# Notes

Consider the model

$d X_t = \alpha \left(\theta - X_t\right)dt + \sigma dB_t$

By Ito Isometry, we have:

$\langle X \rangle_t = E_t(X) = \theta \left(1 - e^{-\alpha t} \right) + X_0 e^{-\alpha t}$

and

$\langle X^2 \rangle_t - \langle X \rangle_t^2 = V_t(X) = \frac{\sigma^2}{2 \alpha }\left(1 - e^{-2 \alpha t} \right)$

If we assume discrete, uniform sampling of spacing $$\tau$$, then we have,

$P(\theta | X; \alpha, \sigma) = \sqrt{\frac{1}{2\pi V_{\tau} }}^{T-1} \exp\left(-\frac{\sum_t^{T-1} \left(X_{t+1} - E\right)^2 }{2V}\right)$

or more explicitly (taking $$\tau_i = 1$$ without loss of generality),

$P(\theta | X; \alpha, \sigma) = \sqrt{\frac{\tfrac{\alpha}{\sigma}}{\pi (1- e^{-\alpha t}) }}^{T-1} \exp\left( - \frac{\sum_t^{T-1} \left(X_{t+1} - ( \theta (1 - e^{-\alpha}) + X_0 e^{-\alpha}) \right)^2 }{\tfrac{\sigma^2}{\alpha}(1-e^{-2\alpha})}\right)$

To integrate out $$\theta$$, $$P(X | \alpha, \sigma) = \int P(X | \theta, \alpha, \sigma ) P(\theta) d\theta$$, we’ll make this look like a Gaussian in $$\theta$$ by completing the square. First, let us introduce a more compact notation to manipulate terms independent of $$\theta$$:

$A_t := X_{t+1} - X_0e^{-\alpha}$ $B := 1-e^{-\alpha}$

Our sum in the exponent can then be written more succinctly:

$\sum_t^{T-1} (A_t - \theta B)^2$

Squaring out inside the sum and distributing the summation operator (by linearity) and extracting or summing terms constant in $$t$$, we have:

$\sum A_t^2 - 2 \theta B \sum A_t + \theta^2 B^2 (T-1)$

For which we complete the square in $$\theta$$,

$B^2 (T-1) \left( \left(\theta - \frac{\sum_t A_t}{B(T-1)}\right)^2 + \frac{\sum A_t ^2 - (\sum A_t)^2}{B^2 (T-1)} \right)$

Which lets us write $$P(\theta | X; \alpha, \sigma)$$ as a normal distribution in $$\theta$$ times some constant terms:

$P(\theta | X; \alpha, \sigma) = \exp \left(\frac{-\sum A_t^2 + (\sum A_t)^2 }{V_{\tau} }\right) \sqrt{\frac{1}{2\pi V_{\tau} }}^{T-1} \int d\theta\exp\left( -\frac{ \left(\theta - \frac{\sum_t A_t}{B(T-1)} \right)^2 }{\tfrac{2V_{\tau}}{B^2 (T-1)}}\right)$

In which we recognize the integral as Gaussian with mean $=$ and variance $$\nu = V_{\tau}/(B^2(T-1))$$, and thus can replace the integral with $$\sqrt{2 \pi \nu} = \sqrt{\frac{2 \pi V_{\tau} }{B^2(T-1)}}$$,

$P(\theta | X; \alpha, \sigma) = \exp \left(\frac{-\sum A_t^2 + (\sum A_t)^2 }{V_{\tau} }\right) \sqrt{\frac{1}{2\pi V_{\tau} }}^{T-1} \sqrt{\frac{2 \pi V_{\tau} }{B^2(T-1)}}$

Collecting common terms in $$V_t$$,

$P(\theta | X; \alpha, \sigma) = \exp \left(\frac{f(A_t)}{V_{\tau}}\right) \sqrt{\frac{1}{2\pi}}^{T-1} \sqrt{\frac{2 \pi }{B^2(T-1)}} V_{\tau}^{\tfrac{1}{2}} V_{\tau}^{\tfrac{1-T}{2}}$

$= \exp \left(\frac{f(A_t)}{V_{\tau}}\right) \sqrt{\frac{(2 \pi)^{2-T}}{B ^ 2(T-1)}} V_{\tau}^{1 - \tfrac{T}{2}}$

where

$f(A_t) = - \sum A_t^2 + \left( \sum A_t \right)^2$

Then substituting in $$V_t = \tfrac{\sigma^2}{2\alpha} \left(1 - e^{-2\alpha}\right)$$

$= \exp \left(\frac{\tfrac{2\alpha f(A)}{1 - e^{-2\alpha}} }{\sigma^2}\right) \sqrt{\frac{(2 \pi)^{2-T}}{B ^ 2(T-1)}} \left(\tfrac{1}{2\alpha} \left(1 - e^{-2\alpha}\right)\right)^{1 - \tfrac{T}{2}}\sigma^{2(1 - T/2)}$

Which can be written to look like an Inverse Gamma integral in $$\sigma^2$$

$= \mathcal{N} e^{\tfrac{-\beta}{\sigma^2}} (\sigma^2)^{-\gamma - 1}$

where

$\gamma = T/2 -2,$

$\beta = \frac{2 \alpha \left( \left( \sum A_t \right)^2- \sum A_t^2 \right) }{1 - e^{-2\alpha}}$ and

$\mathcal{N} = \sqrt{\frac{(2 \pi)^{2-T}}{B ^ 2(T-1)}} \left(\tfrac{1}{2\alpha} \left(1 - e^{-2\alpha}\right)\right)^{1 - \tfrac{T}{2}}$

Which we can integrate as Inverse Gamma (again assuming uniform prior, now in $$\sigma^2$$),

$= \mathcal{N} \int e^{\tfrac{-\beta}{\sigma^2}} (\sigma^2)^{-\gamma - 1} d\sigma^2 = \mathcal{N} \frac{\beta^{\gamma}}{\Gamma(\gamma)}$

with $$\beta$$, $$\gamma$$, and $$\mathcal{N}$$ as defined above.