Bayesian Filtering

and how it connects to Kalman Filtering.

Preliminaries

Let \(P(E)\) denote the probability of event \(E\) occuring. The probability of multiple events \(E_1, E_2, ...\) occuring simultaneously is denoted \(P(E_1, E_2, ...)\).

The probability of \(E_1\) occuring given that \(E_2\) occured is denoted \(P(E_1 \| E_2)\).

This notation can easily be extended. For example \(P(E_1, E_2 \| E_3) = P(E_1 , (E_2 \| E_3))\) should be read as the probability of (\(E_1\) occuring) and (\(E_2\) occuring given \(E_3\)).

For a continuous random variable \(X\), we write

\[F(x) = P(X \leq x)\]

where \(F : \mathbb{R} \to \mathbb{R}\) is the cummulative distribution of \(X\), and

\[f(x) = \lim_{\delta \to 0} P(x \leq X \leq x + \delta) = \frac{dF}{dx}(x)\]

where \(f: \mathbb{R} \to \mathbb{R}\) is the probability density function of \(X\).

Bayes’ Rule:

For any two events \(A, B\),

\[P(A, B) = P(A\|B) P(B)\]

This can be used to derive the following:

\[P(A\|B) = \frac{P(A, B)}{P(B)} = \frac{P(B, A)}{P(B)} = \frac{P(B\|A) P(A)}{P(B)}\]

Chain Rule:

For any three events \(A, B, C\),

\[P(A, B, C) = P(A\|B, C) P(B\|C) P(C)\]

Bayesian Inference:

The goal is to determine the distribution

\[p(x_t |\ y_{1:t})\]

i.e., the distribution of the state \(x_t\) at time \(t\) conditioned on all the measurements \(y_{1:t}\) obtained upto and including time \(t\).

We assume that we have access to

• \(p(y_t \| x_t)\) i.e., the distribution of the output conditioned on the state, and
• \(p(x_{t-1} \| y_{1:t-1})\) i.e., the distribution at the previous timestep, and
• \(p(x_t \| x_{t-1})\) i.e., a prediction model for the distibution of new state given the prior state.

Further, we make the following assumptions:

• \(y_t\) is only dependent on \(x_t\)
• \(x_t\) is only dependent on \(x_{t-1}\)

Then, we can write the two steps of inference:

Prediction

Notice,

\[\begin{align} p(x_t \| y_{1:t-1}) & \stackrel{(a)}{=} \int p(x_t, x_{t-1} \| y_{1:t-1}) dx_{t-1}\\ &\stackrel{(b)}{=} \int p(x_t \| x_{t-1}, y_{1:t-1}) p(x_{t-1} | y_{1:t-1}) dx_{t-1}\\ &\stackrel{(c)}{=} \int p(x_t \| x_{t-1}) p(x_{t-1} \| y_{1:t-1}) dx_{t-1} \end{align}\]

where

• (a) makes the dependence on \(x_{t-1}\) explicit
• (b) follows from \(P(A, B) = P(A\|B)P(B)\), and
• (c) is because \(x_t\) is independent of \(y_{1:t-1}\).

Correction

When the \(k\)-th measurement is available, we can now update the posterior distribution:

\[\begin{align} p(x_t \| y_{1:t}) &\stackrel{(a)}{=}\frac{p(x_t, y_{1:t})}{p(y_{1:t})}\\ &\stackrel{(b)}{=}\frac{p(x_t, y_t, y_{1:t-1})}{p(y_{1:t})}\\ &\stackrel{(c)}{=}\frac{p(y_t, x_t, y_{1:t-1})}{p(y_{1:t})}\\ &\stackrel{(d)}{=}\frac{p(y_t \| x_t) p(x_t \| y_{1:t-1}) p(y_{1:t-1})}{p(y_{1:t})}\\ &\stackrel{(e)}{=}\frac{p(y_t \| x_t) p(x_t \| y_{1:t-1}) p(y_{1:t-1})}{p(y_t, y_{1:t-1})}\\ &\stackrel{(f)}{=}\frac{p(y_t \| x_t) p(x_t \| y_{1:t-1}) p(y_{1:t-1})}{p(y_t|y_{1:t-1}) p(y_{1:t-1})}\\ &\stackrel{(g)}{=}\frac{p(y_t \| x_t) p(x_t \| y_{1:t-1}) }{p(y_t|y_{1:t-1})}\\ &\propto p(y_t \| x_t) p(x_t \| y_{1:t-1}) \end{align}\]

