Krdytkyu

Let $X_1,...,X_n$ be a random sample from $N(0,sigma_X^2)$ and let $Y_1,...,Y_m$ be a random sample from $N(0,sigma_Y^2)$. Define $alpha := sigma_Y^2/sigma_X^2$. Find the level $alpha$ LRT of $H_0 : alpha = alpha_0$ versus $H_1 : alpha ne alpha_0$. Express the rejection region of the LRT in terms of an $F(n,m)$ random variable. (Hint: $F$ can be obtained as the ratio of scaled $chi^2$ distributions, i.e. $F(n,m) = frac{chi^2_n/n}{chi_m^2/m}$.)

First of all, I find it a little bit confusing to define $alpha$ as $sigma_Y^2/sigma_X^2$. This $alpha$ is not the same $alpha$ as the level of the LRT, right?

Anyway, I determined that the LRT is $$lambda(X,Y) = frac{sup_{sigma_X^2,sigma_Y^2:frac{sigma_Y^2}{sigma_X^2} = alpha_0}L(sigma_X^2|X)L(sigma_Y^2|Y)}{sup_{sigma_X^2,sigma_Y^2}L(sigma_X^2|X)L(sigma_Y^2|Y)}$$

Calculating where the suprema are taken and substituting that gave me $$lambda(X,Y)=frac{(n+m)^{(n+m)/2}alpha_0^{n/2}big(sum X_i^2big)^{n/2}big(sum Y_i^2big)^{m/2}}{n^{n/2}m^{m/2}big(alpha_0sum X_i^2+sum Y_i^2big)^{(n+m)/2}}le c$$

where $c$ still needs to be determined to ensure we have a level $alpha$ test. However, to do so I would need to know the distribution of this monstrous expression. I know that I can rescale everything a bit to get that e.g. $sum X_i^2$ is $chi_n^2$-distributed, but I still do not know what happens if such a distribution is taken to some power, or multiplied by something, etc.

Furthermore, it is not clear to me how I should express the rejection region using this random variable $F$, but maybe this will become clear when I know how to solve the level $alpha$ LRT. Thank you for any help in clearing things up for me.

Here is a somewhat heuristic argument without going into details of a likelihood ratio test:

Suppose $theta=sigma_Y^2/sigma_X^2$, and we are to test $H_0:theta=theta_0$ versus $H_1:theta=theta_1,(ne theta_0)$.

Recall that the statistics $s_1^2=frac{1}{n}sumlimits_{i=1}^n X_i^2$ and $s_2^2=frac{1}{m}sumlimits_{i=1}^m Y_i^2$ are unbiased and sufficient for $sigma_X^2$ and $sigma_Y^2$ respectively. Moreover, $frac{ns_1^2}{sigma_X^2}simchi^2_n$ and $frac{ms_2^2}{sigma_Y^2}simchi^2_m$ are independently distributed.

Then we readily have

$$F=frac{ns_1^2/nsigma_X^2}{ms_2^2/msigma_Y^2}=frac{s_1^2}{s_2^2}thetasim F_{n,m}$$

So a test statistic for testing $H_0$ would be $$F=frac{s_1^2}{s_2^2}theta_0$$

We can say that expected value of the observed $F$ statistic is $$E(F)=frac{m}{m-2}approx 1$$

So it could be argued that the decision rule is "Reject $H_0$ if observed $F<c_1$ or observed $F>c_2$", where $c_1,c_2$ are so chosen that $$P_{H_0}(F<c_1)+P_{H_0}(F>c_2)=alpha$$

I haven't made much progress with the LR test specifically, but I am pretty sure you would end up with a test of the above form.