}\\
m_{t+1}=\frac{m_{t}h_{t}+rp_{t+1}p_{t}}{h_{t+1}}
\end{cases}$$</span></p>

CC BY-SA 4.0
2024-05-04T08:21:04.610Z
2024-05-04T08:21:04.610Z
optimal control
<p>In a paper by Cyert and Degroot (1974) (<a href="https://www.jstor.org/stable/1829843" rel="nofollow noreferrer">Rational Expectations and Bayesian Analysis</a>, in Journal of Political Economy), authors use Bayesian update for an uncertain parameter. They have a model for pricing as follows</p>
<p><span class="math-container">$$p_{t+1}=ap_{t}+v_{t+1}$$</span></p>
<p>where <span class="math-container">$v_1$</span>, <span class="math-container">$v_2$</span>,... form a sequence of iid error terms. The parameter on which there is a Bayesian update is <span class="math-container">$a$</span>.
<span class="math-container">$r$</span> is the known precision of the signal (from the pricing equation). They find the following updating rule for the mean and the variance for <span class="math-container">$a$</span></p>
<p><span class="math-container">$$m_{t+1}=\frac{h_{t}m_{t}+rp_{t}p_{t+1}}{h_{t}+r\left(p_{t}\right)^{2}}$$</span></p>
<p>and</p>
<p><span class="math-container">$$h_{t+1}=h_{t}+r\left(p_{t}\right)^{2}$$</span></p>
<p>I have some trouble to find these values. <strong>Here is what I have tried</strong>; They say they use prior conjugates to find these updating rules. So,the posterior</p>
<p><span class="math-container">$$p\left(a\mid p_{t+1}\right)\propto p\left(a\right)p\left(p_{t+1}\mid a\right)$$</span><span class="math-container">$
$</span></p>
<p>where</p>
<p><span class="math-container">$$p\left(a\right)=\left(2\pi\sigma_{0}^{2}\right)^{-\frac{1}{2}}\text{exp}\left(-\frac{1}{2\sigma_{0}^{2}}\left(a-a_{0}\right)^{2}\right)$$</span></p>
<p>and</p>
<p><span class="math-container">$$p\left(p_{t+1}\mid a\right)=\left(2\pi\sigma_{c}^{2}\right)^{-\frac{1}{2}}\text{exp}\left(-\frac{1}{2\sigma_{c}^{2}}\left(p_{t+1}-a\right)^{2}\right)$$</span></p>
<p>So by using these last two expressions, I write</p>
<p><span class="math-container">$$p\left(a\right)p\left(p_{t+1}\mid a\right)\propto exp\left(-\frac{1}{2\sigma_{0}^{2}}\left(a-a_{0}\right)^{2}-\frac{1}{2\sigma_{c}^{2}}\left(p_{t+1}-a\right)^{2}\right)$$</span></p>
<p>where the known and constant precision <span class="math-container">$\frac{1}{\sigma_{c}^{2}}=r$</span> and the time varying precision <span class="math-container">$\frac{1}{\sigma_{0}^{2}}=h_{0}$</span></p>
<p>and I end up with this</p>
<p><span class="math-container">$$p\left(a\right)p\left(p_{t+1}\mid a\right)\propto exp\left(-\frac{h_{0}}{2}\left(a^{2}-2aa_{0}+a_{0}^{2}\right)-\frac{r}{2}\left(p_{t+1}^{2}-2p_{t+1}a+a^{2}\right)\right) \overset{\text{def}}{=}\text{exp}\left(-\frac{h_{1}}{2}\left(a-a_{1}\right)^{2}\right)$$</span></p>
<p>I cannot end up with terms like <span class="math-container">$p_t^2$</span> and <span class="math-container">$p_t p_t+1$</span>. What do I miss? Any hints/suggestions or solution is appreciated, thanks!</p>

