# Why is the false acceptance probability not improving with increasing sample size?

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?

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

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,

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

• 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 $\Phi^{-1}\left(1-\alpha\right)$. Nov 5, 2018 at 1:56