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Why fit a deterministic process model?

Benchmarking

The likelihood for a deterministic latent process

Suppose that the latent process is deterministic. In our POMP notation, this lets us write the latent process as \[X_{n}=x_n(\theta),\] so that the latent process is a known and non-random function of \(\theta\) for each \(n\). What is the likelihood?

Since the probability of the observation, \(Y_n\), depends only on \(X_n\) and \(\theta\), and since, in particular \(Y_{m}\) and \(Y_{n}\) are independent given \(X_{m}\) and \(X_{n}\), we have \[{\mathcal{L}}(\theta) = \prod_{n} f_{Y_n|X_n}\big(y_n^*;x_n(\theta),\theta\big)\] or \[\ell(\theta) = \log{\mathcal{L}}(\theta) = \sum_{n} \log f_{Y_n|X_n}\big(y_n^*;x_n(\theta),\theta\big).\] The following diagram illustrates this.

In this diagram, \(\hat y_n\) refers to the model prediction, \(\hat y_n = {\mathbb{E}\left[{Y_n \vert X_n=x_n(\theta)}\right]}\), and \(y_n^*\) is data.


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