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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.