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In the nonlinear Bayesian Tracking problem, if we consider the noise exists only in the state equation : x[k] = f(x[k-1],v[k-1]) where vk-1 here is an iid process noise sequence

And we suppose that the measurement is directly and noise free observed, that means z[k] = h(x[k])

What is the likelihood function p( z[k] | x[k] ) in this case?

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A likelihood function is typically a function of a parameter. It is not clear what the parameter in your setting is.

However, likelihood functions in Bayesian models often corresponds with a conditional probability density functions, and so perhaps you are asking what the conditional density of $z_k$ given $x_k$ is. This would be the first step in preparing to use Bayes rule to calculate the conditional distribution of $x_k$ given $z_k$. Unfortunately, your problem will not have a conditional probability density if the set of values $H$ that $h(x_k)$ can take is uncountable.

This is because the conditional probability of $z_k$ given $x_k$ satisfies $\Pr[z_k|x_k]= \delta_{h(x_k)}$ a.s., and the set of measures $\{\delta_y\}_{y \in H}$ is not dominated when $H$ is uncountable, i.e., there is no $\sigma$-finite measure $\nu$ such that $\nu(A) = 0 \implies \delta_y(A) = 0$ for all measurable sets $A \subseteq H$ and points $y \in H$. If $h$ takes values in $\mathbb R$, and $z_k$ were to be corrupted by, say, independent additive Gaussian noise, then Lebesgue measure would dominate the family of measures underlying the conditional distribution, and so you would have a conditional density.

When the family of probability measures underlying a conditional distribution is not dominated, then you cannot even use Bayes rule. Of course, there may be a rather simple model for the conditional distribution of $x_k$ given $z_k$ which you could check satisfies the definition of a conditional distribution of $z_k$ given $x_k$. Namely, $x_k$ is distributed according to its marginal distribution, but restricted to the set $h^{-1}(z_k)$.

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The likelihood really isn't relevant without noise, but probably the right answer, in some sense, and assuming you mean h to be a known function, is that the likelihood is 1 if z[k] = h(x[k]) and 0 otherwise.

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