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Following on from my previous question, kindly answered by @peter-k, I have a related one...

I have, in a small interval $a \leq x \leq b$, a function $q(x) := q(x;\rho,\gamma)$ whose exact form is unknown. However, a Taylor expansion around a point $x=\rho_0$ shows that, to the first order, we have $$ q(x;\rho_0,\gamma) \approx c(x;\rho_0,\gamma) = \frac{1}{\pi}\left[ \frac{\gamma}{(x-\rho_0)^2 + \gamma^2} \right], $$ where in general, $c(x) := c(x;\rho,\gamma)$ is the density function of the Cauchy probability distribution with location $\rho$ and scale $\gamma$. This is illustrated in the figure. enter image description here

From prior analysis we also have that $$ \arg\max_\rho q(x;\rho,\gamma) = \arg\max_\rho c(x;\rho,\gamma) = \rho_0 $$ and $\gamma > 0$ is a (small) known constant.

My question therefore is what method(s) are available to 'fit' $c(x)$ to $q(x)$ in the interval $[a,b]$ so as to discover the value of $\rho_0$ and hence the value of $x \in \mathbb{R}$ at which $q(x)$ attains its maximum?

I have been experimenting with the discretised convolution of $q(x)$ with $c(x)$ but so far have not achieved the desired result. If this is indeed an applicable approach, any comments or guidance would be much appreciated.

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I think a least square error fit should do the trick. Assuming $c$ is discrete at the values $x_i$ you can define your error as:

$$E = \sum \left[ \frac{1}{\pi} \frac{\gamma}{(x_i-\rho)^2 + \gamma^2} - c_i\right]^2 $$

and you can get $\rho_0$ by minimizing the error, i.e. solving for

$$\frac{\partial E}{\partial \rho} = 0$$

That's not going to be a pretty equation but since it's only one variable and your function is well behaved with a single smooth minimum, a simple binary search may do the trick. Since you can calculate the partial derivatives you can also do steepest descent or maybe conjugate gradient search.

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  • $\begingroup$ Thanks for this ... yes, we can discretise $c$ at points $x_i$. I actually started my approach of attacking this problem with least-squares and you're right the equation is pretty nasty! $\mathcal{O}(\rho^5)$ in the denominator (without looking back at my notes). But as you say the function is well-behaved so it is an option and I'm glad to have found your support in the approach. I'd like to leave the question open for a while to see what other ideas may be suggested, but otherwise I will persist with this one. $\endgroup$
    – RickyBoy
    Commented Jul 22, 2022 at 14:42

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