I'm start to learn Decision Theory and I'm trying to solve (analytically) the exemple 2 from Berger, pag. 5-6 (James O. Berger - Statistical Decision Theory - 1980).
I can't understand the result (how the "n" - the sample size) appear on the result). I'm using the priori given from the text.
Summarized, the enunciated is:
A random sample of "n" transistors is chosen from the shipment and tested. Based upon X, the number of defective transistors in the sample, the shipment will be accepted or rejected. Thus there are two possible actions: $a_1$ accept the shipment or $a_2$ reject the shipment. If "n" is small compared to the shipment size, X can be assumed to have a $\mathcal B(n, \theta)$ distribution, where $\theta$ is the proportion of defective transistors in the shipment The company determines that their loss function is: $$ L(\theta,a_i) = \begin{cases} 10\theta & if & \delta(a_1) \\ 1) & if & \delta(a_2) \\ \end{cases} $$ Based on the past, the radio company data reveals that $\theta$ was distributed according to a $\mathcal Be(0.05, 1)$ distribution. Hence $\pi(\theta)=(0.05)\theta^{-0.95}I_{[0,1]}(\theta)$.
My question is: What is the best Bayes rule?
The answer from the text given is: $$ \delta(x) = \begin{cases} a_1 & if & \frac{x}{n} \leq 0.05 \\ a_2 & if & \frac{x}{n} > 0.05 \\ \end{cases} $$ I tryed: $ $$ \rho(\pi,a) = E_{\pi} L(\theta,a)= \int_0^x L(\theta,a)\pi(\theta)d\theta = \int_0^x 10(\theta) (0.05)\theta^{-0.95}I_{[0,1]}(\theta) d\theta $$
But doesn't have "n" here?
Thanks