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I observe draws of some random variable $Y$ over time where $Y_{t} = aY_{t-1} + \epsilon_{t}$.

$\epsilon \sim N(0, 1/\rho_\epsilon)$ and $a$ is an unknown parameter with prior distribution $a \sim N(\mu_0, \Sigma_0)$.

Since both the noise and the prior are normal, after we observe $Y_t$, the posterior of $a$ is also normal and follows an updating process: $$ \mu_t = (\mu_{t-1}+\Sigma_{t-1}\rho_{\epsilon} Y_{t-1}Y_t)/(1+\Sigma_{t-1}\rho_{\epsilon}Y_{t-1}^2)\\ \Sigma_t = \Sigma_{t-1}/(1+\Sigma_{t-1}\rho_{\epsilon}Y_{t-1}^2) $$

Moving two periods into the future, it's easy to see that: $$ Y_{t+2}= a(\underbrace{a Y_t + \epsilon_{t+1}}_{Y_{t+1}}) + \epsilon_{t+2} $$

Given that I'm at time $t$, I'm looking to evaluate the variance of $Y_{t+2}$.

Since $\epsilon_{t+1}$ and $\epsilon_{t+2}$ affect future draws of $Y$, they're independent of our current beliefs on the distribution of $a$. The variance is then: $$ var(Y_{t+2}) = var(b Y_{t+1}) + 1/\rho_{\epsilon} $$

Is there a way to calculate the first variance term conditional on being at time $t$? I suppose I could take the covariance of $b$ and $T_{t+1}$ to obtain the joint distribution and then, if I wanted to solve this numerically, do the integration for the variance.

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  • $\begingroup$ Where did $b$ come from? $\endgroup$ – Dave Feb 28 '14 at 17:07
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$$ var(aY_{t+1}) = E[a^2 Y_{t+1}^2] - E[a Y_{t+1}]^2 $$ where $$ E[a^2 Y_{t+1}^2] = E[a^2 (a^2 Y_t^2 + 2aY_t \epsilon_{t+1} + \epsilon_{t+1}^2)] = E[a^4] Y_t^2 + E[a^2]/\rho_\epsilon $$ $$ E[a Y_{t+1}] = E[a (a Y_t + \epsilon_{t+1})] = E[a^2] Y_t $$

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