$f(x;\theta)=\frac{1}{\pi[1+(x-\theta)^2]}$; $-\infty<x<\infty,\quad-\infty<\theta<\infty$
$\log f(x;\theta)=\log (\frac{1}{\pi[1+(x-\theta)^2]})$
$\Rightarrow \log f(x;\theta)=\log(1)-\log(\pi[1+(x-\theta)^2])$
$\Rightarrow \log f(x;\theta)=0-\log(\pi)-\log[1+(x-\theta)^2]$
$\Rightarrow \log f(x;\theta)=-\log(\pi)-\log[1+(x-\theta)^2]$
Now
$\frac{\partial}{\partial\theta}\log f(x;\theta)$ $=0-\frac{1}{[1+(x-\theta)^2]}2(x-\theta)(-1)$ $=\frac{2(x-\theta)}{[1+(x-\theta)^2]}$
$\frac{\partial^2}{\partial\theta^2}\log f(x;\theta)$
$=2\frac{[1+(x-\theta)^2]\frac{\partial}{\partial\theta}(x-\theta)-(x-\theta)\frac{\partial}{\partial\theta}[1+(x-\theta)^2]}{[1+(x-\theta)^2]^2}$
$=2\frac{[1+(x-\theta)^2](-1)-(x-\theta)2(x-\theta)(-1)}{[1+(x-\theta)^2]^2}$
$=2\frac{(x-\theta)^2-1}{[1+(x-\theta)^2]^2}$
I have tried to take the Expectation of $[\frac{\partial}{\partial\theta}\log f(x;\theta)]^2$ and $\frac{\partial^2}{\partial\theta^2}\log f(x;\theta)$ and simplify the resulting equations,but i couldn't.
$\mathbb E[\frac{\partial}{\partial\theta}\log f(X;\theta)]^2=\mathbb E[\frac{2(x-\theta)}{[1+(x-\theta)^2]}]^2$
$\mathbb E[\frac{\partial^2}{\partial\theta^2}\log f(X;\theta)]=\mathbb E[2\frac{(x-\theta)^2-1}{[1+(x-\theta)^2]^2}]$
please compute the
$\mathbb E[\frac{\partial}{\partial\theta}\log f(X;\theta)]^2$ and $\mathbb E[\frac{\partial^2}{\partial\theta^2}\log f(X;\theta)]$
We wish to find the Expectation of The Function of The Cauchy Random Variable.
Let $f_X(x;\theta) = \frac{1}{\pi(1+(x-\theta)^2)}$, with $x\in(-\infty,+\infty)$ and $\theta\in(-\infty,+\infty)$.
$$ \begin{aligned} log f_X(x;\theta) & = (-1)\log\pi(1+(x-\theta)^2) \\ & = (-1)[\log \pi + \log(1+(x-\theta)^2)] \\ & = -\log \pi - \log(1+(x-\theta)^2) \end{aligned} $$ $$ \begin{aligned} \frac{\partial \log f_X(x;\theta)}{\partial \theta} & = 0 + (-1)\frac{1}{(1+(x-\theta)^2)}\frac{\partial}{\partial \theta}(1+(x-\theta)^2) \\ & = (-1)\frac{1}{(1+(x-\theta)^2)}\cdot 2 (x-\theta)^1 \cdot (-1) \\ & = (+1)\cdot \frac{2(x-\theta)}{1+(x-\theta)^2} = \frac{2x-2\theta}{1+(x-\theta)^2} \end{aligned} $$ where we have used the facts: $$ \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \log x & = \frac1x \\ \frac{\mathrm{d}}{\mathrm{d}x} \log f(x) & = \frac{1}{u(x)} \frac{\mathrm{d}}{\mathrm{d}x} u(x) \end{aligned} $$ $$ \begin{aligned} \frac{\partial^2 \log f_X(x;\theta)}{\partial \theta^2} & = \frac{(-2)(1+(x-\theta)^2) - 2(x-\theta)^1 (-1)(2x-2\theta)}{(1+(x-\theta)^2)^2} \\ & = \frac{-2\cdot(1+(x-\theta)^2)+2(x-\theta)\cdot2(x-\theta)}{(1+(x-\theta)^2)^2} \\ & = \frac{-2-2(x-\theta)^2 + 4(x-\theta)^2}{(1+(x-\theta)^2)^2} \\ & = \frac{-2 + 2(x-\theta)^2}{(1+(x-\theta)^2)^2} \\ & = \frac{2\cdot((x-\theta)^2 - 1)}{(1+(x-\theta)^2)^2} \end{aligned} $$
Let: $$ \begin{aligned} A & = \left(\frac{\partial \log f_X(x;\theta)}{\partial \theta}\right)^2 = \frac{4(x-\theta)^2}{(1+(x-\theta)^2)^2} \\ B & = \frac{\partial^2 \log f_X(x;\theta)}{\partial \theta^2} = \frac{2((x-\theta)^2 - 1)}{(1+(x-\theta)^2)^2} \end{aligned} $$
As a hint, we will only be solving $E_X[A]$. The same mathematical principles can be used to solve $E_X[B]$.
