# Maximum likelihood estimator of Cauchy distribution

The following exercise is from Bickel and Doksum, volume one.

Let $$g(x) = 1/[\pi (1+x^2)]$$, $$x \in \mathbb{R}$$, be the Cauchy density, let $$X_1$$ and $$X_2$$ be i.i.d. with density $$g(x-\theta)$$, $$\theta \in \mathbb{R}$$. Let $$x_1$$ and $$x_2$$ be observations and set $$\Delta = \frac{1}{2} (x_1 - x_2)$$. Let $$\hat \theta = \arg \max L_X (\theta)$$ be "the" MLE. Show that if $$|\Delta| \leq 1$$, then the MLE exists and is unique. Give the MLE when $$|\Delta| \leq 1$$.

The likelihood function is

$$L(\theta) = \frac{1}{\pi^2 \cdot \prod_{i = 1}^{2} (1 + (x_i - \theta)^2) }$$ I tried taking the log-likelihood and differentiating it w.r.t $$\theta$$, which resulted in $$\hat \theta = \overline{X}$$. I do not understand, however, why we need $$|\Delta| \leq 1$$ for uniqueness of the maximum likelihood estimator, or how to establish uniqueness of the MLE.

Check the derivative of the log-likelihood. It's true that the derivative $$\ell'(\theta)$$ equals zero at $$\hat\theta:=\frac{x_1+x_2}2$$. Compute the second derivative to establish that $$\hat\theta$$ leads to a local maximum.
However, you'll find that setting the derivative of the log-likelihood to zero yields additional solutions for $$\theta$$ exactly when $$|\Delta|>1$$. In that case it's not obvious which of these solutions is in fact the maximizer.
ADDED: For $$\ell'(\theta)$$ I get a numerator of $$[1+(\theta-x_1)(\theta-x_2)](\theta-x_1 + \theta-x_2).$$ The quantity in square brackets on the left is a quadratic in $$\theta$$. Under what conditions does this quadratic, when set to zero, yield a solution? (Hint: check the discriminant $$b^2-4ac$$)
• Thank you so much. I have one last question : when $b^2 - 4ac < 0$, we have two complex solutions. Shouldn't the MLE be unique only for $b^2 - 4ac = 0$ then? Which would result in $|\Delta| = 1$? Commented May 26, 2023 at 2:46
• Complex solutions don't count, because the parameter set consists of $\theta\in{\mathbb R}$. So when the discriminant is negative there is only one real solution to $\ell'(\theta)=0$. When the discriminant is zero the additional root coincides with $\hat\theta:=\frac{x_1+x_2}2$, as you can check. Commented May 26, 2023 at 4:04