CC BY-SA 4.0
2024-05-03T17:11:35.807Z
optimal control : I think this should help. https://people.eecs.berkeley.edu/~jordan/courses/260-spring10/lectures/lecture5.pdf
mark leeds
CC BY-SA 4.0
2024-05-04T09:49:56.240Z
I'm glad if it helps. It doesn't show every single detail but I bet you can figure out from it where the difference lies.
mark leeds
<statistics><bayesian><learning>
Conjugate priors and Bayesian update
optimal control
# Conjugate priors and Bayesian update

(Asked by: username_0 on 2024-05-03)

In a paper by Cyert and Degroot (1974) ([Rational Expectations and Bayesian Analysis](https://www.jstor.org/stable/1829843), in Journal of Political Economy), authors use Bayesian update for an uncertain parameter. They have a model for pricing as follows

$$p\_{t+1}=ap\_{t}+v\_{t+1}$$

where $v\_1$, $v\_2$,... form a sequence of iid error terms. The parameter on which there is a Bayesian update is $a$.
$r$ is the known precision of the signal (from the pricing equation). They find the following updating rule for the mean and the variance for $a$

$$m\_{t+1}=\frac{h\_{t}m\_{t}+rp\_{t}p\_{t+1}}{h\_{t}+r\left(p\_{t}\right)^{2}}$$

and

$$h\_{t+1}=h\_{t}+r\left(p\_{t}\right)^{2}$$

I have some trouble to find these values. **Here is what I have tried**; They say they use prior conjugates to find these updating rules. So,the posterior

$$p\left(a\mid p\_{t+1}\right)\propto p\left(a\right)p\left(p\_{t+1}\mid a\right)$$$
$

where

$$p\left(a\right)=\left(2\pi\sigma\_{0}^{2}\right)^{-\frac{1}{2}}\text{exp}\left(-\frac{1}{2\sigma\_{0}^{2}}\left(a-a\_{0}\right)^{2}\right)$$

and

$$p\left(p\_{t+1}\mid a\right)=\left(2\pi\sigma\_{c}^{2}\right)^{-\frac{1}{2}}\text{exp}\left(-\frac{1}{2\sigma\_{c}^{2}}\left(p\_{t+1}-a\right)^{2}\right)$$

So by using these last two expressions, I write

$$p\left(a\right)p\left(p\_{t+1}\mid a\right)\propto exp\left(-\frac{1}{2\sigma\_{0}^{2}}\left(a-a\_{0}\right)^{2}-\frac{1}{2\sigma\_{c}^{2}}\left(p\_{t+1}-a\right)^{2}\right)$$

where the known and constant precision $\frac{1}{\sigma\_{c}^{2}}=r$ and the time varying precision $\frac{1}{\sigma\_{0}^{2}}=h\_{0}$

and I end up with this

$$p\left(a\right)p\left(p\_{t+1}\mid a\right)\propto exp\left(-\frac{h\_{0}}{2}\left(a^{2}-2aa\_{0}+a\_{0}^{2}\right)-\frac{r}{2}\left(p\_{t+1}^{2}-2p\_{t+1}a+a^{2}\right)\right) \overset{\text{def}}{=}\text{exp}\left(-\frac{h\_{1}}{2}\left(a-a\_{1}\right)^{2}\right)$$

I cannot end up with terms like $p\_t^2$ and $p\_t p\_t+1$. What do I miss? Any hints/suggestions or solution is appreciated, thanks!

**Tags:** statistics, bayesian, learning

> **Comments:**
> **username_1**: username_0 username_0 : I think this should help. https://people.eecs.berkeley.edu/~jordan/courses/260-spring10/lectures/lecture5.pdf  
> **username_1**: I'm glad if it helps. It doesn't show every single detail but I bet you can figure out from it where the difference lies.  

