Kalman Filter

Kalman filtering (and filtering in general) considers the following setting: we have a sequence of states  x_t, which evolves under random perturbations over time. Unfortunately we cannot observe x_t, we can only observe some noisy function of  x_t, namely,  y_t. Our task is to find the best estimate of x_t given our observations of y_t.

Consider the equations

where \epsilon_t \sim \mathcal N ( 0, \Sigma^\epsilon_t) ,  \eta_t ) \sim \mathcal N ( 0, \Sigma^\eta_t) and \epsilon_t and \nu_t are independent. (We let \Sigma^{\epsilon} be the sub-matrix of the covariance matrix corresponding to \epsilon and so forth…)

The Kalman filter has two update stages: a prediction update and a measurement update. These are

Screenshot 2019-05-14 at 07.12.37.pngandScreenshot 2019-05-14 at 07.12.41.png

where

The matrix K_t is often referred to as the Kalman Gain. Assuming the initial state x_0 is known and deterministic P_{0|0} =0 in the above.


We will use the following proposition, which is a standard result on normally distributed random vectors, variances and covariances,

Prop 1. Let u be normally distributed vector with mean \bar{u} and covariance \Sigma_{ u}, i.e.

i) For any matrix A and (constant) vector c, we have that

ii) If we take  u = ( v,  w) then w conditional on v gives

iii) Var ( A u ) = A\Sigma_u A^\top, Cov ( Au , Bu) = A \Sigma_u B^\top.


We can justify the Kalman filtering steps by proving that the conditional distribution of x_{t+1} is given by the Prediction and measurement steps. Specifically we have the following.

Theorem 1.

where y_{[0:t]} := (y_0,...,y_{t}) and a_{[0:t]} := (a_0,...,a_{t}).

Proof. We show the result by induction supposing that

Since x_{t+1} is a linear function of x_t, we have that

where, by Prop 1ii), we have that

Given y_{t+1} = C_t x_{t+1} + \eta_t, we have by Prop 1iii) that Var(y_{t+1} | y_{[0,t]}, a_{[0,t]}) = C_t P_{t+1|t} C^\top_t and Cov(x_{t+1}, y_{t+1} | y_{[0,t]}, a_{[0,t]}) = P_{t+1|t} C^\top_t . Thus

Thus applying Prop 1ii), we get that

$ \square$

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