0
$\begingroup$

The question is the following. Say I have observations of a Gaussian stochastic process ($\{x_i\}_{i=1}^n$) for which is convenient to use the state space formalism (and Kalman recursions) to describe it. Let its log-likelyhood function be $f(x;\theta)$, which depends on some parameter(s) $\theta$. In order calculate the log-likelyhood, we use the Kalman recursions.

Let $\hat{\theta}$ be the maximum likelyhood estimator of $\theta$. We use the Kalman prediction and $\hat{\theta}$ to get the prediction $\hat{x}_{n+1}$.

Le us consider now the following procedure. Let us assume $x_{n+1}$ and consider it as a parameter together with $\theta$. Then, we calculate the Kalman recursions but taking $x_{n+1}$ as given, that is, instead of using the innovations we use the smoothing solutions to compare with $x_1, x_2,...x_n$. Then we maximize likelyhood respect to $\theta$ and $x_{n+1}$, and we take the latter as the estimators of $\theta$ and $x_{n+1}$, respectively.

  1. Is this a sensible procedure to calculate a likelyhood function with $x_{n+1}$ as a parameter?
  2. In case it is, are these two ways of estimating $x_{n+1}$ and $\theta$ the same?
$\endgroup$
  • $\begingroup$ After studying a bit more, it seems that under gaussian hypotheses the Kalman prediction actually gives you the conditional expectation of $x_{n+1}$ given the previous data. This is the best possible estimator (in min-squared sense) based on functions of the previous variables. Therefore, maximizing the likelyhood of the observed data $\{x_i\}_{i=1}^n$ given $x_{n+1}$ - which would be obtaining the MLE of $x_{n+1}$ - is necessarily worse or equally good as the Kalman prediction. $\endgroup$ – Enredanrestos Aug 7 '18 at 5:20
1
$\begingroup$

This is not a complete answer, but a short example which shows that: (i) the two approaches are not equivalent, (ii) the Kalman prediction gives a better solution. Let $x_1$ and $x_2$ observations a zero-centered, mean reverting Gaussian process. Say, $$ x_1\sim N(0,\sigma^2)$$ $$ x_{n+1}|x_{n} \sim N(a x_n,(1-a^2)\sigma^2),$$ with $0<a<1$ and $\sigma>0$ given. Then, the Kalman prediction of the second term based on one observation is $x_{2|1}=ax_1$. The PDF of $(x_1,x_2)$ is $$ f(x_1,x_2)=\frac{1}{2\pi\sigma^2\sqrt{1-a^2}}\exp\left(-\frac{1}{2}\left(\frac{x_1^2}{\sigma^2}+\frac{(x_2-ax_1)^2}{(1-a^2)\sigma^2}\right)\right)~~.$$ This way of writing the density function hides a bit the fact that it is symmetrical in $x_1$ and $x_2$. In any case, it is not difficult to prove that \begin{align} \ln f(x_2|x_1)+\text{ct.}&\propto -(x_2-a x_1)^2\\ \ln f(x_1|x_2)+\text{ct.}&\propto -(x_1-a x_2)^2~~.\tag{1}\\ \end{align} Therefore $\mathbb{E}(x_2|x_1)=a x_1$ is the best estimator of $x_2$ based on $x_1$, and it corresponds to the Kalman prediction (side note: $\mathbb{E}(x_1|x_2)=a x_2$, which is rather surprising. I think that it has to do with this stochastic process being reversible).

However, as derived from $(1)$, treating $x_2$ as a parameter of the likelyhood function based on a single observation ($x_1$) and maximizing it, gives $\hat{x}_2=x_1/a$, which is certainly worse and unintuitive compared with the Kalman result. What is the main maximum-likelyhood hypothesis which is failing here? The MLE does not seem to get any better with more terms in the series.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.