# Finding UMP test for $\text{Cauchy}(0, \theta)$ distribution

If $$X$$ follows $$\text{Cauchy}(0, \theta)$$ distribution, construct a UMP size $$\alpha$$ test for testing $$H_0:\theta=\theta_0$$ against $$H_1:\theta>\theta_0$$

My attempt:

Let us take any $$\theta_1>\theta_0$$. First we have to find a most powerful test of size $$\alpha$$ for testing $$H_0:\theta=\theta_0$$ against $$H'_1:\theta=\theta_1$$

Now, $$f_\theta (x)=\frac{\theta}{\pi(x^2+\theta^2)}$$, $$x\in\mathbb R, \theta>0$$

So, $$\lambda(x)=\frac{f_{\theta_1}(x)}{f_{\theta_0}(x)}=\frac{\theta_1(x^2+\theta_0^2)}{\theta_0(x^2+\theta_1^2)}$$

$$\frac{d\lambda(x)}{dx^2}>0$$, so $$\lambda(x)$$ is an increasing function of $$x^2$$ or $$|x|$$, i.e, $$\lambda(x)>k \iff |x|>c$$

Therefore, by N-P lemma, a most powerful test of size $$\alpha$$ for testing $$H_0:\theta=\theta_0$$ against $$H'_1:\theta=\theta_1$$ is given by,

$$\phi(x)= 1$$ if $$|x|>c$$

$$\phi(x)= 0$$ if $$|x|

where c is such that $$E_{\theta_0}\phi(x)=\alpha$$, i.e, $$P_{\theta_0}(|X|>c)=\alpha$$

I'm getting stuck here. I cannot understand how to find the value of c. If $$X$$ follows $$\text{Cauchy}(0, \theta_0)$$, then what does $$|X|$$ follow?

Any kind of hints and suggestions are appreciated. Thanks in advance.

• Is the likelihood ratio function a monotonic function of $x$? Commented Feb 1, 2021 at 15:46
• The test is correct. You don't need to know the distribution of $|X|$ to find $c$, just integrate using the distribution of $X$. Commented Feb 1, 2021 at 16:54
• @StubbornAtom Thanks. Commented Feb 5, 2021 at 12:03
• @Henry : You're mistaken. The hypothesis test is for a scale parameter, and the test statistic can vary in either direction from the known median. Commented Feb 5, 2021 at 23:52
• @MichaelHardy On reviewing this, you seem to be correct Commented Feb 6, 2021 at 0:00

$$P_{\theta_0}(|X|>c)=\alpha$$

$$\displaystyle \Rightarrow P_{\theta_0}(X>c)+P_{\theta_0}(X<-c)=\alpha$$

$$\displaystyle \Rightarrow \frac{\theta_0}{\pi}\int_c^{\infty}\frac{dx}{x^2+{\theta_0}^2}+\frac{\theta_0}{\pi}\int_{-\infty}^{-c}\frac{dx}{x^2+{\theta_0}^2}=\alpha$$

$$\displaystyle \Rightarrow 1-\frac{2}{\pi}\tan^{-1}\left(\frac{c}{\theta_0}\right)=\alpha$$

$$\displaystyle \Rightarrow c=\theta_0\tan\frac{\pi\left(1-\alpha\right)}{2}$$