4
$\begingroup$

Question:

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\theta = {x'}^d$. $\endgroup$ – Yuri Nenakhov Apr 6 '15 at 18:34
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $x$ 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. $\endgroup$ – Antoine Bassoul Apr 9 '15 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.