# Approximating a Gaussian Process

Suppose that $\theta_t$ is an exogenous variable with known Gaussian process. Next, suppose that for any index $i\in [0,1]$, $$a_{i,t} = (1-\beta)\mathbb E[\theta_t|\mathcal I_{i,t}]+\beta \mathbb E[a_t|\mathcal I_{i,t}],$$ where $a_t=\int_0^1 a_{i,t} di$, and $$\mathcal I_{i,t} = \{\dots,\theta_{t-1}+\epsilon_{i,t-1},a_{t-2},\theta_{t}+\epsilon_{i,t},a_{t-1} \}.$$ Finally, for all $i$ and $t$, $\epsilon_{i,t} \sim \mathcal N(0,\sigma^2)$.

I am trying to find an algorithm to numerically approximate the stochastic process for $a_t$. I have tried many different ways of applying the Kalman filter to do this but none of them converge.

Is there an algorithm that is likely (or better yet guaranteed) to converge?

• Your notation seems a bit off (to me anyway), do you mean to say that the sum of the $\alpha$'s is expected to be 1? Also, is $i$ state and $t$ time? Why is $\alpha$ part of the $I$ set and why does it contain both $\theta$s and $\alpha$s? In other words, if you were to send $I$ to a function, would you use two different indexes (odd, even) to address the two things it seems to be holding? – A_A Feb 7 '18 at 12:59
• Perhaps the $\alpha$ looked too close to the $a$ so I changed it to $\beta$. You can think of $i$ as indexing an agent whose action in period $t$, $a_{i,t}$, is a weighted average of their best forecast about $\theta_t$ and the average action $a_t$. In their information set they have the past history of average actions $\{a_\tau\}_{\tau=-\infty}^{t-1}$ and noisy signals about the fundamentals, $\{\theta_\tau+\epsilon_{i,\tau}\}_{\tau=-\infty}^{t}$. – mzp Feb 7 '18 at 13:12
• Thanks for the clarification. Is the process time-invariant? – A_A Feb 7 '18 at 13:23
• Yes, sorry for not mentioning this earlier. – mzp Feb 7 '18 at 17:17
• Can I please ask if this was resolved? – A_A Feb 15 '18 at 11:22

You can use an ARMA model for $\alpha_t$. But, there is an equivalence between ARMA models and Kalman filters.

Therefore, what I am thinking is that maybe the formulation of the problem is a little bit off and instead of having to produce one estimate, you really have to estimate two. Here is why:

You present an $\alpha_{i,t} = (1-\beta) \mathbb{E}[\theta_t|\mathcal{I}_{i,t}] + \beta \mathbb{E}[\alpha_t|\mathcal{I}_{i,t}]$. What does this say? It says that $\alpha_{i,t}$ is a "mixture" of the average value of two processes:

1. The expectation of $\theta$ given the current $\mathcal{I}$
2. The expectation of $\alpha$ given the current $\mathcal{I}$

With $\beta$ controlling the mix.

It is then added (as a constraint (?)) that the value of $\alpha_t$ is known to be the sum of all $\alpha_i$ (here, two of them) at the current time instance. So, is $\mathbb{E}[a_t|\mathcal{I}_{i,t}]$ different than the constraint on $\alpha_t$ or the same thing? Because, both seem to determine $\alpha_t$ (?).

Then comes the definition of $\mathcal{I}_{i,t}$ and I think that this one should really be "split" into two different things. In its current form it is specified as $\mathcal I_{i,t} = \{\dots,\theta_{t-1}+\epsilon_{i,t-1},a_{t-2},\theta_{t}+\epsilon_{i,t},a_{t-1} \}$. Notice here that $\mathcal{I}$ contains both $\theta$ and $\alpha$ elements.

This is the bit that I don't get. Do $\theta, \alpha$ have the same units / dynamic range? Are they "similar"? If they are, why don't we call them all the same name / type? Is $\mathcal{I}$ meant to say that $\theta,\alpha$ are considered together for the same time instance? That's "fine" to a point but it would not work from the point of view of estimation. Imagine, for example, if you had to get a derivative of $\mathcal{I}$. It would work across the $t$s, but not across the $i$s because you would be differentiating different things, in different "references" ($\frac{\alpha - \theta}{(what?)}$)

Therefore, given what is presented in the question, my recommendation(s) would be that:

1. $\mathcal{I}$ is split into two parts, one for the $\theta$s and one of the $\alpha$s.

2. These are then plugged back into the estimation of $\alpha_{i,t}$ but essentially what we get here is two things that need to be estimated prior to obtaining the final value for $\alpha_t$.

3. The constraint (?) on $\alpha_t$ is clarified. If $\alpha_t$ is known to be this integral, then there is no point in also deriving the expectation of $\alpha_t$ from $\mathcal{I}$. It is either one or the other. Except if it is a constraint that means to say "whatever you work out your $\alpha$s to be, their sum will have to be unity"...but at the moment it does not look like this (?).

Finally, if you do have some vectors of the quantities you are trying to estimate, it would be good to provide one or two plots, just to see what you are dealing with and what this "non-convergence" looks like.

Hope this helps.