How to estimate for an exponential in WGN

Question

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.

Answer 1

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.

Answer 2

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.