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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 $|x|<c$

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.

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  • $\begingroup$ Is the likelihood ratio function a monotonic function of $x$? $\endgroup$
    – Henry
    Commented Feb 1, 2021 at 15:46
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    $\begingroup$ The test is correct. You don't need to know the distribution of $|X|$ to find $c$, just integrate using the distribution of $X$. $\endgroup$ Commented Feb 1, 2021 at 16:54
  • $\begingroup$ @StubbornAtom Thanks. $\endgroup$
    – user587389
    Commented Feb 5, 2021 at 12:03
  • $\begingroup$ @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. $\endgroup$ Commented Feb 5, 2021 at 23:52
  • $\begingroup$ @MichaelHardy On reviewing this, you seem to be correct $\endgroup$
    – Henry
    Commented Feb 6, 2021 at 0:00

1 Answer 1

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

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

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

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

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

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