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

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.
