0
$\begingroup$

In the context of hypothesis test $$H_0:\theta\in \Theta_0$$ $$H_1:\theta\notin \Theta_0$$. Find the relationship between the 0-1 loss defined by $$L(\theta,\delta)= \begin{cases} 1-\delta & \theta \in \Theta_0 \\ \delta & \theta \notin \Theta_0 \end{cases}$$ and the type I and II errors in the Neyman-Pearson approach

What I found:

In the Neyman-Pearson perspective the testing problem is formalized trough a decision space $\mathbb{D}$ restricted to {yes,no} or equivalently $(1,0)$. But what is the direct relationship with type I and II errors?

$\endgroup$
0
$\begingroup$

The direct relationship is as follows, the 0-1 loss is given by $L(y,z) = I(y \not= z)$ where $I$ is the indicator function. If you follow the table here you can see that there are two ways for $z \not= y$ to transpire, corresponding to whether or not the answer is supposed to be ${0,1}$ and the prediction being the alternative one. So, calling these $T_1(y,z)$ and $T_2(y,z)$ we would have that $L(y,z) = T_1(y,z) + T_2(y,z)$ where $T_1(y,z) = 1$ if $y=1, z=0$ and $T_2$ defined similarly.

$\endgroup$

Your Answer

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