# Likelihood Ratio Test Variance of Normal Distribution

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 $$\lambda(X,Y)=\frac{(n+m)^{(n+m)/2}\alpha_0^{n/2}\big(\sum X_i^2\big)^{n/2}\big(\sum Y_i^2\big)^{m/2}}{n^{n/2}m^{m/2}\big(\alpha_0\sum X_i^2+\sum Y_i^2\big)^{(n+m)/2}}\le c$$

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.

• I agree using $\alpha$ to define the ratio of the variances is confusing (and not appropriate) when $\alpha$ is also used to refer to the size of the test. This is likely a typo or an oversight on the author/instructor. – Just_to_Answer Jan 4 at 20:25
• Three other notes: (1) Since $\sum X^2_i / \sigma^2_X$ has a $\chi^2$ distribution, it might be beneficial to simplify the likelihood ratio keeping that in mind; (2) If the left-side of the inequality can be written inside a single power, by taking an appropriate root the power can be moved over to the other side and call it a new "constant"; (3) similar idea for any lingering constants on the left side. – Just_to_Answer Jan 4 at 20:38

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 $$s_1^2=\frac{1}{n}\sum\limits_{i=1}^n X_i^2$$ and $$s_2^2=\frac{1}{m}\sum\limits_{i=1}^m Y_i^2$$ are unbiased and sufficient for $$\sigma_X^2$$ and $$\sigma_Y^2$$ respectively. Moreover, $$\frac{ns_1^2}{\sigma_X^2}\sim\chi^2_n$$ and $$\frac{ms_2^2}{\sigma_Y^2}\sim\chi^2_m$$ are independently distributed.

$$F=\frac{ns_1^2/n\sigma_X^2}{ms_2^2/m\sigma_Y^2}=\frac{s_1^2}{s_2^2}\theta\sim F_{n,m}$$
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 $$F or observed $$F>c_2$$", where $$c_1,c_2$$ are so chosen that $$P_{H_0}(Fc_2)=\alpha$$