This is Theorem 8.3.12 (Neyman-Pearson Lemma) in George Casella stat inference. Consider testing $H_0: \theta=\theta_0$ versus $H_1: \theta=\theta_1$, where the pdf pr pmf corresponding to $\theta_i$ is $f(\textbf{x}|\theta_i),i=0,1.$, using a test with rejection region R that satisfies (8.3.1)

\begin{align} \begin{matrix} x\in R & \text{if}& f(\textbf{x}|\theta_1)>kf(\textbf{x}|\theta_0)\\ x\in R^c & \text{if} & f(\textbf{x}|\theta_1)<kf(\textbf{x}|\theta_0) \end{matrix}\tag{8.3.1}\label{8.3.1} \end{align} for some $k \geq 0$, and \begin{align} \alpha=P_{\theta_0}(\textbf{x}\in R)\tag{8.3.2}\label{8.3.2} \end{align}

Then (Sufficiency): Any test that satisfies the above is a UMP level $\alpha$ test.

(Necessity): If there exists a test satisfying (8.3.1) and (8.3.2) with k >0, then every UMP level $\alpha$ test is a size $\alpha$ test (satisfies (8.3.2)) and every UMP level $\alpha$ test satisfies (8.3.1) except perhaps on a set A satisfying $P_{\theta_0}(\textbf{x}\in A)=P_{\theta_1}(\textbf{x}\in A)=0.$

I feel difficulty to understand this Necessity. In the proof, it is clear that any test satisfying (8.3.2) is a size $\alpha$ test. So I don't know why in the Necessity part, we say it again:

If there exists a test satisfying (8.3.1) and (8.3.2) with k >0, then every UMP level $\alpha$ test is a size $\alpha$ test (satisfies (8.3.2))

. And the expressions are different. It says we need to satisfy (8.3.1) and (8.3.2). But don't the truth be we only need to satisfy (8.3.2)?

  • $\begingroup$ @OliverDiaz I am sorry. I tried to see your answer this morning, but then it showed you deleted the answer. Now I am starting to see your answer. $\endgroup$
    – Mariana
    Apr 13, 2021 at 21:01
  • $\begingroup$ @OliverDiaz Thank you. I am still seeing your solution and have asked you 2 questions in chat. $\endgroup$
    – Mariana
    Apr 13, 2021 at 23:01

1 Answer 1


Before trying sketching any arguments, let me introduce some notation and remarks.

Let $T_*(x)$ denote the test described by $(8,3,1)$, that is \begin{align} T_*(x)=\left\{\begin{matrix} 1& \text{if}& f(\textbf{x}|\theta_1)>kf(\textbf{x}|\theta_0)\\ \gamma &\text{if} & f(\textbf{x}|\theta_1)=kf(\textbf{x}|\theta_0)\\ 0 & \text{if} & f(\textbf{x}|\theta_1)<kf(\textbf{x}|\theta_0) \end{matrix} \right. \end{align} where $k$ and $\gamma$ is taken so that $$E_{\theta_0}[T_*(X)]=P_{\theta_0}\big[f(X|\theta_1)>k f(X|\theta_0\big]+\gamma P_{\theta_0}\big[f(X|\theta_1)=k f(X|\theta_0)\big]=\alpha $$ Notice that The function $t\mapsto P_{\theta_0}[f(X|\theta_1)>t f(X|\theta_0)]$ is positive, monotone nonincreasing and right continuous, so such $k$ and $\gamma$ exists and are unique. If one is dealing with continuous distributions, then the bit $\gamma P_{\theta_0}\big[f(X|\theta_1)=k f(X|\theta_0)\big]$ does not appear.

To show necessity, that is, that any UMP test $T_u(x)$ at level $\alpha$ satisfies $T_u(x)=T_*(x)$ in $\{x:f(x|\theta_1)\neq f(x|\theta_0)\}$, we should understand first what is so special about $T_*(X)$. Once this is established, necessity follows from some basic measure theoretic arguments that I will only explain towards the end of my answer.

First we reproduce the arguments that show why $T_*(x)$ is a UMP at level $\alpha$, that is, we show that for any other test $T(x)$ with $E_{\theta_0}[T(X)]\leq\alpha$, we have that $E_{\theta_1}[T_*(X)]\geq E_{\theta_1}[T(X)]$.

Here is more or less how the argument works.

Let $T(x)$ by any other test with power at most $\alpha$, that is $E_{\theta_0}[T(X)]\leq\alpha$. Recall that tests take only values between $0$ and $1$. Notice that:

  1. If $T_*(x)-T(x)>0$, then $T_*(x)>0$ (for $T(x)\geq0$ for all $x$); hence $f(\textbf{x}|\theta_1)\geq kf(\textbf{x}|\theta_0)$.
  2. If $T_*(x)-T(x)<0$, then $T_*(x)<1$ (for $T(x)\leq1$ for all $x$); hence $f(\textbf{x}|\theta_1)\leq kf(\textbf{x}|\theta_0)$.

