After centering, 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 $ $=0$


$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?

  • $\begingroup$ Please, explain why did you split x into -x and +x? This is my homework and I'm getting stuck at that step. I guess you applied Newton's Raphson Method to it. But I'm not getting how to apply it. Will please tell me? $\endgroup$ – user89929 Sep 20 '15 at 16:09

There is a math typo in your calculations. The first order condition for a maximum is:
\begin{align} \frac {\partial L}{\partial \theta}= 0 &\Rightarrow \frac {2(x+\theta)}{ 1+(x+\theta)^2} - \frac{2(x-\theta)}{ 1+(x-\theta)^2}&=0 \\[5pt] &\Rightarrow (x+\theta)+(x+\theta)(x-\theta)^2 - (x-\theta)-(x-\theta)(x+\theta)^2&=0 \\[3pt] &\Rightarrow 2\theta +(x+\theta)(x-\theta)\left[x-\theta-(x+\theta\right]&=0 \\[3pt] &\Rightarrow2\theta -2\theta(x+\theta)(x-\theta) =0\Rightarrow 2\theta -2\theta(x^2-\theta^2)&=0 \\[3pt] &\Rightarrow2\theta(1-x^2+\theta^2)=0 \Rightarrow 2\theta\big(\theta^2+(1-x^2)\big)&=0 \end{align}

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, apart from the candidate point $\theta =0$ you also 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.


For $x =\pm 0.5$ the graph of the log-likelihood is enter image description here

while for $x =\pm 1.5$ the graph of the log-likelihood is, enter image description here

Now all you have to do is to prove it algebraically and then wonder "fine -now which of the two should I choose?"

  • $\begingroup$ Thanks! I can't see why $\theta=0$ would no longer be an MLE though $\endgroup$ – user123965 May 16 '14 at 21:09
  • $\begingroup$ Work the 2nd order condition for a maximum, or evaluate the likelihood at the candidate solutions $\endgroup$ – Alecos Papadopoulos May 16 '14 at 21:17
  • 2
    $\begingroup$ +1 great answer. Also, this might be interesting: wolframalpha.com/share/… wolframalpha.com/share/… $\endgroup$ – random_user May 17 '14 at 0:35
  • $\begingroup$ @random_user Thanks! - I took the liberty to incorporate the plot in the answer. $\endgroup$ – Alecos Papadopoulos May 17 '14 at 0:47
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    $\begingroup$ 2nd derivative positive so indeed a local minimum $\endgroup$ – Alecos Papadopoulos May 17 '14 at 9:24

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