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If I have a relationship:

$ y_t = a + \theta b_t \epsilon_t,$

where I observe $y_t$ and $b_t$. $a$ is a known parameter, $\theta$ is an unknown parameter with prior distribution at time $t$: $\theta \sim logN(\mu_t, \sigma_t^2)$, and $\epsilon_t$ is an IID random variable with distribution: $\epsilon_t \sim logN(0,\sigma_\epsilon^2)$.

Is it possible to obtain an analytic log-normal posterior for $\theta$? If not, do there exist other possible conjugate prior distributions with non-negative support (and whatever distribution on the noise term $v_t$ that's necessary)?

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  • $\begingroup$ So $y_t$ is intended to be some kind of shifted lognormal? $\endgroup$ – Glen_b Jun 2 '14 at 23:40
  • $\begingroup$ If the prior for $\theta$ depends on $t$, then $\theta$ itself depends on $t$. So the question decomposes into a separate problem for each $\theta_t$. $\endgroup$ – Tom Minka Oct 3 '14 at 17:39
  • $\begingroup$ If $a$ is known, why not write $z_t = \log(y_t-a)$ and write a linear model in $z_t$? $\endgroup$ – Glen_b Mar 26 '15 at 9:14
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We have to calculate the posterior distribution of $\theta$ for the sample $D = \{(b_t, y_t)\}_{t = 1}^N$ $$ p(\theta | D) = \frac{p(D|\theta)p(\theta)}{p(D)}. $$

We are not interesting in the term in denominator. We need to select $p(\theta)$ such that posterior is log-normal.

Let's take a look at $p(D|\theta)$: $$ p(D|\theta) = \prod_{t = 1}^N p(y_t| b_t) = \\=\prod_{t = 1}^N \frac{\theta b}{\sqrt{2 \pi} \sigma_{\varepsilon} (y_t - a)} \exp\ \left(-\frac{1}{2 \sigma^2_{\varepsilon} } \left(\ln (y_t - a)- \ln b_t - \ln \theta \right)^2 \right) = \\ \frac{C_1}{\sqrt{2 \pi} \sigma_\theta} \theta^N \exp \left(-\frac{1}{2 \sigma_\theta} \left(\ln \theta - C_2 \right)^2 \right). $$

One can see that if we select an improper prior of the form $\frac{1}{\theta^{N + 1}} [\theta > 0]$ we get the log-normal distribution as a posterior distribution. Here $[\cdot]$ is the indicator function i.e. it equals $1$ if statement in the brackets holds and $0$ vice versa.

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  • $\begingroup$ I think we might need more information about the specification for $\theta$: it seems like user6600 wants it to have a different distribution at different points in time. $\endgroup$ – user44764 Jun 3 '14 at 13:03

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