Finding the distribution of the test statistic under the null hypothesis $H_0$

Suppose we have a random sample $$X_1, \dots, X_n$$ from $$N(\mu , \sigma^2 )$$, with known mean $$\mu$$. For the hypotheses $$H_0: \sigma^2 = \sigma_0^2$$ and $$H_A : \sigma^2 = \sigma_1^2$$, where $$\sigma_1^2 > \sigma_0^2$$, derive the Likelihood Ratio Test with significance level $$0<\alpha <1$$. Find the distribution of the test statistic under $$H_0$$.

$$\textbf{My approach so far}$$

I obtained the likelihood ratio

$$R = {\bigg(}\sqrt{\frac{\sigma_0^2 }{\sigma_1^2 }}{\bigg)}^n \text{exp} {\bigg\{} \frac{-1}{2} (\sigma_0^{-2} - \sigma_1^{-2}) \sum_{i=1}^n (X_i - \mu)^2 {\bigg\}}$$

which by the Neyman-Pearson lemma, the most powerful test for this type of hypothesis for the given data will depend only on $$\sum_{i=1}^n (x_i - \mu)^2$$.

For the next part, when $$\sigma_1^2 > \sigma_0^2$$, we have that $$R$$ is a decreasing function of $$\sum_{i=1}^n (x_i - \mu)^2$$. In which case we should reject $$H_0$$ when $$\sum_{i=1}^n (x_i - \mu)^2$$ is sufficiently large. So there is some constant $$K$$ such that $$\sum_{i=1}^n (x_i - \mu )^2 > K$$, where the threshold for rejection depends on the size of the test.

Here's wehre I am stuck. To answer the second part, I believe I need to find (please correct me if I am wrong) $$K$$ where

$$P(R \leq K | H_0 ) = P(\sum_{i=1}^n (x_i - \mu )^2 > K | H_0 ) = \alpha$$

I'm trying to figure out how I can convert the middle term to have the form $$\frac{X_i - \mu }{\sigma_0}$$ for e.g., so that I can obtain the value of $$K$$.

Am I on the right track and how can I obtain $$K$$ from the above equation (if it is indeed correct)?

Suppose $$L(\sigma^2\mid x_1,\ldots,x_n)$$ is the likelihood function given the sample $$(x_1,\ldots,x_n)$$.

The unrestricted MLE of $$\sigma^2$$ is $$\hat\sigma^2=\frac1n \sum\limits_{i=1}^n(x_i-\mu)^2$$

The likelihood ratio statistic is defined as

\begin{align} \Lambda(x_1,\ldots,x_n)&=\frac{\sup_{\sigma^2=\sigma_0^2}L(\sigma^2\mid x_1,\ldots,x_n)}{\sup_{\sigma^2>\sigma_0^2}L(\sigma^2\mid x_1,\ldots,x_n)} \\&=\frac{L(\sigma_0^2\mid x_1,\ldots,x_n)}{L(\tilde\sigma^2\mid x_1,\ldots,x_n)}\,, \end{align}

where $$\tilde\sigma^2$$ is the restricted MLE of $$\sigma^2$$ when $$\sigma^2>\sigma_0^2$$.

It can be argued that

$$\tilde\sigma^2=\begin{cases}\sigma_0^2 &,\text{ if }\hat\sigma^2\le \sigma_0^2 \\ \hat\sigma^2&,\text{ if }\hat\sigma^2> \sigma_0^2\end{cases}$$

Therefore,

$$\Lambda(x_1,\ldots,x_n)=\begin{cases}1 &,\text{ if }\hat\sigma^2\le \sigma_0^2 \\ \frac{L(\sigma_0^2\mid x_1,\ldots,x_n)}{L(\hat\sigma^2\mid x_1,\ldots,x_n)}&,\text{ if }\hat\sigma^2> \sigma_0^2\end{cases}$$

When $$\Lambda=1$$, we trivially fail to reject $$H_0$$. For the other case, we reject $$H_0$$ for small values of $$\Lambda$$.

So the critical region is of the form $$\Lambda for some $$c$$ when $$\hat\sigma^2>\sigma_0^2$$.

You would find that $$\Lambdak$$ where $$k$$ is such that $$P_{H_0}\left(\sum\limits_{i=1}^n(X_i-\mu)^2>k\right)=\alpha$$

Since $$\frac{1}{\sigma_0^2}\sum\limits_{i=1}^n(X_i-\mu)^2\sim \chi^2_n$$ under $$H_0$$, we must have $$k=\sigma_0^2\cdot\chi^2_{\alpha,n}$$ where $$\chi^2_{\alpha,n}$$ is the $$(1-\alpha)$$th quantile of a $$\chi^2_n$$ distribution. So the likelihood ratio test rejects $$H_0$$ when $$\sum\limits_{i=1}^n (X_i-\mu)^2>\sigma_0^2\cdot\chi^2_{\alpha,n}$$

This is in fact the same test you would get while searching for a most powerful test using Neyman-Pearson lemma. But if you are asked to derive a likelihood ratio test, you should stick to its own method.

• Thank you for the detailed breakdown and explanation. I was pondering over your earlier comment till your answer came along.. Give me some time to digest it and I'll be back :) – Stoner Nov 10 '19 at 14:16
• Thanks! After reading your answer and my notes again, it makes sense to me now. :) – Stoner Nov 11 '19 at 15:38