# Comparison of variances: symmetric F-test is likelihood ratio test?

Suppose $$X_i\stackrel{IID}\sim N(\mu,\sigma^2)$$ for $$i=1,...,n$$, where $$\mu$$ is known. We want to apply the likelihood ratio test to decide between the hypotheses $$H_0: \sigma=\sigma_0 \\ H_1: \sigma\neq \sigma_0$$ for a given fixed $$\sigma_0$$. Doing some computation and writing $$t=\sum_{i=1}^n (x_i-\mu)^2/n\sigma_0^2$$, we find that $$2\log L(H_0,H_1)=n(t-1-\log t).$$ This quantity has a minimum in $$t=1$$ and is monotonic in $$(0,1)$$ and $$(1,+\infty)$$, and hence the likelihood ratio test of size $$\alpha$$ consists in rejecting $$H_0$$ for $$t\in C:= (0,a)\cup(b,+\infty)$$, where $$a-\log a=b-\log b$$ and $$\mathbb{P}(\chi_n^2\in C)=\alpha$$ (using that $$\sum_{i=1}^n (X_i-\mu)^2/\sigma_0^2\sim \chi_n^2$$ under $$H_0$$).

Now my question is, how can we prove - if it is even true - that we must have $$a=q(\chi_n^2,\alpha/2)$$ and $$b=q(\chi_n^2,1-\alpha/2)$$ ($$q$$ is the quantile function here), that is that the likelihood ration test is exactly a symmetric two-tailed F-test? I understand that picking $$a$$ and $$b$$ this way gives rise to a size-$$\alpha$$ test, but I want to find exactly the likelihood ratio test, i.e. the one that comes from taking $$C=\{L(H_0,H_1)>k\}$$ for some $$k$$.

• I'm sure this has been addressed in other posts here. Jun 19, 2023 at 12:28
• @utobi the only discussion about this topic I could find is contained in the final comments to the answer to this question stats.stackexchange.com/questions/189153/… (there is also a comment from you). If you know of any other posts where this issue is explicitly discussed please share the links, I searched for a long time but couldn't find any. Jun 19, 2023 at 21:22
• You need to meditate over the expression for $C$, giving the rejection region. It is an equation that is formulated for $L$. Try to transform it into an equation for $t$. You may want to graph $f(t) =t-\log(t)$ to get an idea why the test becomes two tailed.
– Ute
Jun 20, 2023 at 10:58
• @Ute I've got no problems with the fact that the likelihood ratio test is two-tailed, my problem is with the fact that it is symmetric (which I actually believe not to be the case). Jun 20, 2023 at 13:00
• I remember that I once had the same question, Titti, but I don't remember if I found a satisfactory answer - more likely not. The test has the correct size, but could it be made more powerful? Not, as long as we have a compound alternative: which of the many possible alternatives $\sigma_1^2$ would we sharpen the test towards? I believe it is a compromise to have good power for each side of the alternative - even then it sounds slightly murky. It is a justified question, and you are not the only one who gets doubts :-)
– Ute
Jul 18, 2023 at 11:31