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First, we have a state space model with mean reversion and $\mu$ is unknown

$y(t )= F* x_t +e_t$
$x_t- \mu = G* (x_{t-1}-\mu) +n_t$

There is a option to add unknown parameters to the state vector and pass them on with zero error to estimate them.

The state equation can be writen as

$$\pmatrix{x_t \\ \mu_t} = \pmatrix{G & I-G \\ 0 & I}\pmatrix{x_{t-1} \\\mu_{t-1}} + \pmatrix{n_t \\ 0}.$$

After that I can run the Kalman filter and get an estimate for $\mu$. The question is, which estimate is the one that would be equivalent to a maximum likelihood estimate in advance.

Is it the estimate that would result in the last time step?

Or to put it another way: $\mu_t$ changes over time. Which $\mu_t$ then represents the estimate, so to speak, if, for example, the estimate of $\mu$ is asked for? Is it the estimate from the last time step in the Kalman filter?

Would be very grateful for help.

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  • $\begingroup$ There isn't a $\mu_t$ in your state space model, so why is it in your state equation? I suspect that $\mu_t$ should be in both, in which case the question is moot. Note, however, that in your state equation $\mu_t = \mu_{t-1}$, which may mean it should be in neither, i.e., $\mu_t$ is a notational error and $\mu$ doesn't evolve over time. $\endgroup$
    – jbowman
    Aug 28 at 20:55
  • $\begingroup$ No time index is given above, as this signals that this is a constant value. Below a time index is given, but since the variance is 0, this is to express that the following value should be equal to the previous one. However, through the Kalman filter, the value still changes through the Kalman gain and converges to the ML estimator. The question is, which value is that exactly? The last filtered value? $\endgroup$
    – Marv91
    Aug 30 at 20:34
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    $\begingroup$ Does this Q&A help? stats.stackexchange.com/questions/511618/… $\endgroup$ Aug 30 at 21:41
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Suppose that the model does not involve any parameter other than $\boldsymbol{\mu}$. You can regard $\mu$ as a random quantity in a Bayesian approach. Then, if the prior on $\boldsymbol{\mu}$ is Gaussian, the posterior is also Gaussian and is given by the last step of the Kalman Filter. The filtered mean $\widehat{\boldsymbol{\mu}}_{T\vert T}$ for the last time $T$ gives the posterior mean which also is the posterior mode, and the posterior variance is the filtered variance. Moreover, you can use a diffuse (improper) prior for $\boldsymbol{\mu}$; provided that $\boldsymbol{\mu}$ can be identified, the posterior will be proper, and the mode will be the value of $\mu$ that maximises the likelihood. Note that since $\boldsymbol{\mu}$ does not depend on $t$, a better notation would be something like $\widehat{\boldsymbol{\mu}}_{\bullet \vert T}$.

Things are more complex in the case where the model involves more parameters forming a vector $\boldsymbol{\theta}$.

  • The Bayesian approach would then lead to regard $\boldsymbol{\theta}$ as random as well. With mild conditions the full conditional $p(\boldsymbol{\mu} \vert \mathcal{Y}, \, \boldsymbol{\theta})$ will be Gaussian where $\mathcal{Y}$ is the whole series. This can be of some help.

  • In a frequentist approach, you can maximise the profile-likelihood that depends only of $\boldsymbol{\theta}$. Indeed, the value of $\boldsymbol{\mu}$ which maximises the likelihood for a given $\boldsymbol{\theta}$ is given by the Kalman filter as above. Remind however that you have to use the diffuse prior.

In the second case, the estimate of $\boldsymbol{\mu}$ is provided by the Kalman filter at a very small cost.

Note that depending on the dimensions of the system, it can be better to use an alternative state vector $\mathbf{x}_t^\star := \mathbf{x}_t - \boldsymbol{\mu}$ along with $\boldsymbol{\nu} := \mathbf{F} \boldsymbol{\mu}$ and the state space model \begin{align*} \mathbf{x}_t^\star &= \mathbf{G} \mathbf{x}^\star_{t-1} + \mathbf{n}_t\\ \mathbf{y}_t &= \mathbf{F} \mathbf{x}^\star_ t + \boldsymbol{\nu} + \mathbf{e}_t. \end{align*} The augmented state form would then take $\boldsymbol{\nu}$ as augmented part with a block diagonal transition matrix. The motivation is that $\boldsymbol{\nu}$ can be of smaller length than $\boldsymbol{\mu}$, especially when the observation is a scalar $y_t$. Note also that as in in this question linked in the comment by @Cam.Davidson.Pilon, we can cope similarly with the case where $\boldsymbol{\nu}$ is replaced by $\mathbf{z}_t^\top \boldsymbol{\beta}$ for a vector $\mathbf{z}_t$ of known covariates. Then we would put the vector $\boldsymbol{\beta}$ of unknown regression coefficients in the augmented state.

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  • $\begingroup$ Sorry for the late reply. But thank you for the extremely comprehensive response! $\endgroup$
    – Marv91
    Sep 11 at 8:22
  • $\begingroup$ Mind that in the case when the dimension of the state is >1 your representation lacks the controllability property, so you may encounter serious numerical problems. You should use the alternative representation of my answer. $\endgroup$
    – Yves
    Sep 13 at 7:41

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