$$ \begin{aligned} E_X[A] & = \int_{-\infty}^{+\infty} A\cdot f_X(x;\theta)\,\mathrm{d}x \\ & = \int_{-\infty}^{+\infty} \frac{4\cdot(x-\theta)^2}{(1+(x-\theta)^2)^2} \cdot \frac{1}{\pi(1+(x-\theta)^2)}\,\mathrm{d}x \\ & = \int_{-\infty}^{+\infty} \frac{4}{\pi}\cdot \frac{(x-\theta)^2}{(1+(x-\theta)^2)^3} \, \mathrm{d}x \end{aligned} $$ Using u-substitution, let $u = x-\theta$, $\mathrm{d}u = \mathrm{d}x$, $u_\mathrm{lower} = -\infty - \theta = -\infty$, $u_\mathrm{upper} = +\infty - \theta = +\infty$. Then, $$ E_X[A] = \int_{-\infty}^{+\infty} \frac{4}{\pi} \frac{u^2}{(1+u^2)^3} \, \mathrm{d}u $$ Now, recall the Trigonometric Identities: $\sin^2 x + \cos^2 x = 1$, $1 + \tan^2 x = \sec^2 x$, $\sec x = \frac{1}{\cos x}$. Furthermore, recall the following Derivatives of Trigonometric Functions: $\frac{\mathrm{d}}{\mathrm{d}x}\sin x = \cos x$, $\frac{\mathrm{d}}{\mathrm{d}x}\cos x = -\sin x$, and $\frac{\mathrm{d}}{\mathrm{d}x}\tan x = \sec^2 x$.
Applying the Trigonometric Substitution Technique, let $u = \tan \theta$, $\mathrm{d}u = \sec^2 \theta \mathrm{d}\theta$. This implies $\theta = \tan^{-1}u$. Then, $\theta_\textrm{lower} = \tan^{-1}(-\infty) = -\frac\pi 2$ and $\theta_\textrm{upper} = \tan^{-1}(+\infty) = +\frac\pi 2$.
$$ \begin{aligned} E_X[A] & = \int_{-\frac \pi 2}^{+\frac \pi 2} \frac{4}{\pi} \cdot \frac{\tan^2 \theta}{(1+\tan^2 \theta)^3} \sec^2 \theta \, \mathrm{d}\theta \\ & = \int_{-\frac \pi 2}^{+\frac \pi 2} \frac{4}{\pi} \cdot \frac{\tan^2 \theta (\sec^2 \theta)^1}{(\sec^2 \theta)^3} \, \mathrm{d} \theta \\ & = \int_{-\frac \pi 2}^{+\frac \pi 2} \frac{4}{\pi}\cdot \frac{\tan^2 \theta}{(\sec^2 \theta)^2}\, \mathrm{d}\theta \\ & = \int_{-\frac \pi 2}^{+\frac \pi 2} \frac{4}{\pi} \sin^2 \theta \cos^2 \theta \, \mathrm{d}\theta \\ & = \int_{-\frac \pi 2}^{+\frac \pi 2} \frac{4}{\pi} (1-\cos^2 \theta)\cos^2 \theta \, \mathrm{d}\theta \\ & = \int_{-\frac \pi 2}^{+\frac \pi 2} \frac{4}{\pi}(\cos^2 \theta - \cos^4 \theta)\,\mathrm{d}\theta \end{aligned} $$
Recall the following Properties of Even Functions: If $f(x)$ is even, then $\int_{-A}^{+A}f(x)\,\mathrm{d}x = 2\int_{0}^{+A}f(x)\,\mathrm{d}x$. If $f(x)$ is even, then $[f(x)]^n$ is even.
