# MLE of a Cauchy distribution

After centring, the two measurements x and −x can be assumed to be independent observations from a Cauchy distribution with probability density function:

$f(x :\theta) =$ $1\over\pi (1+(x-\theta)^2)$ $, -∞ < x < ∞$

Show that if $x^2≤ 1$ the MLE of $\theta$ is 0, but if $x^2>1$ there are two MLE's of $\theta$, equal to ±$\sqrt {x^2-1}$

I think to find the MLE I have to differentiate the log likelihood:

$dl\over d\theta$ $=\sum $$2(x_i-\theta)\over 1+(x_i-\theta)^2 = 2(-x-\theta)\over 1+(-x-\theta)^2 + 2(x-\theta)\over 1+(x-\theta)^2 =0 So, 2(x-\theta)\over 1+(x-\theta)^2 = 2(x+\theta)\over 1+(x-\theta)^2 which I then simplified down to 5x^2 = 3\theta^2+2\theta x+3 Now I've hit a wall. I've probably gone wrong at some point, but either way I'm not sure how to answer the question. Can anyone help? - ## 1 Answer There is a math typo in your calculations. The first order condition for a maximum is$$\frac {\partial L}{\partial \theta}= 0 \Rightarrow \frac {2(x+\theta)}{ 1+(x+\theta)^2} - \frac{2(x-\theta)}{ 1+(x-\theta)^2}=0\Rightarrow (x+\theta)+(x+\theta)(x-\theta)^2 - (x-\theta)-(x-\theta)(x+\theta)^2=0\Rightarrow 2\theta +(x+\theta)(x-\theta)\left[x-\theta-(x+\theta\right]=0\Rightarrow2\theta -2\theta(x+\theta)(x-\theta) =0\Rightarrow 2\theta -2\theta(x^2-\theta^2)=0\Rightarrow2\theta(1-x^2+\theta^2)=0 \Rightarrow 2\theta\big(\theta^2+(1-x^2)\big)=0$$If x^2\leq 1 then the term in the parenthesis cannot be zero (for real solutions of course), so you are left only with the solution \hat \theta =0. If x^2 >1 you have 2\theta\big[\theta^2-(x^2-1)\big]=0 so you can get$$\frac {\partial L}{\partial \theta}= 0,\;\; \text{for}\;\;\hat \theta = \pm\sqrt {x^2-1}$$You also have to justify why in this case$\hat \theta =0$is no longer an MLE. ADDENDUM For$x =\pm 0.5$the graph of the log-likelihood is while for$x =\pm 1.5$the graph of the log-likelihood is, Now all you have to do is to prove it algebraically and then wonder "fine -now which of the two should I choose?" - Thanks! I can't see why$\theta=0\$ would no longer be an MLE though –  user123965 May 16 at 21:09
Work the 2nd order condition for a maximum, or evaluate the likelihood at the candidate solutions –  Alecos Papadopoulos May 16 at 21:17
+1 great answer. Also, this might be interesting: wolframalpha.com/share/… wolframalpha.com/share/… –  random_user May 17 at 0:35
@random_user Thanks! - I took the liberty to incorporate the plot in the answer. –  Alecos Papadopoulos May 17 at 0:47
2nd derivative positive so indeed a local minimum –  Alecos Papadopoulos May 17 at 9:24