Formulating a Maximum likelihood estimator:

So, the likelihood will be

$p(y;\mathbf{h}) = \frac{1}{{(2 \pi \sigma_\eta^2)}^{T/2}} \exp{(-(y - y_0(t))^2)/ 2\sigma_\eta^2}$.

Then, I need to differentiate w.r.t the unknowns and equate to zero.

This is a nonlinear equation in $x$ and cannot be solved directly. Newton-Raphson is a method but it works only for very close initial guess to $x$ and $d$. So, Was thinking how to apply Expectation - Maximization.

I don't know if my approach is correct or not.


Maximum $d$ you may get is a total number $N$ of $y_n$'s you have - 1. Larger $d$'s yield curvatures that can't be represented by N points.

Solve a matrix equation $A\vec{t}=\vec{y}$ where $\vec{y}$ is a measurements vector (size $N \times 1$) and $A$ is a normalized version of matrix $X$ (size $N \times N$), where each column $A_j$ is calculated as:

$A_{ij} = \frac {X_{ij}} {\sum \limits_j (X_{ij})^2}$

$X = \left [ \begin{matrix} x_0^0 & \cdots &x_0^{N-1} \\ \vdots & \ddots & \vdots \\ x_{N-1}^0 & \cdots & x_{N-1}^{N-1} \end{matrix} \right ]$

d you are interested in is a position of largest coefficient in a vector $\vec{t}$ (size $N \times 1$ starting from zero). Your approximation would be $y = t_d \cdot x^d$.

You can achieve better results if your solver finds solution optimal in $L^1$ sense.

This is a sort of exhaustive search, yes.

  • $\begingroup$ You can assume $x_n = 0, 1,\ldots N-1$. After you find the approximation $y_n = t_d \cdot x_n^d$, you may find $x_n' = \frac{x_n}{t_d}$. This will give you $\theta = {x'}^d$. $\endgroup$ Apr 6 '15 at 18:34

In your observation model it is not clear what does $x$ represant, how is it related to time ?

You could look at your observation model on a logarithmic scale.

$$ \log(y(t))=d\log(x) $$

If you havn't any a priori on $x$ you cannot estimate $d$ and $x$ because there for any $d$ there is always a $x$ time series that would suit your observation. Thus the system is inobservable and estimation isn't possible.

  • $\begingroup$ $x$ is the sum of distances, distances between what ? So you have distances between "stuff". Let's say at each time step you have $K$ distances, thoses distance are all independant and each one follow a given pdf. Then $x$ is, at each time step, the sum of these distance ? Your problem's statment isn't clear AT ALL. It looks like you're missing the main part of the problem, please be more precise. $\endgroup$ Apr 9 '15 at 18:11

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