# Conjugate priors and Bayesian update

In a paper by Cyert and Degroot (1974) (Rational Expectations and Bayesian Analysis, 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!

• optimal control : I think this should help. people.eecs.berkeley.edu/~jordan/courses/260-spring10/lectures/… Commented May 3 at 17:11
• 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. Commented May 4 at 9:49

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

$$\begin{cases} h_{t+1}=h_{t}+r_{t}p_{t}^{2}\\ m_{t+1}=\frac{m_{t}h_{t}+rp_{t+1}p_{t}}{h_{t+1}} \end{cases}$$