Let $Y = +1(-1)$ if a person succeeds (fails) in the exam. Suppose $Y$ depends only on the two variables $A$ and $B$, where $A$ is the hours of partying time per week and $B$ is an intangible unknown quantity. Suppose that $A$ and $B$ are independent uniform [0,4] random variables and $Y$ = 1 if and only if $AB \leq 4$
(a) find the Bayes decision if no variable is available and if only $A$ is available.
I know that the Bayes Decision looks like this:
\begin{equation} P(Y=1) = \frac{1}{16}\int_{0}^{4} \int_{0}^{min(4, 4/a)} 1dadb \end{equation}
Can someone, please, explain why are we integrating over 1?
In the book "A Probabilistic Theory of Pattern Recognition" the author is integrating over the regressor function; as I understand in our case it is $A*B$ ? Could someone clarify these points and provide a resource where Bayes decision and Loss calculations are explained in a more detailed way?