Sign up ×
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

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?

share|improve this question
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? – user89929 Sep 20 at 16:09

1 Answer 1

up vote 21 down vote accepted

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?"

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.