# In what sense is the Kalman filter optimal?

The Kalman filter is a minimum mean-square error estimator. The MSE is defined as $$E\left(||\hat{x}_k-x_k||^2\right)$$ where $$x$$ is the state and $$\hat{x}$$ is the estimate. When $$x$$ is a vector, for example, a vector that contains distance and velocity, is the MSE equals to distance MSE plus velocity MSE? If so, the base units of distance and velocity are different. Does the MSE have any physical meaning?

In the academic sense, where dimensions are not allowed into the room, the Kalman filter minimizes the expected MSE of the state vector, as you stated.

You mentioned dimensions, and I thought "uh oh, this is a conundrum". But for a properly-constructed Kalman filter* the states are uncorellated, i.e. $$\mathrm E \left \lbrace x_k \cdot x_n \right \rbrace \ 0\ \forall \ n \ne k$$. This means that for any weighting vector $$\mathbf w$$, the Kalman minimizes $$\mathbf w^T x$$. So you can choose any values for the elements of $$\mathbf w$$ that make the dimensions work out, and the resulting error will be minimized.

To answer your direct question: the Kalman is optimal in the sense that minimizes the expected error of each state. It just happens that in the process (because it also decorrelates the errors) it minimizes any global weighted sum of the states, regardless of the weighting you choose.

* "Properly constructed" in this case means that the model the Kalman was designed to actually matches the system whose states you're estimating**.

** Which really never happens in practice. It takes a lot of work to get close enough so that you can ignore the difference. In actual practice a very few folks do that, but more often they either design a Kalman using informed guesses about the system dynamics and the process and measurement noises then iterate on a solution, or they design some robust variant of the Kalman, such as an H-infinity filter.

• $H_\infty$ for the win! ;-)
– Peter K.
Commented Oct 4, 2021 at 15:22
• Thanks a lot for your enlightening answer. My comprehension is that all the states are linked according to the state transition model; Therefore, to achieve the optimal linear estimation, the estimation of every single state must be optimal. Commented Oct 5, 2021 at 6:58
• The states are thusly linked. But the Kalman assumes a perfect model, and if the model is perfect, then the state errors are uncorrelated. Commented Oct 5, 2021 at 15:01

This is a harder question to answer than I thought. Here's a statement from Chapter 3, page 49 of Anderson and Moore, Optimal Filtering:

Here $$x_0$$ is the initial state, and $$v_k$$ and $$w_k$$ are the measurement noise and the process noise respectively.

• Now I want to know where I can find out what they mean by "smallest" in terms of matrix theory. Commented Oct 4, 2021 at 16:26
• It's a known thing that if the optimality criterion is to minimize the MSE error, then the Kalman filter is the best linear filter, for measurement and process noise that are zero-mean but otherwise any probability distribution. With emphasis on the MSE and linear -- there are some PDFs (i.e., strongly bimodal) where the least-MSE error may be the worst possible estimate in real life (e.g., the optimal path if you're driving straight toward an obstacle is either the left or right -- if you split the difference you'll be sorry). Commented Oct 4, 2021 at 16:30
• In link, the optimality of the Kalman innovation gain is interpreted in the sense that it minimizes the trace of the a posteriori estimate covariance matrix Commented Oct 5, 2021 at 7:08