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.

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jan 4, 2019 at 20:25
  • $\begingroup$ 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. $\endgroup$ Commented Jan 4, 2019 at 20:38

1 Answer 1


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.

Then we readily have

$$F=\frac{ns_1^2/n\sigma_X^2}{ms_2^2/m\sigma_Y^2}=\frac{s_1^2}{s_2^2}\theta\sim F_{n,m}$$

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 $F<c_1$ 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.


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