# Alternates to the Neyman-Pearson Lemma?

Let $$X_1, X_2, \ldots, X_n$$ be a random sample from a distribution with unknown parameter $$\theta$$. Consider two hypotheses:

• Null hypothesis $$H_0$$: $$\theta = \theta_0$$
• Alternative hypothesis $$H_1$$: $$\theta = \theta_1$$ ( $$\theta_1 \neq \theta_0$$)

Let $$f(x|\theta)$$ denote the probability density function (or probability mass function) of $$X_1, X_2, \ldots, X_n$$.

The Neyman-Pearson lemma states that the likelihood ratio test (LRT) is the most powerful test at level $$\alpha$$, where the LRT rejects $$H_0$$ if and only if:

$$\frac{L(\theta_1 | \mathbf{x})}{L(\theta_0 | \mathbf{x})} > k,$$

where $$\mathbf{x} = (x_1, x_2, \ldots, x_n)$$ is the observed sample, $$L(\theta|\mathbf{x}) = \prod_{i=1}^n f(x_i|\theta)$$ is the likelihood function, and $$k$$ is a constant chosen such that the test has level $$\alpha$$.

My Question: This might sound like a rhetorical question - but what other possible tests could exist? The Neyman-Pearson Lemma is said to be very important as it shows us what is the most powerful statistical test - but it seems to me that the test indicated by the Neyman-Pearson is practically the only test that exists. What other tests can be considered as competitors?

Thanks!

• Any function that maps $x\in \mathcal{X}^n \mapsto\{\text{accept, reject}\}$ is a test, so a lot of tests exist. Commented Feb 24, 2023 at 6:55
• At any level $\alpha$ any statistic $T(X)$ with $E_{\theta_0}[T(X)]\leq \alpha$ is a valid test. It just so happened that if $T_p(X)$ Pearson's then $E_{\theta_1}[T_p(X)]\geq E_{\theta_1}[T(X)]$. Commented Feb 24, 2023 at 13:41
• @ user51547: thank you for your reply! For argument sake - could you please give me an example of such a test? Thanks! Commented Feb 24, 2023 at 15:37
• @stats_noob e.g. the test where you always accept the null regardless of the data....not a very good test, but a test nonetheless! Commented Feb 25, 2023 at 9:09
• @stats_noob You could look at $T(X)=\bar X$, and decide to reject the null depending on how close it is to $E_{\theta_1} X$ compared to $E_{\theta_0} X$ (provided these exist and are different). Commented Feb 25, 2023 at 17:40

This is an belated answer to this common question (in Statistics). Recall that if $$\{P_\theta:\theta\in \Theta\}$$ is a population (family of probability distributions on a common space $$(\mathbb{X},\mathscr{B})$$, then a testing statistic $$T(X)$$ for the test $$H_0: \theta\in\Theta_0$$ v.s. $$H_1: \theta\in\Theta_1$$, where $$\Theta=\Theta_0\cup\Theta_1$$, $$\emptyset=\Theta_0\cap \Theta_1$$, is a uniformly most powerful test(UMP) of size $$0\leq\alpha\leq 1$$ if

1. $$\sup_{\theta\in\Theta_0}\mathbb{E}_\theta[T(X)]=\alpha$$
2. If $$\tilde{T}(X)$$ is another statistic with $$0\leq \tilde{T}(X)\leq 1$$ and $$\sup_{\theta\in\Theta_0}\mathbb{E}_\theta[\tilde{T}(X)]\leq\alpha$$, then $$\mathbb{E}_{\theta}[\tilde{T}(X)]\leq \mathbb{E}_\theta[T(X)]$$ for all $$\theta\in\Theta_1$$.

In particular, if $$\Theta=\{\theta_0,\theta_1\}$$, and $$P_{\theta_j}=f_j\,d\mu$$, $$j\in\{0,1\}$$, $$f_0\leq f_1$$, where $$\mu$$ is a $$\sigma$$-finite measure on $$(\mathbb{X},\mathscr{B})$$ and $$1\leq \alpha\leq 1$$, Neyman-Paerson's theorem states that the test statistic $$\psi(X)$$ given by \begin{align} \psi(X) = \left\{\begin{array}{lcr} 1 & \text{if} & f_1(X)>k f_0(X)\\ \gamma & \text{if} & f_1(X)=k f_0(X)\\ 0 & \text{if} & f_1(X) where $$k$$ and $$\gamma$$ are chosen so that $$\alpha=\mathbb{E}_0[\psi(X)]$$, is a UMP test of level $$\alpha$$ for the test $$H_0: \theta=\theta_0$$ versus $$H_1:\theta=\theta_1$$. Furthermore, if $$T(X)$$ is another statistic with $$0\leq T(X)\leq 1$$, we have that $$T(X)$$ is UMP of level $$\alpha$$ iff $$\mathbb{E}_0[T(X)]=\alpha$$ and on $$\{f_1(X)\neq k f_0(X)\}$$ $$T(X)=\psi(X)$$ $$\mu$$-almost everywhere.

