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I´m preparing for a lecture in decision theory and I´m a little bit confused by the notation used by my prof.

On the first slide under remark 3.2 point v) its written, that $\beta(\varphi)$ is equal to the power of the test $\varphi$, so the probability of correctly rejecting a false null.

Consequently, $1 - \beta(\varphi)$ was defined as the type II error, so failing to reject a false null.

But isn't it the other way around, that $1-\beta$ is the power of the test and and $\beta$ the type II error? On wikipedia for example its written, that the power is defined as $1-\beta$.

Also, there seems to be a error regarding the graph she wrote on the slide, where $\alpha$ (type I error) was on the x axis and $1-\beta$ on the y axis. The graph shows the convex set of risk points and she was discussing the points $(0,1)$ and $(1,0)$. She basically said something like

"The point $(0,1)$ with type I error $\alpha$ is $0$ and type II error $1-\beta$ is $1$ means the test $\varphi = 0$ and the power of the test is 1"

"Regarding the other situation: The point $(1,0)$ means type I error is $1$ and power of the test is $0$"

Thats basically what she said. The first statement cant be right, since if the power is 1 type II error has to be zero.

I think she messed up the notation at some point and probaly meant, that:

  • $\alpha:$ type I error
  • $1-\alpha:$ condidence level
  • $\beta:$ type II error
  • $1-\beta:$ power of the test

So the definition on remark 3.2 point v) is false, it should be the other way round, right? Consequently, the interpretation of the graph would be following:

"The point $(0,1)$ with

  • type I error is $0$
  • type II error is $0$
  • power is $1$"

"Regarding the other situation: The point $(1,0)$ means

  • type I error is $1$
  • and power of the test is $0$
  • so type II error is $1$"

I hope someone can clear that up for me, since I dont want to learn this wrong. Thanks in advance and happy new year!

slide1 [slide22

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1 Answer 1

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Here is a generic discussion:

Consider a randomized test $\delta$ that assigns $\mathbf x \mapsto (1-\varphi(\mathbf x), \varphi(\mathbf x))$ with $\varphi(\mathbf x)\in [0,1].$ That is, given $\mathbf X=\mathbf x,~\delta$ rejects $\mathcal H_0$ (action $a_1$) with probability $\varphi(\mathbf x)$ and accepts $\mathcal H_0$ (action $a_0$) with probability $1-\varphi(\mathbf x).$

The loss functions are

$$\begin{align}L(\theta, a_0) &= \mathbb I_{\Theta_1}(\theta)=\begin{cases}1, &\theta\in\Theta_1,\\0,&\theta\in\Theta_0,\end{cases}\\ L(\theta, a_1) &= \mathbb I_{\Theta_0}(\theta)\end{align}.$$

The risk function can be formulated as $$R(\theta, \delta) = L(\theta, a_0) + \mathbb E_\theta\varphi(\mathbf X)[L(\theta,a_1)- L(\theta, a_0)]\tag{ 1.1} $$

or succinctly,

$$R(\theta,\delta) = \begin{cases}\alpha_\delta(\theta) := \mathbb E_\theta\varphi(\mathbf X), &\theta\in\Theta_0\\\beta_\delta(\theta) := 1-\mathbb E_\theta\varphi(\mathbf X), &\theta\in\Theta_1\end{cases}.\tag{1.2} $$

The power of the test is $\gamma_\delta(\theta) = \mathbb E_\theta\varphi(\mathbf X)$ for $\theta\in\Theta_1.$

The risk set is given as $S:= \{(R(\theta_0, \delta),~ R(\theta_1, \delta)):\delta\in \mathfrak D\}.$

$\mathbb E_{\theta_0}\varphi(\mathbf X)$ is the size of the test $\delta$ while $\mathbb E_{\theta_1}(1-\varphi(\mathbf X))$ is the probability of error of second kind.

When taken as a function of $\theta,\gamma_\delta(\theta)= \mathbb E_\theta\varphi(\mathbf X)$ is the power function of the test $\delta.$

Some authors use $\beta(\theta)$ for the power function instead (cf. $\rm [III]$).


References:

$\rm [I]$ A Course in Mathematical Statistics and Large Sample Theory, Rabi Bhattacharya, Lizhen Lin, Victor Patrangenaru, Springer-Verlag, $2016,$ sec. $5.1,$ pp. $67-68.$

$\rm[II]$ Mathematical statistics: A Decision Theoretic Approach, Thomas S. Ferguson, Academic Press, $1967,$ sec. $5.1,$ pp. $198-201.$

$\rm [III]$ Testing Statistical Hypotheses, E. L. Lehmann, Joseph P. Romano, Springer Science$+$Business Media, $2005,$ sec. $3.1,$ pp. $57-59.$

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