# Likelihood ratio test and sample statistics

Given a sample $$\mathbf X =(X_1,...,X_n)$$ from a parent random variable $$X$$, Neyman-Pearson's test for two point hypotheses $$H_0$$ and $$H_1$$ is the one defined by the critical region $$C=\left\{\mathbf x \in \mathbb R^n \; \colon\; \lambda(\mathbf x)= \prod_{i=1}^n \frac{f_X(x_i|H_0)}{f_X(x_i|H_1)} \le k_\alpha \right\}$$ where $$k_\alpha$$ is determined by $$\alpha = \mathbb P (\lambda(\mathbf X)\le k_\alpha)\,.$$

Now, in practice, $$C$$ is often defined in terms of a statistic $$T=g(\mathbf X)$$. For example, if \begin{align}H_0 \,\colon \;X\sim N(\mu_0,\sigma^2)\\H_1 \,\colon \;X\sim N(\mu_1,\sigma^2)\end{align} with $$\mu_1>\mu_0$$, then $$C=\left\{\mathbf x \in \mathbb R^n \; \colon\; \bar x = \frac{\textstyle \sum_i x_i}{n}\ge \mu_0 + z_\alpha \frac \sigma {\sqrt n} \right\}$$ where $$z_\alpha$$ is the $$\alpha$$-quantile of the normal distribution. Here the statistic in play is the sample average $$\bar X$$.

I have found that we can get to the same $$C$$ in another way: letting $$f(\cdot;\mu,\sigma^2/n)$$ be the pdf of $$\bar X\sim N(\mu,\sigma^2/n)$$, we can construct the likelihood ratio of $$\bar X$$ via $$\lambda'(\cdot)=\frac {f(\cdot;\mu_0,\sigma^2/n)}{f(\cdot;\mu_1,\sigma^2/n)}$$ and find the critical region as $$C=\left\{ \mathbf x \in \mathbb R^n \; \colon\; \lambda'(\bar x)\le k'_\alpha \right\}$$ with $$\alpha = \mathbb P (\lambda '(\bar X) \le k'_\alpha)\,.$$

Analogously, for the hypotheses \begin{align}H_0 \,\colon \;X\sim N(\mu,\sigma_0^2)\\H_1 \,\colon \;X\sim N(\mu,\sigma_1^2)\end{align} considering the statistic $$Q=\sum_i (X_i-\mu)^2$$ and using the fact that, under $$H_i$$, $$Q/\sigma_i^2 \sim \chi^2(n)$$, we come to the same critical region as the standard Neyman-Pearson definition.

So, is this just luck?

Given any statistic $$T=g(\mathbf X)$$, is the likelihood ratio test constructed for $$T$$ the same as the original one?

Of course not. $$g=0$$ breaks this pattern...

Ok, so are there always some $$g$$'s for which this pattern holds?

This would mean, that in the case of point hypotheses, if you're lucky enough to get the genius idea about what $$g$$ to use, you can use $$T$$ instead of $$\mathbf X$$ to construct the test. This doesn't seem so useful though, because Neyman-Pearson's lemma says your test can't be better than that, so you're just being as good as Neyman-Pearson.

And what about the general case (non-point hypotheses)? Are there cases where the likelihood ratio test constructed by some statistic is even better than the "ordinary" likelihood ratio test?