I have a three part question:

1) If I have a $Cauchy(\theta, 1)$ with density:

$p(x-\theta) = \frac{1}{\pi\{1+{(x-\theta)^2}\}}$

and $x_1, ..., x_n$ forms i.i.d sample. I see on Wiki that this will make a likelihood function of:

$l(\theta) = -nlog\pi - \sum_{i=1}^{n} log(1+{(x_i-\theta)^2})$

But how is that derived?

2) What derivative rules would I use to show that:

$l'(\theta) = -2 \sum_{i=1}^{n} \frac{x_i-\theta}{1+(x_i-\theta)^2}$

(I feel this is probably straight-forward, but I am missing it)

3) How would I obtain that the Fisher information is equal to $I(\theta)=\frac{n}{2}$?

I started by:

$\frac{\partial^2f(x;\theta)}{2\theta^2} = \frac{8(x-\theta)^2}{\pi[1+(x-\theta)^2]^3} - \frac{2}{\pi[1+(x-\theta)^2]^2}$

I think that I next need to find the integral of:

$I(\theta) = -E[\frac{\partial^2f(x;\theta)}{2\theta^2}]$

But I cannot get this last step of reducing this information to $\frac{n}{2}$.

  • 1
    $\begingroup$ See wwwf.imperial.ac.uk/~das01/MyWeb/M3S3/Handouts/… Nothing special - just integration. $\endgroup$ – Alexey Zaytsev Apr 6 '15 at 14:40
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    $\begingroup$ 3) Fisher information is defined as$$I(\theta) = -\mathbb{E}_\theta\left[\frac{\partial^2 \log f(X;\theta)}{\partial\theta^2}\right]$$ $\endgroup$ – Xi'an Apr 6 '15 at 14:41
  • $\begingroup$ @Alexey: Thanks, that sure is a lot of integration. I am pretty familiar with tan/cos/sin integration. However, I am unsure how to integrate my specific case, where I have $l(\theta) = -nlog\pi - \sum_{i=1}^{n} log(1+{(x_i-\theta)^2})$. Could I begin by saying $I(\theta) = 2n \int_{-\infty}^\infty \frac{1-(x-\theta)^2}{(1+(x-\theta)^2)^2} \frac{1}{\pi} \frac{1}{1+(x-\theta)^2}dx$ ? I am not sure how to deal with the summation of n in the easiest way possible. $\endgroup$ – user84756 Apr 6 '15 at 15:39
  • $\begingroup$ To clarify, my question is how to get the n constant out of the way before all the integration! Thanks again. $\endgroup$ – user84756 Apr 6 '15 at 15:45
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    $\begingroup$ Could you please correct the mistakes in the question, i.e., $l^\prime(\theta)=-2...$ which should be $l^\prime(\theta)=+2...$ and the wrong resolution you copied from this forum? $\endgroup$ – Xi'an Apr 7 '15 at 7:34

The Fisher information for one observation is given by\begin{align*}I(\theta) &= -\mathbb{E}_\theta\left[\frac{\partial^2 \log f(X;\theta)}{\partial\theta^2}\right]\\ &=\mathbb{E}_\theta\left[ \frac{\partial^2 \log \{1+(X-\theta)^2\}}{\partial\theta^2}\right]\\ &=2\mathbb{E}_\theta\left[ -\frac{\partial }{\partial\theta}\frac{(X-\theta)}{1+(X-\theta)^2}\right]\\ &=2\mathbb{E}_\theta\left[\frac{1}{1+(X-\theta)^2}-\frac{2(X-\theta)^2}{[1+(X-\theta)^2]^2}\right]\\ &= \frac{2}{\pi}\int_\mathbb{R} \frac{1}{[1+(x-\theta)^2]^2}-\frac{2(x-\theta)^2}{[1+(x-\theta)^2]^3} \text{d}x\\ &= \frac{2}{\pi}\int_\mathbb{R} \frac{1}{[1+x^2]^2}-\frac{2x^2}{[1+x^2]^3} \text{d}x\\ &= \frac{2}{\pi}\int_\mathbb{R} \frac{1}{[1+x^2]^2}-\frac{2}{[1+x^2]^2}+\frac{2}{[1+x^2]^3} \text{d}x\\ &= \frac{2}{\pi}\int_\mathbb{R} \frac{-1}{[1+x^2]^2}+\frac{2}{[1+x^2]^3} \text{d}x \end{align*} because the integral (and the information) is translation invariant.

Now it is easy to establish a recurrence relation on$$I_k=\int_\mathbb{R} \frac{1}{[1+x^2]^k}\text{d}x$$Indeed \begin{align*} I_k &= \int_\mathbb{R} \frac{1+x^2}{[1+x^2]^{k+1}}\text{d}x\\ &= I_{k+1} + \int_\mathbb{R} \frac{2kx}{[1+x^2]^{k+1}}\frac{x}{2k}\text{d}x\\ &= I_{k+1} + \frac{1}{2k} \int_\mathbb{R} \frac{1}{[1+x^2]^{k}}\text{d}x = I_{k+1} + \frac{1}{2k} I_k \end{align*} by an integration by parts. Hence $$I_1=\pi\quad\text{and}\quad I_{k+1}=\frac{2k-1}{2k}I_k\quad k>1$$ which implies $$I_1=\pi\quad I_2=\frac{\pi}{2}\quad I_3=\frac{3\pi}{8}$$ and which leads to the Fisher information: $$I(\theta)=\frac{2}{\pi}\left\{-I_2+2I_3\right\}=\frac{2}{\pi}\left\{\frac{-\pi}{2}+\frac{3\pi}{4}\right\}=\frac{1}{2}$$

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    $\begingroup$ +1. You could also note that differentiating both sides of $$\int_\mathbb{R}\frac{dx}{t+x^2}=\pi t^{-1/2}$$ twice in succession with respect to $t$ and setting $t=1$ easily yields the answer. $\endgroup$ – whuber Apr 6 '15 at 21:32
  • $\begingroup$ (+1) Even nicer! $\endgroup$ – Xi'an Apr 8 '15 at 19:30
  • $\begingroup$ Working with $I(\theta)=\text{E}_{\theta}\left[\frac{\partial}{\partial\theta}\ln f_{\theta}(X_1)\right]^2$ makes calculations simpler here as we end up with a Beta integral of the second kind. $\endgroup$ – StubbornAtom Aug 10 '18 at 14:20

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