where

• (a) is Bayes’ rule
• (b) makes explicit that \(p(y_{1:t}) = p(y_{t}, y_{1:t})\)
• (c) reorders the terms in the numerator
• (d) applies Chain Rule
• (e) expands the denominator using the same property as in (b)
• (f) applies Bayes’ rule to the denominator
• (g) cancels the \(p(y_{1:t-1})\) term from the numerator and denominator

Connection to Kalman Filtering

In Kalman Filtering, the filter maintains the state of \(\hat x\) and \(P\), defined in terms of the distributions of \(x\), the random variable that is the state.

We assume the system model

\[\begin{align} x_{t} &= Fx_{t-1} + G w_{t-1}\\ y_{t} &= H x_t + v_t \end{align}\]

where \(w_{t} \sim \mathcal{N}(0, Q)\), \(v_t \sim \mathcal{N}(0, R)\).

Define \(\hat x_{t\|t}\), \(P_{t\|t}\), \(\hat x_{t-1\|t}\) and \(P_{t\|t-1}\) such that

\[\begin{align} \hat x_{t-1|t-1} &= \mathbb{E}[ x_{t-1} \| y_{1:t-1}]\\ \hat x_{t|t-1} &= \mathbb{E}[ x_{t} \| y_{1:t-1}]\\ \hat x_{t|t} &= \mathbb{E}[ x_{t} \| y_{1:t}] \end{align}\] \[\begin{align} P_{t-1|t-1} &= \mathbb{V}[ x_{t-1} \| y_{1:t-1}]\\ P_{t|t-1} &= \mathbb{V}[ x_{t} \| y_{1:t-1}]\\ P_{t|t} &= \mathbb{V}[ x_{t} \| y_{1:t}] \end{align}\]

where \(\mathbb{E}[x]\) is the expected value of \(x\), and \(\mathbb{V}[x]\) is the variance of \(x\).

Then, we must have

\[p(x_{t-1} \| y_{1:t-1}) = \mathcal{N}( \mathbb{E}[x_{t-1} \| y_{1:t-1}], \mathbb{V}[x_{t-1} \| y_{1:t-1}]) = \mathcal{N}(\hat x_{t-1\|t-1}, P_{t-1\|t-1})\] \[p(x_{t} \| y_{1:t-1}) = \mathcal{N}( \mathbb{E}[x_{t} \| y_{1:t-1}], \mathbb{V}[x_{t} \| y_{1:t-1}]) = \mathcal{N}(\hat x_{t\|t-1}, P_{t\|t-1})\] \[p(x_{t} \| y_{1:t}) = \mathcal{N}( \mathbb{E}[x_{t} \| y_{1:t}], \mathbb{V}[x_{t} \| y_{1:t}]) = \mathcal{N}(\hat x_{t\|t}, P_{t\|t})\]

Now, we are given \(\hat x_{t-1\|t-1}\), \(P_{t-1\|t-1}\), and the measurement \(y_{t}\), and the goal is to determine \(\hat x_{t\|t}\) and \(P_{t\|t}.\)

Prediction

We can directly apply the Bayesian rules established above:

\[\begin{align} p(x_{t} \| y_{1:t-1}) &= \int p(x_t \| x_{t-1}) p(x_{t-1} \| y_{1:t-1}) d x_{t-1}\\ &= \int p((F x_{t-1} + G w_{t-1}) \| x_{t-1}) p(x_{t-1} \| y_{1:t-1}) d x_{t-1}\\ &= \int (p((F x_{t-1}) \| x_{t-1}) + p((G w_{t-1}) \| x_{t-1})) p(x_{t-1} \| y_{1:t-1}) d x_{t-1} \end{align}\]

Now notice \(\begin{align} p((F x_{t-1}) \| x_{t-1}) &= \frac{p(F x_{t-1}, x_{t-1})}{p(x_{t-1})}\\ &= \end{align}\)