Question:
Hello,
I am attempting all parts of the attached question. I have done part a, b and c.
I have 2 questions.
For part b)
I am not quite sure what it means to "argue" that there is a UMP test of size alpha? I know to exhibit its structure we just do the usual 1 if T(x)> some constant and 0 if T(x) < some constant.
For part d)
I did $$P\left(\sum_{i=1}^{n}\ln{(X_i)^2}>k\right)=P\left(\sum_{i=1}^{n}Y_i^2>k\right)=P(\chi_n^2>k)$$ And this makes $$k=\chi_{n,\alpha}^2 $$ So my test is $$\psi=\begin{cases}1 \text{ if} \sum_{i=1}^{n}Y_i>\chi_{n,\alpha}^2 \\ 0 \text { if } \sum_{i=1}^{n}Y_i<\chi_{n,\alpha}^2 \end{cases}$$
Now my question is, have i done question d right? Also, I am not sure how to calculate the power function and subsequently sketch it.
Thanks.
Since there is monotone likelihood ratio in $T=\sum\limits_{i=1}^n (\ln X_i)^2$, it can be argued by Karlin-Rubin theorem that $\phi=\begin{cases}1&,\text{ if }T>c \\ 0&,\text{ otherwise }\end{cases}$ is a UMP test for testing $H_0$ against $H_1$.
Now $T/\theta^2\sim \chi^2_n$ and size of the test is $\alpha$, so that
$$E_{\theta_0}\,\phi=P_{\theta_0}[T>c]=P_{\theta_0}\left[\frac{T}{\theta_0^2}>\frac{c}{\theta_0^2}\right]=P\left[\chi^2_n>\frac{c}{\theta_0^2}\right]=\alpha$$
This gives $$\frac{c}{\theta_0^2}=\chi^2_{n,\alpha}\,,\text{ i.e. }\, c=\theta_0^2\chi^2_{n,\alpha}$$
As for the power function, you have
\begin{align} E_{\theta}\,\phi&=P_{\theta}\left[T>\theta_0^2\chi^2_{n,\alpha}\right] \\&=P_{\theta}\left[\frac{T}{\theta^2}>\frac{\theta_0^2\chi^2_{n,\alpha}}{\theta^2}\right] \\&=P\left[\chi^2_n>\frac{\theta_0^2\chi^2_{n,\alpha}}{\theta^2}\right] \end{align}
Try to show that this is an increasing function of $\theta$. Further argue that $E_{\theta}\,\phi<\alpha$ for $\theta<\theta_0$ and $E_{\theta}\,\phi>\alpha$ for $\theta>\theta_0$. Using this information, you can provide a rough sketch for power function.