This means that $$ \big(T_*(x)-T(x)\big)\big(f(x|\theta_1)-kf(x|\theta_0)\big)\geq0 $$ for all $x$. Integration gives \begin{align} \int_X\big(T_*(x)-T(x)\big)\big(f(x|\theta_1)-kf(x|\theta_0)\big)\,dx\geq0 \end{align} simplifying the expression on the left-hand side gives \begin{align} E_{\theta_1}[T_*(X)-T(X)]&=\int_X\big(T_*(x)-T(x)\big)f(x|\theta_1)\,dx\\ &\geq k\int_X (T_*(x)-T(x)\big)f(x|\theta_0)\,dx=kE_{\theta_0}[T_*(X)-T(X)]\\ &=k\big(\alpha-E_{\theta_0}[T(X)]\big)\geq0 \end{align} Hence $$ E_{\theta_1}[T_*(X)]\geq E_{\theta_1}[T(X)]$$

In other words, $T_*$ is more powerful that $T$.

Observation: The key parts of the argument above are contained (1) and (2).

We are now ready to argue for necessity, that is, that any test $T_u$ that is UMP at level $\alpha$, must be equal to $T_{*}(X)$ in $\{x: f(x|\theta_1)\neq k f(x|\theta_0)\}$.

Suppose now that that $T_u$ is another UMP of power $\alpha$; that is $E_{\theta_0}[T_u(X)]=\alpha$, and $E_{\theta_1}[T_u(X)]\geq E_{\theta_1}[T(X)]$ for any other feasible test $T$. Then, since $T_*(X)$ is UMP, we must have that $E_{\theta_1}[T_*(X)]=E_{\theta_1}[T_u(X)]$. Consider the set $$ A:=\{x:T_*(x)\neq T_u(x)\}\cap\{x:f(x|\theta_1)\neq k f(x|\theta_0)\}$$ The arguments used in (1) and (2) imply that \begin{align} \begin{matrix} \big(T_*(x)-T_{u}(x)\big)\big(f(x|\theta_1)-k f(x|\theta_0)\big)>0&\text{if} &x\in A\\ \big(T_*(x)-T_{u}(x)\big)\big(f(x|\theta_1)-k f(x|\theta_0)\big)=0&\text{if} &x\in X\setminus A \end{matrix} \end{align} Integration gives \begin{align} \int_A\big(T_*(x)-T_{u}(x)\big)\big(f(x|\theta_1)-k f(x|\theta_0)\big)\,dx &=\int_X\big(T_*(x)-T_{u}(x)\big)\big(f(x|\theta_1)-k f(x|\theta_0)\big)\,dx\\ &=E_{\theta_1}[T_*(X)-T_{u}(X)]-k E_{\theta_0}[T_*(X)-T_{u}(X)]\\ &=0 \end{align} Since $\big(T_*(x)-T_{u}(x)\big)\big(f(x|\theta_1)-k f(x|\theta_0)\big)>0$ for all $x\in A$, then it must be that $A$ is negligible (i.e. $\int_A\,dx=0$). Therefore $P_{\theta_0}(A)=P_{\theta_1}(A)=0$, that is $T_*(X)=T_{u}(X)$ $\{P_{\theta_1},P_{\theta_0}\}$-almost surely.

The last bit is based on a couple of basic measure theory facts:

  1. If $f\geq0$ and $\int f\,d\mu=0$, then $\mu(\{x:f(x)>0\})=0$. That is $f$ must be $0$ almost surely (with respect to the measure $\mu$).
  2. If $\mu$ is a finite measure with a density function respect to another ($\sigma$-finite) measure $\nu$, then $\nu(A)=0$ implies that $\mu(A)=0$. (this is related to a deep result called Radon-Nikodym theorem).
  • $\begingroup$ At the very beginning, to find out $\gamma$, I think the notation should also include $E_{\theta_0, \theta_1}[T_*(X)]$, and also $P_{\theta_0, \theta_1}$. $\endgroup$
    – Mariana
    Apr 13, 2021 at 21:10
  • $\begingroup$ But at the very beginning, we don't know $\alpha$ is just the lever. We only know $\alpha$ is the the expectation of $T_*(X)$, which is in terms of $\theta_0, \theta_1$. $\endgroup$
    – Mariana
    Apr 13, 2021 at 21:20
  • $\begingroup$ @Mariana: Basically the whole problem of hypothesis testing is to find a test $T(X)$ that solves the following optimization problem: \begin{align} T^*&=\operatorname{arg}\sup_{\theta\in\Theta_1}E_{\theta}[T(X)]\\ &\quad\text{subject to}\\ &\sup_{\theta_0}E_{\theta_0}[T(X)]\leq\alpha \end{align} $\endgroup$ Apr 13, 2021 at 21:23
  • $\begingroup$ "First we reproduce the arguments that show why $T_*(x)$ is a UMP at level $\alpha$, that is, we show that for any other test $T(x)$ with $E_{\theta_0}[T(X)]\leq\alpha$," Can you explain why the test is of level $\alpha$ is equivalent to saying $E_{\theta_0}[T(X)]\leq\alpha$? I only know a test with power function $\beta(\theta)$ is a level $\alpha$ test if $sup_{\theta \in \Theta_0} \beta(\theta) \le \alpha$ $\endgroup$
    – Mariana
    Apr 13, 2021 at 21:32
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Mariana
    Apr 13, 2021 at 21:34

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.