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I have:

  • a function of the form

    $S=M f(t, \theta)$

where $\theta$ is a variable, $t$ and $M$ are parameters.

  • two observations

    $(\theta_1, S_1)$ and $(\theta_2, S_2)$

in which $S_1$ and $S_2$ are observed values for $t$, and $\theta_1$ and $\theta_2$ are the observed values of $\theta$.

Furthermore, we know that my $S_1$, $S_2$ contains noise, which are normal distribution $N(0, d_1)$ $N(0,d_2)$ with unknown variance $d_1$, $d_2$.

Now I want to estimate $t$.

Mathematically, if $S_1$ and $S_2$ are observed precisely, we have

$$S_1= M f( t, \theta_1)$$ $$S_2= M f( t, \theta_2)$$

Then, by division we have $\frac{S_1}{S_2}=g(t,\theta_1,\theta_2)$ following which the parameter $t$ can be estimated by a Mathematica tool like Findroot w.r.t. the parameter $t$.

However, the problem here is the noise of normal distribution. The goal is to find a $t$ as precise as possible, in the sense of some probability-theoretic characteristics. The preferred way is to use a Matlab or Mathematica function, although I don't think they have a direct solution.

I am totally new to numeric analysis, and totally new to Matlab. My naive idea is that:

  1. Should the first step be removing the parameter $M$?
  2. If so, there are infinitely many ways to do that, among which we can list

    • $\frac{S_1}{S_2} = \frac{f(t, \theta_1)}{f(t,\theta_2)}$
    • $\frac{S_1+S_2}{S_1-S_2} = \frac{f(t, \theta_1)+f(t,\theta_2)}{f(t, \theta_1)-f(t,\theta_2)}$
    • $ \frac{S_1^2 }{S_2^2} = \frac{f(t, \theta_1) ^2 }{f(t,\theta_2) ^2}$

It is clear that all these variations are equivalent mathematically if $S_1$, $S_2$ are observed precisely. However, they should yield different $t$ (because of the Gaussian noise in $S_1$ and $S_2$)

My questions are

  1. Do you think my first step should be to remove $M$ using the foregoing approach?
  2. Is there some kind of combination of $S_1$, $S_2$ that gives better results than others?
  3. What would be the best way to estimate $t$ of this problem? (Maybe in Matlab, or Mathematica)

Thank you. I hope I was clear.

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    $\begingroup$ I don't want to edit your question but it is a little confusing. since you are talking to statisticians. You should say that S1 and S2 are observed values for t and theta1 and theta2 are the observed values of theta. You misuse the term precise. To a statistician precise means small variability where I am pretty sure you mean to say that S1 and S2 are observed exactly (without any noise). You also should not say that after division t and be calculated rather you should say it can be estimated. $\endgroup$ Commented May 12, 2012 at 17:19
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    $\begingroup$ Also English can be a confusing language with plurals, the plural of dog is dogs but the plural of noise is noise not noises. $\endgroup$ Commented May 12, 2012 at 17:20
  • $\begingroup$ A mistake in my comment that I am too late to edit was saying that S1 and S2 were sample values of t. $\endgroup$ Commented May 12, 2012 at 18:10

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To answer your question as I understand it, although all your approaches are mathematical ways to get a value for t when an exact formula is given. The answer will not be exact only because of the use of numerical methods to solve. But in the presence of noise all your approaches could have problems depending on the magnitude of the noise. I assume that the divisions are done to eliminate M from the equation so that you have reduced it to a matter of getting t and then since you can evaluate f(t,theta1) and then given that and S you can solve for M. But in the presence of noise things work differently because say the noise is additive. Then S1= M* f( t, theta1)+e1 S2= M* f( t, theta2) +e2 and S1/S2 = [M* f( t, theta1)+e1]/[M* f( t, theta2)+e2] Because of the noise you really have not eliminated M in the equation. What could be worse is that the error could be large and in opposite sign to M * f and hence the denominator can be zero or near it. The accuracy of an estimate depends on what function of the random variables it is. If the noise is very small probably all the approaches would work with some more than others.

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