Linear Gaussian State Space Models (LGSSMs) are generally expressed in the following way:

\begin{align} {\bf x}_t &= A {\bf x}_{t-1} + {\bf w}_t, &{\bf w}_t \sim \mathcal N ( 0, Q)\\ {\bf y}_t &= B {\bf x}_t + {\bf v}_t, &{\bf v}_t \sim \mathcal N ( 0, R) \end{align} where the states are denoted by ${\bf x}_t$ and the observations are denoted by ${\bf y}_t$.

I am working on a Dynamic Factor Model (DFM) where the observation errors form a Gaussian vector autoregressive processes of order 1: \begin{align} f_t &= \Phi_f f_{t-1} + \epsilon_t, &\epsilon_t \sim \mathcal N (0, \Sigma_f) \\ y_t &= \Lambda f_t + e_t \\ e_t &= \Phi_y e_{t-1} + \varepsilon_t, &\varepsilon_t \sim \mathcal N (0, \Sigma_y), \end{align} where the latent factors are denoted by $f_t$ and the observations are denoted by $y_t$.

While studying AR models I can across A unified view of linear AR(1) models by Grunwald, Hyndman and Tedesco. There it says that "a useful feature of Gaussian AR(1) processes is that the marginal distribution is also normal."

I imagine that the statement holds for vector autoregressive processes as well. So given that $e_0 \sim \mathcal N (0, (\text{Id} - \Phi_y^2)^{-1}\Sigma_y)$ according to the linked paper, I can say that $$e_t \sim \mathcal N(0, (\text{Id} - \Phi_y^2)^{-1}\Sigma_y),$$ or in other words the observation errors are marginally Gaussian.

Question: Does this mean that the DFM above is essentially a LGSSM?

EDIT: I failed to mention that for the DFM, $\Phi_y$ and $\Sigma_y$ are assumed to be diagonal matrices.

  • 1
    $\begingroup$ Yes, it is indeed a linear and normal state-space model. That is how you estimate it by Bayesian methods or using the EM algorithm. Your expression for the variance is not correct, you need to mind the fact that you are now dealing with matrices. $\endgroup$ – hejseb Jan 28 at 21:11
  • $\begingroup$ Okay, thanks for the clarification. $\endgroup$ – SOULed_Outt Jan 28 at 23:33
  • $\begingroup$ Regarding the matrices, I failed to mention something very important... $\Phi_y$ and $\Sigma_y$ are diagonal. What I posted is not true in general. $\endgroup$ – SOULed_Outt Jan 28 at 23:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.