# Are Kalman Filter recursions valid when the state noise has a singular covariance matrix?

Consider a Linear Gaussian State-Space Model where the states are denoted by $$X_t$$ and observations are denoted by $$Y_t$$: \begin{align} X_t &= A X_{t-1} + \epsilon_t, &&\epsilon_t \sim \mathcal{N}(0, Q) \\ Y_t &= B X_t + \nu_t, &&\nu_t \sim \mathcal{N}(0, R) \end{align}

If we let $$\mu_{t|1:t} \equiv \text{E}(X_t | Y_{1:t})$$ and $$P_{t|1:t} \equiv \text{Var}(X_t | Y_{1:t})$$ be the mean and variance of the filtering density at time $$t$$, respectively, then the Kalman Filter recursions can be expressed as

\begin{align} \mu_{t|1:t} &= \mu_{t | 1:t-1} + P_{t|1:t-1} B^\top (B P_{t|1:t-1} B^\top + R)^{-1} (y_t - B \mu_{t | 1:t-1}) \tag{1} \\ P_{t|1:t} &= P_{t|1:t-1} - P_{t|1:t-1}B^\top (B P_{t|1:t-1} B^\top + R)^{-1} B P_{t|1:t-1} \tag{2} , \end{align}

where

\begin{align} \mu_{t | 1:t-1} &= A \mu_{t-1 | 1:t-1} \\ P_{t|1:t-1} &= A P_{t-1 | 1:t-1} A^\top + Q. \end{align}

You can see a derivation of these recursion (with a different naming convention being used) in Lecture 3: Bayesian Optimal Filtering Equations and Kalman Filter by Simo Särkkä

If $$Q$$ is a singular matrix, then $$Q^{-1}$$ does not exist. Furthermore, the density $$p(\epsilon_t)$$ does not exist, which I believe implies that the density $$p(x_t | x_{t-1})$$ does not exist. From the lecture notes you will be able to see how this density is needed to obtain the mean and variance of the filtering density.

Are the Kalman Filter recursions still valid when $$Q$$ is singular? In other words, given that $$Q$$ is singular, are equations $$(1)$$ and $$(2)$$ the correct representations of the mean and variance of the filtering density?

Please understand that my question is not about whether I can code the recursions and run them successfully. I am asking this from a probabilistic viewpoint.

Yes, many common models feature nonsingular state noise. For example, consider the case of an AR(2) with observational noise. You could set $$X_t = (x_t, x_{t-1})^\intercal$$ and make your state equation equal to $$\mathbf{x}_t \overset{\text{def}}{=} \begin{bmatrix} x_t \\ x_{t-1} \end{bmatrix} = \begin{bmatrix} \phi_1 & \phi_2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x_{t-1} \\ x_{t-2} \end{bmatrix} + \begin{bmatrix} \epsilon_t\\ 0 \end{bmatrix}.$$

In this case $$Q$$ is $$2 \times 2$$ and has all elements as zero, except for the one in the top left corner.

Whether or not the Kalman filter code will run depends on the invertibility of the one matrix you have to invert. You do not invert the state covariance matrix; rather you invert something has $$Q$$ added to it. This matrix will usually be invertible because it's positive-definite in most cases.

Regarding your question about densities, you are correct that $$p(\mathbf{x}_t \mid \mathbf{x}_{t-1})$$ cannot be a density, but $$p(x_t \mid x_{t-1}, x_{t-2})$$ is.

Regarding your edit, it would probably be better to ask a separate question instead of changing the nature of an existing one so drastically. If densities do not exist, then you would work with measures $$(P)$$ and kernels ($$K$$). You could replace, for example, this

\begin{align} p(x_t | y_{1:t-1}) &= \int p(x_t, x_{t-1} | y_{1:t-1}) \; dx_{t-1} \tag{1} \\ &= \int p(x_t | x_{t-1}) p(x_{t-1} | y_{1:t-1}) \; dx_{t-1} \tag{2} \end{align}

with this:

\begin{align*} P(\text{d}x_t | y_{1:t-1}) &= \int P( \text{d}x_t, \text{d} x_{t-1} | y_{1:t-1}) \\ &= \int K( \text{d}x_t | x_{t-1}) P( \text{d} x_{t-1} | y_{1:t-1}) . \end{align*} Note that I'm writing the kernels in a right-to-left way, instead of in the customary, left-to-right way. Following along the AR(2) model, our transition kernel would be written as $$K(\text{d} x_t \mid x_{t-1}) = \delta_{x_{t-1}^1}(\text{d}x_t^2 ) p(x_t^1 ; \mu, \sigma^2) \text{d}x_t^1,$$ where $$p(x_t^1 ; \mu, \sigma^2)$$ is a normal density, and $$\delta$$ is the Dirac function that evaluates to either $$0$$ or $$1$$, depending on whether $$x_{t-1}^1 \in \text{d}x_t^2$$.

• I would try to figure out why $Q_t$ is not invertible. The solution provided by Taylor is interesting but a different problem is being solved using his solution procedure. In the case of an AR(2), ( where the noise is a scalar ) his solution works. But, in the general case where the observational equation is not AR and is truly multivariate, a different problem is being solved. – mlofton Jul 13 '19 at 16:57
• @mlofton I’m not sure what you mean by “why,” but yes, this small example has only proved it’s not a problem always. You would like a set of criteria that would guarantee it’s never a problem. Point taken. – Taylor Jul 13 '19 at 17:21
• @Taylor: I understand that the recursions may not directly take the inverse of $Q$. I guess the wording of my question is poor, so my apologies. I am wondering whether the recursions are valid because the density $p(x_t | x_{t-1})$ is necessary, but in the case where $Q$ is singular it doesn't exist. See the edit in the OP. – user235125 Jul 13 '19 at 19:39
• @Taylor: I am not familiar with measures and transition kernels, but do we know that the recursions would look the same by using them? – user235125 Jul 13 '19 at 20:42
• @mlofton: Yes, in that case we can still say that $X_t \sim \mathcal{N}(0, Q_t)$ but $X_t$ does not have a probability density function. Some would call it a degenerate Normal distribution. – user235125 Jul 17 '19 at 3:32