Question 11.14: Determine the LR test for testing the hypothesis H0: θ = 0 (......
Determine the LR test for testing the hypothesis H_{0}\colon\theta=0 (against the alternative H_{A}\colon\theta\neq0.) at level of significance α on the basis of one observation from the p.d.f. f(x;\theta)={\frac{1}{\pi}}\times{\frac{1}{1+(x-\theta)^{2}}},\ x\in\Re,\ \theta\gt 0 (the Cauchy p.d.f.)
First, f(\cdot;\theta) is a p.d.f., since
\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{d x}{1+(x-\theta)^{2}}=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{d y}{1+y^{2}}~~~ (by setting x − θ = y)
={\frac{1}{\pi}}\int_{-\pi/2}^{\pi/2}d t=1
Next, clearly, L(\theta\mid x)(=f(x;\theta)) is maximized for θ = x, so that \lambda={\frac{1}{\pi}}\times{\frac{1}{1+x^{2}}}/ \textstyle{\frac{1}{\pi}}={\frac{1}{1+x^{2}}},{~{\mathrm{and}}~}\lambda\,\lt {{\lambda_{0}}}, if and only if x^{2}\gt {\frac{1}{\lambda_{0}}}-1=C,or x < −C or x > C, where C is determined through the relation: P_{0}(X\lt -C\ \mathrm{or}\ X\gt C)=\alpha, or P(X\gt C)={\frac{\alpha}{2}} due to the symmetry (around 0) of the p.d.f. f(x;0). But
P(X\gt C)=\int_{C}^{\infty}{\frac{1}{\pi}}\times{\frac{d x}{1+x^{2}}}={\frac{1}{\pi}}\int_{\tan^{-1}C}^{\pi/2}d t={\frac{1}{\pi}}\left({\frac{\pi}{2}}-\tan^{-1}C\right)={\frac{\alpha}{2}},
or \tan^{-1}C\;=\;{\frac{(1-\alpha)\pi}{2}}, and hence C\;=\;\tan(\frac{(1-\alpha)\pi}{2}). So, \textstyle H_{0} is rejected whenever x\lt -\tan({\frac{(1-\alpha)\pi}{2}}){\mathrm{~or~}}x\gt \tan({\frac{(1-\alpha)\pi}{2}}). For example, for \alpha\,=\,0.05,\,C\,= \tan(0.475\pi)\simeq12.706, and \textstyle H_{0} is rejected when x < −12.706 or x > 12.706.