# Retrodiction / Specific filter to obtain initial state

## Problem

I have a system that is measured at regular intervals. The state of the system at those times is given by the vector $$\vec x=(x_0, x_1, x_2,\cdots)$$. In between each measurement, a random variable $$z$$, drawn from a Gaussian distribution with zero mean and standard deviation $$\sigma_z$$ is added onto the state and there is some decay, i.e., $$x_i=(1-\gamma)x_{i-1}+z_i$$. (The underlying model is a damped harmonic oscillator in a thermal environment, probed at multiples of its period.)

My measurements are imperfect, i.e., I measure $$\vec y=(y_0,y_1,y_2,\cdots)$$, where $$y_i=x_i+d_i$$. Again, the $$d$$ are drawn from a Gaussian distribution of zero mean, with standard deviation $$\sigma_d$$. Note that this setup is a close variant of an example on the Wikipedia page for the Kalman filter. But as far as I understand, the Kalman filter does not take the whole measurement record into account, but provides an update rule for the best guess. In particular, it's estimate for $$x_0$$, which I am particularly interested in, is bad.

I would like to find $$\vec x$$ given some measurement record $$\vec y$$ (parameters $$\sigma_d, \sigma_z$$, $$\gamma$$ are known). I believe I should use some maximum likelihood analysis, but I have trouble finding the right probability distributions for $$P(\vec y|\vec x), P(\vec x), P(\vec y)$$ that I need to determine $$P(\vec x|\vec y)$$.

## Attempt

I tried guessing what the right probability distributions should be. This one is fairly easy: $$P(\vec y|\vec x) = \prod_{i=0}^{N-1}\frac{1}{\sqrt{2\pi\sigma_z^2}}\exp\left(-\frac{(y_i-x_i)^2}{2\sigma_z^2}\right).$$

Now for $$P(\vec x)$$, I'm already slightly unsure. I start at equilibrium of the above update rule for $$x_i$$, i.e., when the covariance $$\sigma_{x_i}^2\to(1-\gamma)\sigma_{x_{i-1}}^2+\sigma_d^2$$ has reached a steady state, i.e., $$\sigma_{x_0}=\sigma_d/\sqrt{\gamma}$$. This is my expectation for $$x_0$$. Thus I think I should write $$P(\vec x) = \frac{1}{\sqrt{2\pi\sigma_{x_0}^2}}\exp\left(-\frac{x_0^2}{2\sigma_{x_0}^2}\right)\times\prod_{i=1}^{N-1}\frac{1}{\sqrt{2\pi\sigma_d^2}}\exp\left(-\frac{(x_i-x_{i-1})^2}{2\sigma_d^2}\right).$$

However, the real problem is with $$P(\vec y)$$, and I think this is due to my limited experience with this method. Naively, without thinking about Bayesian updates, my guess for $$P(\vec x|\vec y)$$ would have looked like the product of $$P(\vec x)$$ and $$P(\vec y|\vec x)$$. I think my expectation for $$\vec y$$ should be $$P(\vec y) = \int d\vec x\, P(\vec x)P(\vec y|\vec x).$$ It is possible to do all the Gaussian integrals, but it doesn't seem to give a simple result.

Is there some simpler assumption I can make? And how is it justified? Is there a mistake somewhere else?

## Attempt #2

I'm still not entirely sure about $$P(\vec y)$$ (I think it's correct though), but one can get by without explicitly calculating it.

One mistake above is calling this approach "maximum likelihood". It's not. MLE would mean that I chose $$\vec x$$ such that $$P(\vec y|\vec x)$$ above becomes maximal for a given set of measurement results $$\vec y$$. This is a bad guess, as it neglects what we know about how the state evolves (note that $$\sigma_d$$ does not crop up in $$P(\vec y|\vec x)$$). Instead, what I am doing is Bayesian inference, i.e., finding the conditional probability $$P(\vec x|\vec y)$$.

To do this, we start with the joint distribution $$P(\vec x, \vec y)=P(\vec y|\vec x)P(\vec x)$$. This is a normal distribution $$\begin{pmatrix}\vec x \\ \vec y\end{pmatrix} \sim N\left[\begin{pmatrix}\vec \mu_x \\ \vec \mu_y\end{pmatrix}, \begin{pmatrix}\Sigma_{xx} & \Sigma_{xy} \\ \Sigma_{xy} & \Sigma_{yy}\end{pmatrix}\right],$$ with $$\mu_x,\mu_y=0$$, $$\Sigma_{yy} = \Sigma_{xy} = (1/2\sigma_z^2)\mathbb 1$$, and $$[\Sigma_{xx}]_{ij} = \left( \frac{1}{2\sigma_z^2} + \frac{1}{\sigma_d^2} \right)\delta_{i,j} + \left( \frac{1}{2\sigma_{x_0}^2} - \frac{1}{2\sigma_d^2} \right)\delta_{i,0}\delta_{j,0}+\frac{1}{2\sigma_d^2}(\delta_{i,j+1}+\delta_{i,j-1}).$$ NB: $$\Sigma_{xx}$$ is tri-diagonal apart from the first element, which differs (indexing from 0).

