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We have a normal distribution with zero mean. We have two hypotheses for the variance $\sigma^2$:

$$H_0: \sigma^2=\sigma_0^2$$ $$H_1: \sigma^2=\sigma_1^2$$

We make $n$ independent observations $X_1, X_2, ..., X_n$. The likelihood ratio is

$$L(x)=\left(\frac{\sigma_0}{\sigma_1}\right)^ne^{\sum_{i=1}^nx_i^2\left(\frac1{2\sigma_0^2}-\frac1{2\sigma_1^2}\right)}$$

Comparing $L(x)$ to $\xi$ is the same as comparing $\sum_{i=1}^nx_i^2$ to $\gamma=\frac{n\ln\left(\frac{\sigma_1}{\sigma_0}\right)+\ln\xi}{\frac1{2\sigma_0^2}-\frac1{2\sigma_1^2}}$ so the rejection region is $\{\left(x_1,x_2,...,x_n\right)|\sum_{i=1}^nx_i^2>\gamma\}$. We note that $\Bbb E[X_i^2]=\sigma^2$ and $\text{var}(X_i^2)=2\sigma^4$ so $\Bbb E[\sum_{i=1}^nX_i^2]=n\sigma^2$ and $\text{var}(\sum_{i=1}^nX_i^2)=2n^2\sigma^4$.

Now we set the false rejection probability $\Bbb P\left(\sum_{i=1}^nX_i^2>\gamma;H_0\right)$ to $\alpha$ and standardize to get $\Bbb P\left(\frac{\sum_{i=1}^nX_i^2-n\sigma_0^2}{\sqrt2n\sigma_0^2}>\frac{\gamma-n\sigma_0^2}{\sqrt2n\sigma_0^2};H_0\right)=\alpha$. By the central limit theorem, we have $\Bbb P\left(Z>\frac{\gamma-n\sigma_0^2}{\sqrt2n\sigma_0^2}\right)=\alpha$, where $Z$ is the standard normal random variable. This gives us $\gamma=n\sigma_0^2\left(1+\sqrt2\left(1-\Phi^{-1}\left(1-\alpha\right)\right)\right)$, where $\Phi^{-1}$ is the inverse of the cumulative distribution function of $Z$. We note that $\gamma$ increases linearly with $n$.

The false acceptance probability is $\Bbb P\left(\sum_{i=1}^nX_i^2\le\gamma;H_1\right)$, which again by the central limit theorem equals $\Phi\left(\frac{\gamma-n\sigma_1^2}{\sqrt2n\sigma_1^2}\right)=\Phi\left(\frac{n\sigma_0^2\left(1+\sqrt2\left(1-\Phi^{-1}\left(1-\alpha\right)\right)\right)-n\sigma_1^2}{\sqrt2n\sigma_1^2}\right)=\Phi\left(\frac{\sigma_0^2\left(1+\sqrt2\left(1-\Phi^{-1}\left(1-\alpha\right)\right)\right)-\sigma_1^2}{\sqrt2\sigma_1^2}\right)$.

This means that given $\sigma_0$ and $\sigma_1$, the false acceptance probability depends only on $\alpha$ and does not improve with increasing sample size. How can we explain this?

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  • $\begingroup$ Please check your algebra. Why did the factor of $n/\sqrt{n}$ disappear at the end? $\endgroup$
    – whuber
    Nov 4, 2018 at 15:51

1 Answer 1

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First, the rejection region is $\left\{\left(x_1,x_2,...,x_n\right)|\sum_{i=1}^nx_i^2>\gamma\right\}$ when $\sigma_0^2 \le \sigma_1^2$, otherwise, the rejection region is $\left\{\left(x_1,x_2,...,x_n\right)|\sum_{i=1}^nx_i^2<\gamma\right\}$.

Second: $\text{var}(\sum_{i=1}^nX_i^2)=2n^2\sigma^4$ is incorect. It should be $\text{var}(\sum_{i=1}^nX_i^2)=2n\sigma^4$

Let $Z_0 =\Phi^{-1}(1-\alpha)$ Assume $\sigma_0^2 \le \sigma_1^2$, and follow the same steps,

$$\Phi\left(\frac{\gamma-n\sigma_1^2}{\sqrt{2n}\sigma_1^2}\right)=\Phi\left(\frac{\sigma_0^2}{\sigma_1^2}Z_0 + \frac n{\sqrt{2n}}\left(\frac{\sigma_0^2-\sigma_1^2}{\sigma_1^2} \right)\right)$$

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  • $\begingroup$ Thanks for pointing out. I also found that the formula for $\gamma$ in my question is incorrect, even with $\text{var}\left(\sum_{i=1}^nX_i^2\right)=2n^2\sigma^4$. $1-\Phi^{-1}\left(1-\alpha\right)$ should have been $\Phi^{-1}\left(1-\alpha\right)$. With the correct $\text{var}\left(\sum_{i=1}^nX_i^2\right)=2n\sigma^4$, $\gamma=n\sigma_0^2+\sqrt{2n}\sigma_0^2\Phi^{-1}\left(1-\alpha\right)$. So $Z_0$ in your answer should be $\Phi^{-1}\left(1-\alpha\right)$. $\endgroup$
    – W. Zhu
    Nov 5, 2018 at 1:56

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