The test $$\psi(X)$$ is a randomized test when $$\mu(f_1(X)=k f_0(X))>0)$$. Neyman-Pearson's theorem implies that outside $$\{f_(X)=k f_0(X)\}$$, which could be empty $$\mu$$-almost everywhere, any UMP test at level $$\alpha$$ coincides with $$\psi(X)$$. However this does not imply that the Neyman-Pearson statistic is the only UMP. Here is one example.

Example: Suppose $$0<\theta_0<\theta_1$$ and $$f_{\theta_j}(\mathbf{x})=\frac1{\theta^n_j}\mathbb{1}_{(0,\theta_j)}(x_1)\cdot\ldots\cdot\mathbb{1}_{(0,\theta_j)}(x_n)=\frac1{\theta^n_j}\mathbb{1}_{(0,\theta_j)}(x_{(n)})$$ where $$x_{(n)}=\max_{1\leq k\leq n}x_k$$. This is the case of $$n$$-i.i.d. uniform random variables on $$(0,\theta_j)$$, $$j\in\{0,1\}$$. The dominating measure $$\mu$$ is Lebesgue's measure on the line. Notice that $$\frac{f_1(\mathbf{x})}{f_0(\mathbf{x})}=\Big(\frac{\theta_0}{\theta_1}\Big)^n\mathbb{1}_{(0,\theta_0)}(x_{(n)})+\infty\cdot\mathbb{1}_{[\theta_0,\theta_1)}(x_{(n)})$$ It is easy to check that the size $$\alpha$$ Neyman-Peason statistic \eqref{one} for $$H_0:\theta=\theta_0$$ versus $$H_1:\theta=\theta_1$$ is given by \begin{align} \psi(X) = \left\{\begin{array}{lcr} 1 & \text{if} & X_{(n)}>\theta_0\\ \alpha & \text{if} & X_{(n)}<\theta_0 \end{array} \right. \tag{2}\label{two} \end{align} The power of $$\psi(X)$$ against $$\theta_1$$ is given by $$E_1[\psi(X)]=1-(1-\alpha)n\theta^{-n}_0\int^{\theta_1}_{\theta_0}x^{n-1}\,dx=1-\big(\frac{\theta_0}{\theta_1}\big)^n(1-\alpha)$$

Observe that the ration $$f_1/f_0$$ is a monotone nondecreasing function of the (sufficient) statistic $$X_{(n)}$$. Define the test statistic \begin{align} \psi(X) = \left\{\begin{array}{lcr} 1 & \text{if} & X_{(n)}>c\\ 0 & \text{if} & X_{(n)} where $$\alpha=\mathbb{P}_0[X_{(n)}>c]$$, that is, $$c$$ is a solution to $$\alpha=n\theta^{-n}_0\int^{\theta_0}_c x^{n-1}\,dx$$, that is $$c=(1-\alpha)^{1/n}\theta_0$$. The power of the test $$\tilde{\psi}(X)$$ against $$\theta_1$$ is $$E_1[\tilde{\psi}(X)]=n\theta^{-n}_1\int^{\theta_1}_cx^{n-1}\,dx=1-\big(\frac{\theta_0}{\theta_1}\big)^n(1-\alpha)$$

It follows that $$\tilde{\psi}$$ is also a UPM test. I leave it to the OP to check that $$\psi(X)$$ and $$\tilde{\psi}(X)$$ differ on a set of (Lebesgue) measure $$\mu(\psi(X)\neq \tilde{\psi}(X))>0$$. However, on $$\{f_1(X)\neq kf_0(X)\}=\{X_{(n)}<\theta_0\}$$ we do have that $$\tilde{\psi}(X)=\psi(X)$$ $$\mu$$-a.s. as stated by the theorem.