Maximum likelihood estimator of Cauchy distribution but with a catch I have an exercise to solve that states that we need to find the Maximum likelihood estimator of location parameter of the Cauchy distribution given a set X={x1,x2} , |x1-x2|<2
Now I worked using this approach: Maximum likelihood estimator of location parameter of Cauchy distribution
but I can't get to grasp how " given a set X={x1,x2} , |x1-x2|<2" changes the problem.
Thank you
 A: When you have only two samples $x_1,x_2$ then the likelihood is maximized by finding the maximum of
$$\mathcal{L}(\lambda ; x_1,x_2) \propto\frac{1}{(\gamma^{2}+(x_1-\lambda)^2)}\frac{1}{(\gamma^{2}+(x_2-\lambda)^2)}$$
Which is equivalent to finding the minimum of the polynomial
$$(\gamma^{2}+(x_1-\lambda)^2)(\gamma^{2}+(x_2-\lambda)^2))$$
We can rephrase this in terms of $\bar{x} = \frac{x_1+x_2}{2}$ and $d= \frac{x_1-x_2}{2}$
$$(\gamma^{2}+(\bar{x}-d-\lambda)^2)(\gamma^{2}+(\bar{x}+d-\lambda)^2)$$
without loss of generality we can set $\bar{x}=0$ (it will just shift the solution if $\bar{x}\neq0$)
$$(\gamma^{2}+(d-\lambda)^2)(\gamma^{2}+(d-\lambda)^2) =
\gamma^4 + 2 \gamma^2 (d^2+\lambda^2) + (d^2+\lambda^2)^2 - 4 d^2\lambda^2$$
Whose derivative to $\lambda$ is equal to $0$ in the minimum
$$4(\gamma^2 - d^2+\lambda^2) \lambda = 0$$
In the case of $d^2 < \gamma^2$ then this is equal to zero in only a single point $\lambda = 0$ (or in the general case in $\lambda = \bar{x}$).
Your problem probably uses $\gamma = 1$ such that we have as condition $|d| < 1$ or equivalent $|x_1-x_2| < 2$. For this condition the likelihood is minimised in $\lambda = \frac{x_1+x_2}{2}$. For other cases you have multiple minima.

but I can't get to grasp how " given a set X={x1,x2} , |x1-x2|<2" changes the problem.

So what changes with the condition $|x_1-x_2| < 2$ is that the likelihood function has a single minimum.
A: From symmetry principle location can only be $\frac{x_1+x_2} 2$, when there is a unique solution.
this is regardless of the estimation approach. For instance, a particular estimator, such as MLE, may not even exist in every case. However, if it does exist and it is unique, then the symmetry dictates what should it be.
non MLE estimation
If we don't constrain ourselves with MLE, then we could get the parameters differently. Obviously, the peak (center) is still the same half-way between $x_1$ and $x_2$. So, we only need to figure the other parameter.
In physics we used to characterize Breit-Wigner function with half-width at half-height parameter, which is equal to a shape parameter $\gamma$ of Cauchy distribution definition in statistics: PDF reaches its half height when you step left or right by half-width from the peak. So, naturally, the "width" is $2\gamma$ and intuitively it should be estimated as $2\hat\gamma=|x_1-x_2|$.
The "height" of PDF is at its peak $x_0$:$$H=\frac 1 {\pi\gamma}$$ Solve the PDF for $x$ to equal $H/2$: $$\frac H 2=\frac 1 {2\pi\gamma}=\frac 1 {\pi\gamma\left[1+\left(\frac{x-x_0} \gamma\right)^2\right]}$$ to get $$\frac{x-x_0}\gamma=1$$ So when the variable is $x_0\pm\gamma$ the PDF reaches its half-height
A: To come up with an expression for the conditional likelihood, you express the joint likelihood $L(x_1, x_2)$ as $L(x_2|x_1)L(x_1)$. The unconstrained $x_1$ can be anything of course, so has the usual likelihood
$$f(x_1) = \frac{1}{\pi (1+(x_1-\mu)^2)}$$.
The conditional $x_2$ density can be expressed by assigning likelihood 0 to any value more than 2 units beyond $x_1$, then the density can be "normalized" by the density area within the 2 units, i.e.
$$ f(x_2|x_1) =  \dfrac{f(x_2)}{F(x_1+2) - F(x_1 - 2)} \times \mathcal{I}(|x_1-x_2|<2)$$
I am not sure if the expression $L(x_1, x_2) = f(x_2|x_1) f(x_1)$ has an analytic solution or if it's really necessary to obtain one. But numeric solvers should give you a solution here, and some simulations should show you if the estimator has any nice asymptotic properties.
