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I have a discrete time stochastic process, where at each time the state of the system $X_t$ is given by: $$ X_t = f_\theta(X_{t-1},\epsilon_t), \; \; \text{for} \; t = 1,\dots,T $$ and, for example, the perturbations $\epsilon_t$ are i.i.d. $N(0,\sigma_e)$. So the state of the system at each time is a function (parametrized by $\theta$) of the state at the previous time step $X_{t-1}$ and of the noise $\epsilon_t$.

Suppose that I have observed $X^o_{1:T}$ and that I want to do inference about $\theta$ (by $X^o_{1:T}$ in mean $X^o_1,\dots,X^o_T$). In particular if, instead of observing $X^o_{1:T}$ directly, I was observing $Y_{1:T}$ given by: $$ Y_{t} = X^o_t + Z_t\; \; \text{for} \; t = 1,\dots,T. $$ where the observational noise $z_t$ are i.i.d. from some kernel density $K_{\sigma_o}$ (for example $Z_t \sim N(0,\sigma_o)$), then I would be dealing with a state space model with observations $Y_{1:T}$ and hidden states $X^o_{1:T}$. This means that I could use all the methodologies that have been developed to do inference for state space models (such as particle filters for example).

But the observational noise $Z_{1:T}$ is simply not there in my case, so there are two ways to go:

1) Fit the data as if the observational noise $Z_{1:T}$ was there (that is as if I was observing $Y_{1:T}$, while I'm really observing $X^o_{1:T}$).

2) Add to the the states $X^o_{1:T}$ randomly generated noise $Z_{1:T}$, to obtain the observations $Y_{1:T}$. Then $Y_{1:T}$ has been really generated according to a (artificial) state space model.

Let's consider the first approach. In this case I'm fitting the wrong model, because I'm fitting a model that includes observational noise $Z_{1:T} \sim K_{\sigma_o}$ which is not present in reality. In particular, by treating the model as a state space model, we obtain the likelihood:

$$ \hat{p}_1(X^o_{1:T}|\theta) = \int p(X^o_{1:T},X_{1:T}|\theta) d X_{1:T} = \int K_{\sigma_o}(X^o_{1:T}-X_{1:T})p(X_{1:T}|\theta) d X_{1:T} $$

so $\hat{p}_1(X^o_{1:T}|\theta) = (K_{\sigma_o} \star p(\cdot |\theta))(X^o_{1:T}) \neq p(X^o_{1:T}|\theta)$. The likelihood that we obtain is the convolution of kernel $K_{\sigma_o}()$ with $ p(\cdot |\theta)$ and it is different from the true likelihood $p(X^o_{1:T}|\theta)$. So the resulting ML estimates for $\theta$ are likely to be biased.

The second approach looks more promising. If I add some noise $Z_{1:T}$, distributed according to $K_{\sigma_o}$, to the data then I'm justified to fit state space model. This is because if I define $Y_{1:T} = X^o_{1:T} + Z_{1:T}$, then $X^o_{1:T}$ is really a hidden state. So I would expect the resulting estimates for $\theta$ to be unbiased, even though some efficiency will be lost because I'm adding noise (and thus losing information).

Lets call $\hat{p}_2(X^o_{1:T}|\theta)$ the likelihood given by this procedure. I'm interested in knowing what is it's expected value, conditionally on the observed $X^o_{1:T}$. The expected likelihood over all the possible simulated vectors $Z_{1:T}$, is given by:

$$ E(\hat{p}_2(X^o_{1:T}|\theta)|X^o_{1:T}) = \int \int K_{\sigma_o}(Y_{1:T}-X_{1:T})p(X_{1:T}|\theta) d X_{1:T} K_{\sigma_o}(Z_{1:T}) dZ_{1:T} $$ $$ = \int \bigg ( \int K_{\sigma_o}(Z_{1:T} + X^o_{1:T}-X_{1:T})\, K_{\sigma_o}(Z_{1:T})\, dZ_{1:T} \bigg ) \, p(X_{1:T}|\theta)\, d X_{1:T} $$ $$ = \int K_{\sigma_o} \star K_{\sigma_o}( X^o_{1:T}-X_{1:T}) p(X_{1:T}|\theta)\, d X_{1:T}. $$ $$ = (K_{\sigma_o} \star K_{\sigma_o}) \star \, p(X_{1:T}|\theta)( X^o_{1:T}). $$

So we have the same problem as with the first approach! So if we add noise (generated according to $K_{\sigma_o}$) to the data $X^o_{1:T}$ and we fit $Y_{1:T}$ as a space state model the expected likelihood that we get out is the same that we get I we fit the data without adding noise, using the kernel $K_{\sigma_o} \star K_{\sigma_o}$. This result has to be wrong because we are fitting the right model if we add the noise $Z_{1:T}$!

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  • $\begingroup$ From what I see, your data is generated by a Markov chain. Why don't you use all the work that has been done on this? Fitting your data to a model seems unnatural if the right model exists and is well understood. Markov chains are much easier to do inference on, since you don't have any hidden variables. $\endgroup$ – bayerj Nov 15 '12 at 20:11
  • $\begingroup$ You are right, but my actual problem is more complicated that the one stated here, so it is not straightforward fitting it as a Markov Chain. When I considered the above approach I got the strange results previously described, so I was curious to know how to treat this as a state space model. $\endgroup$ – Gollum Nov 16 '12 at 11:53

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