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## 1 answers

### Answer by username_0 on 2024-05-04. Score: 0

I have figured out the solution. Here is how for the community if one day someone needs it;

The crucial thing to consider is that we have $\mathbb{E}\left(p\_{t+1}\right)=\mathbb{E}\left(ap\_{t}\right)$ since the mean of the error term is $v\_t$ is zero. So, we know that

$$p\left(a\mid p\_{t+1}\right)\propto p\left(a\right)p\left(p\_{t+1}\mid a\right)$$

Then, we write

$$p\left(a\right)=\left(2\pi\sigma\_{t}^{2}\right)^{-\frac{1}{2}}\text{exp}\left(-\frac{1}{2\sigma\_{0}^{2}}\left(a-m\_{t}\right)^{2}\right)$$

and

$$p\left(p\_{t+1}\mid a\right)=\left(2\pi\sigma\_{c}^{2}\right)^{-\frac{1}{2}}\text{exp}\left(-\frac{1}{2\sigma\_{c}^{2}}\left(p\_{t+1}-ap\_{t}\right)^{2}\right)$$

We write $ap\_t$ without any expectation term because this is known once we are in $t+1$. In other words, it becomes something "known". Then,

$$p\left(a\right)p\left(p\_{t+1}\mid a\right)\propto\text{exp}\left(-\frac{1}{2\sigma\_{0}^{2}}\left(a-m\_{t}\right)^{2}-\frac{1}{2\sigma\_{c}^{2}}\left(p\_{t+1}-ap\_{t}\right)^{2}\right)$$

Note that we substitute variance by precision; $\frac{1}{\sigma\_{c}^{2}}=r$ and $\frac{1}{\sigma\_{t}^{2}}=h\_{t}$. By reformulating

$$p\left(a\right)p\left(p\_{t+1}\mid a\right)\propto\text{exp}\left(-\frac{h\_{t}}{2}\left(a^{2}-2am\_{t}+a\_{t}^{2}\right)-\frac{r}{2}\left(p\_{t+1}^{2}-2p\_{t+1}p\_{t}a+a^{2}p\_{t}^{2}\right)\right)\overset{\text{def}}{=}\text{exp}\left(-\frac{h\_{t+1}}{2}\underset{=a^{2}-2am\_{t+1}+m\_{t+1}^{2}}{\underbrace{\left(a-m\_{t+1}\right)^{2}}}\right)$$

We reformulate again, by factorizing both RHS and LHS by $a^2$ and $a$;

$$p\left(a\right)p\left(p\_{t+1}\mid a\right)\propto\text{exp}\left(-\frac{a^{2}}{2}\left(h\_{t}+r\_{t}p\_{t}^{2}\right)-\frac{2a}{2}\left(m\_{t}h\_{t}+rp\_{t+1}p\_{t}\right)-\left(\frac{m\_{t}^{2}h\_{t}+rp\_{t+1}^{2}}{2}\right)\right)\overset{\text{def}}{=}\text{exp}\left(-\frac{a^{2}}{2}\left(h\_{t+1}\right)-\frac{2a}{2}\left(h\_{t+1}m\_{t+1}\right)-h\_{t+1}\frac{m\_{t+1}^{2}}{2}\right)$$.

So by mathcing coefficients of both RHS and LHS, we see easily

$$\begin{cases}
-\frac{a^{2}}{2}\left(h\_{t}+r\_{t}p\_{t}^{2}\right)=-\frac{a^{2}}{2}\left(h\_{t+1}\right)\\
-\frac{2a}{2}\left(m\_{t}h\_{t}+rp\_{t+1}p\_{t}\right)=-\frac{2a}{2}\left(h\_{t+1}m\_{t+1}\right)
\end{cases}$$

From here, it is easy to write

$$\begin{cases}
h\_{t+1}=h\_{t}+r\_{t}p\_{t}^{2}\\
m\_{t+1}=\frac{m\_{t}h\_{t}+rp\_{t+1}p\_{t}}{h\_{t+1}}
\end{cases}$$


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