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$X_1, \ldots, X_n$ iid $\sim N(0, \sigma^2)$. I want to deduce the chi-square test

$H_0 : \sigma^2 = \sigma_0^2, ~ H_1 : \sigma^2 \neq \sigma_0^2$: Decline $H_0$, if $ \displaystyle \frac{(n-1) \hat{S_n}^2}{\sigma_0^2} > \chi_{n-1, 1- \frac{\alpha}{2}}^2$ or $< \chi_{n-1, \frac{\alpha}{2}}^2$

as a likelihood ratio test. I've calculated the likelihood quotient

$\lambda(x) = \displaystyle \left( \frac{\sum_{i=1}^n x_i^2}{n \sigma_0^2} \right) ^{ \displaystyle \frac{n}{2}} \text{exp} \left(-\frac{1}{2\sigma_0^2} \sum_{i=1}^n x_i^2 + \frac{n}{2} \right)$,

but I can't see how to transformate it to get the result I want.

Can anyone help?

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1 Answer 1

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You know that $$\lambda(x)=\frac{sup_{\Theta_0}L(\theta_0|x)}{sup_{\Theta}L(\theta|x)}$$ and $$-2\log \lambda(x)\rightarrow^{Distribution}_{n\rightarrow\infty} \chi^2_1$$

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  • $\begingroup$ This approximation is not what the question is asking for. $\endgroup$ Commented Feb 3, 2020 at 18:42

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