I have this sample exam question and no model answer.

Question.

Let $\displaystyle Y_1,...Y_n$ be a random sample from a distribution from a density function

$\displaystyle f_Y(y;\theta)=\frac{2y}{\theta}e^{\frac{-y^2}{\theta}}, y>0$

$\displaystyle f_Y(y;\theta)=0, otherwise$

where $\displaystyle \theta>0$ is an unknown parameter.

i. Show that the uniformly most powerful test (UMPT) for testing hypotheses

$\displaystyle H_0:\theta=1, H_1:\theta>1$

will reject Ho if and only if T>K for some constant K>0, where $\displaystyle T=\Sigma_{i=1}^nY_i^2$.

ii. Find the distribution of the test statistic T under Ho. [You may use the fact that the $\displaystyle \chi^2$ distribution with k degrees of freedom has moment generating function $\displaystyle (1-2t)^{-k/2}$.

Answer.

i. The likelihood function is

$\displaystyle L_Y=(\frac{2}{\theta})^n\prod_{i=1}^n*y_i*e^{-\frac{1}{\theta}\Sigma_{i=1}^ny_i^2}$

The likelihood ratio from Neyman-Person lemma is

$\displaystyle \frac{L_Y(theta_1;y)}{L_Y(\theta_0;y)}=\frac{(\fra c{2}{\theta})^n\prod_{i=1}^ny_ie^{-\frac{1}{\theta}\Sigma_{i=1}^ny_i^2}}{2^n\prod_{i= 1}^ny_ie^{\Sigma_{i=1}^ny_i^2}}=\theta^{-n}e^{{\Sigma_{i=1}^ny_i^2(-\frac{1}{\theta}-1)}}=\theta^{-n}e^{T(-\frac{1}{\theta}-1)}$

If this ratio is higher than some number k (determined by required confidence level), then we will reject Ho:

$\displaystyle -\theta^{-n}e^{T(-\frac{1}{\theta}-1)}>k$

take logs

$\displaystyle ln(-\theta^{-n})+T(-\frac{1}{\theta}-1)>ln(k)$

$\displaystyle T\frac{-1-\theta}{\theta}>ln(k)-ln(-\theta^{-n})$

$\displaystyle T>\frac{\theta}{1+\theta}(ln(-\theta^{-n}-ln(k))=K$

and the required T>K is the above inequality.

does it make sense?

ii. Distribution of T - let me think it over a bit longer. I am thinking to find MGF of y^2 (with theta=1 the density is simplifed) and to show that it is a form of MGF of the chi squared.