It's actually straightforward to derive the conditional distribution for $$\vec x$$ from the joint distribution [2]. It is given through $$\vec x \sim N(\mu_x+\Sigma_{xy}\Sigma_{yy}^{-1}(\vec y-\vec\mu_y), \Sigma_{xx}-\Sigma_{xy}\Sigma_{yy}^{-1}\Sigma_{xy}).$$ However, in our case, this reduces to $$\vec x\sim N( \vec y, \Sigma_{xx}-1/2\sigma_z^2 ),$$ which is not at all what I am expecting. All values of $$\vec y$$ should be used to determine a specific $$x_i$$. The problem lies in the fact that $$\Sigma_{yy}$$ is diagonal. What am I missing?

• Am I misunderstanding your notation or is what you've called $P(\vec x|\vec y)$ not simply the result of the Kalman smoother? Aug 25, 2019 at 18:53
• Thanks for your comment, Chris. I'm not sure, I will check it out. If it is it should be derivable as I tried, right? The discussion on Wikipedia is not very clear to me, it would be nice to get some insight by deriving it for this example. Aug 26, 2019 at 11:36
• Yes, but it's easiest to derive recursively rather than all at once, first by computing forward each filtering distribution $p(x_t|y_1,...,y_t)$ from $p(x_{t-1}|y_1,...,y_{t-1})$, and then backwards the state posteriors $p(x_{t-1}|y_1,...,y_T)$ from $p(x_t|y_1,...,y_T)$. To be sure, do you really have an observation $y_0$ that corresponds to the state $x_0$ that you care about (as you've written)? Typically the "initial state" is the time right before the first observation, and inference about it is not quite the same. Aug 26, 2019 at 23:56
• a) I've tried to understand the derivation of the Kalman smoother, but in vain. In the end, I need an analytical formula for $x_0$, which is why I was trying to derive it myself. b) I have added a second attempt to solve the problem, which seems to almost get me there, but I must be missing something small. Maybe you can spot it? c) And yes, I do have an observation $y_0$. It doesn't strike me as unusual. The difference is small anyway, waiting a while to make the first measurement just changes the expected initial variance. Aug 27, 2019 at 10:33

## 1 Answer

The posted attempt is almost correct, but it incorrectly identifies the covariance of the joint normal distribution.

## Joint Normal Distribution

Given $$P(\vec y|\vec x) = \prod_{i=0}^{N-1}\frac{1}{\sqrt{2\pi\sigma_z^2}}\exp\left(-\frac{(y_i-x_i)^2}{2\sigma_z^2}\right).$$ and $$P(\vec x) = \frac{1}{\sqrt{2\pi\sigma_{x_0}^2}}\exp\left(-\frac{x_0^2}{2\sigma_{x_0}^2}\right)\times\prod_{i=1}^{N-1}\frac{1}{\sqrt{2\pi\sigma_d^2}}\exp\left(-\frac{(x_i-x_{i-1})^2}{2\sigma_d^2}\right),$$ we can write $$P(\vec x,\vec y)=P(\vec x)P(\vec y|\vec x)\propto \exp\left(-\frac12 (\vec x^\top,\vec y^\top)\mathsf Q\begin{pmatrix}\vec x\\\vec y\end{pmatrix}\right),$$ where $$\mathsf Q$$ has the entries that I thought above where the covariance matrix, i.e., $$\mathsf Q_{yy} = \mathsf Q_{xy} = (1/2\sigma_z^2)\mathbb 1$$, and $$[\mathsf Q_{xx}]_{ij} = \left( \frac{1}{2\sigma_z^2} + \frac{1}{\sigma_d^2} \right)\delta_{i,j} + \left( \frac{1}{2\sigma_{x_0}^2} - \frac{1}{2\sigma_d^2} \right)\delta_{i,0}\delta_{j,0}+\frac{1}{2\sigma_d^2}(\delta_{i,j+1}+\delta_{i,j-1}).$$

## Covariance Matrix

In contrast to what is stated above, the covariance matrix $$\mathsf\Sigma=\mathsf Q^{-1}$$. This is easily done with block matrix inversion. We find $$\mathsf\Sigma=\begin{pmatrix}\mathsf\Sigma_{xx} & -\mathsf\Sigma_{xx} \\ -\mathsf\Sigma_{xx} & 2\sigma_z^2+\mathsf\Sigma_{xx}\end{pmatrix},$$ where $$\mathsf\Sigma_{xx}=(\mathsf Q_{xx}-1/(2\sigma_z^2))^{-1}.$$

## Conditional distribution

Given this, the conditional distribution for $$\vec x$$ can be derived from the joint distribution [2] $$\vec x \sim N(\mu_x+\mathsf\Sigma_{xy}\mathsf\Sigma_{yy}^{-1}(\vec y-\vec\mu_y), \mathsf\Sigma_{xx}-\mathsf\Sigma_{xy}\mathsf\Sigma_{yy}^{-1}\mathsf\Sigma_{xy}).$$

## Analytical solution

I think the matrix $$\mathsf Q_{xx}$$ can be inverted analytically, or at least to a very good approximation. I will include this once I have done so.