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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 $k$.

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    $\begingroup$ I'm sure this has been addressed in other posts here. $\endgroup$
    – utobi
    Jun 19, 2023 at 12:28
  • $\begingroup$ @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. $\endgroup$
    – No-one
    Jun 19, 2023 at 21:22
  • $\begingroup$ 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. $\endgroup$
    – Ute
    Jun 20, 2023 at 10:58
  • $\begingroup$ @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). $\endgroup$
    – No-one
    Jun 20, 2023 at 13:00
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    $\begingroup$ 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 :-) $\endgroup$
    – Ute
    Jul 18, 2023 at 11:31

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