Then, $$ E_X[A] = \frac{8}{\pi}\int_{0}^{+\frac \pi 2}(\cos^2 \theta - \cos^4 \theta)\,\mathrm{d}\theta. $$
Now, we take a slight diversion and calculate a Useful Integral. $$ C(n) = \int_{0}^{+\frac \pi 2}\cos^n x \, \mathrm{d}x = \int_{0}^{+\frac \pi 2} (\cos^1 x)(\cos^{n-1}x)\, \mathrm{d}x. $$ Recall Integration by Parts: $\int_{a}^{b}u\, \mathrm{d}v = uv|_{a}^{b} - \int_{a}^{b}v\,\mathrm{du}$. Let $u = \cos^{n-1}x = (\cos x)^{n-1}$, $\mathrm{d}v = \cos x\, \mathrm{d}x$. Then, $\frac{\mathrm{d}}{\mathrm{d}x}u = \frac{\mathrm{d}}{\mathrm{d}x}(\cos x)^{n-1} = (n-1)(\cos x)^{n-2}(-1)(\sin x)$, and $v = \sin x$. The calculations below hold for $n \neq 1$. (If $n = 1$, we have the expression $0^0$, which is undefined, so we should calculate the $n = 1$ case separately.) $$ \begin{aligned} C(n) & = \int_{0}^{+\frac \pi 2} \underbrace{(\cos^{n-1} x)}_{u}\underbrace{(\cos^1 x)\, \mathrm{d}x}_{\mathrm{d}v} \\ & = \cos^{n-1}x \sin x |_{0}^{+\frac \pi 2} - \int_{0}^{+\frac \pi 2}\sin x (n-1) (\cos x)^{n-2} (-1)\sin x \, \mathrm{d}x \\ & = \underbrace{\left(\cos \frac \pi 2\right)}_{0} \sin \frac \pi 2 - (\cos 0)^{n-1}\underbrace{\sin 0}_0 + \int_{0}^{+\frac \pi 2}(\sin^2 x) (n-1)(\cos x)^{n-2}\, \mathrm{d}x \\ & = \int_{0}^{+\frac \pi 2} (\sin^2 x)(n-1)(\cos^{n-2}x)\, \mathrm{d}x \\ & = \int_{0}^{+\frac \pi 2} (1-\cos^2 x)(n-1)(\cos^{n-2}x)\, \mathrm{d}x \\ & = (n-1)\int_{0}^{+\frac \pi 2} (1-\cos^2 x)\cos^{n-2}x\,\mathrm{d}x \\ & = (n-1) \int_{0}^{+\frac \pi 2}(\cos^{n-2}x - \cos^n x)\, \mathrm{d}x \\ & = (n-1)\left[\int_{0}^{+\frac \pi 2}\cos^{n-2} x\, \mathrm{d}x - \int_{0}^{+\frac \pi 2}\cos^n x \, \mathrm{d}x\right] \\ & = (n-1)[C(n-2) - C(n)] \end{aligned} $$ Then, $$ \begin{aligned} C(n)+(n-1)C(n) = (n-1)C(n-2) & \Rightarrow C(n)(1 + (n-1)) = (n-1)\cdot C(n-2) \\ & \Rightarrow C(n) = \frac{(n-1)C(n-2)}{n} \end{aligned} $$ Plugging in the relevant values of $n$, $$ \begin{aligned} C(0) & = \int_{0}^{+\frac \pi 2} (1)\,\mathrm{d}x = + \frac \pi 2 \\ C(2) & = \frac12 \cdot C(0) = \frac12 \cdot \frac \pi 2 = +\frac \pi 4 \\ C(4) & = \frac34 \cdot C(2) = \frac34 \cdot \frac \pi 4 = \frac{3}{16}\pi. \end{aligned} $$
Returning to where we left off, $$ \begin{aligned} E_X[A] & = \frac{8}{\pi}\int_{0}^{+\frac \pi 2} (\cos^2 \theta - \cos^4 \theta)\, \mathrm{d}\theta \\ & = \frac{8}{\pi}\left[\int_{0}^{+\frac \pi 2} \cos^2 \theta \, \mathrm{d}\theta - \int_{0}^{+\frac ]pi 2}]cos^4 \theta \, \mathrm{d}\theta\right] \\ & = \frac{8}{\pi}[C(2)-C(4)] \\ & = \frac{8}{\pi}\left[\frac{\pi}{4} - \frac{3\pi}{16}\right] \\ & = \frac{8}{\pi}\left[\frac{4\pi}{16} - \frac{3\pi}{16}\right] = \frac{8}{\pi}\cdot \frac{\pi}{16} = \frac12. \end{aligned} $$