Thanks. I don't understand how to calculate the integral for a Bayesian Expected Loss. The problem is from Berger 1985 Stat Decision Theory and Bayesian Analysis page 8.
Example 1. Assume no data is obtained, so that the believed distribution of $\theta_2$ is simply $\pi(\theta_2)=10 I_{0.1,0.2} (\theta_2) d\theta_2$ . Then $$\rho (\pi^*, a)=E^{\pi^*} L(\theta,a)= \int_{\Theta} L(\theta,a) dF^{\pi^*} (\theta) $$
$$=\int^a_0 2(a-\theta_2) 10 I_{0.1,0.2} (\theta_2)d\theta_2 +\int^1_a (\theta_2-a) 10 I_{0.1,0.2} (\theta_2)d\theta_2 $$ The resulting is a step function: $$0.15-a, \text{ if } a\le0.1,$$ $$15a^2-4a+0.3, \text{ if } 0.1\le a\le 0.2,$$ $$2a-0.3, \text{ if } a\ge 0.2$$
The part I am confused is how to go from the integral to the step function. I have done Stieljes-R. Integrals before, but I am not quite sure how to address this. I have tried, unsuccessfully, to solve it, but I am not sure how to proceed.
Thank you very much.
Since $I_{0.1,0.2}$ is zero outside of $(0.1,0.2)$, and $1$ inside this range, you can change the limits of the integrals and drop the $I_{0.1,0.2}$ in the integrands. Thus for example, when $a < 0.1$, $$\int^a_0 2(a-\theta_2) 10 I_{0.1,0.2} (\theta_2)d\theta_2 +\int^1_a (\theta_2-a) 10 I_{0.1,0.2} (\theta_2)d\theta_2 \\= 0 + \int^{0.1}_{0.2} (\theta_2-a) 10 I_{0.1,0.2} (\theta_2)d\theta_2 \\= 10\left[ \frac{\theta_2^2}{2}-a\theta_2 \right]^{0.2}_{0.1} \\= 10\left[ \frac{0.04}{2} - 0.2a - \frac{0.01}{2} + 0.1a \right] \\= 0.15-a$$ Then when $0.1 \leq a \leq 0.2$, $$\int^a_0 2(a-\theta_2) 10 I_{0.1,0.2} (\theta_2)d\theta_2 +\int^1_a (\theta_2-a) 10 I_{0.1,0.2} (\theta_2)d\theta_2 \\= 0 + \int^{0.1}_{0.2} (\theta_2-a) 10 I_{0.1,0.2} (\theta_2)d\theta_2 \\= \int_{0.1}^a 20(a-\theta_2)d\theta_2 + \int_a^{0.2} 10(\theta_2 - a) d\theta_2 \\= \left[ 20 a \theta_2 - 10 \theta_2^2 \right]^a_{0.1} + \left[ 5 \theta_2 - 10a\theta_2 \right]^{0.2}_a \\= 20a^2-10a^2-2a+0.1+0.2-2a-5a^2+10a^2 \\= 15a^2 - 4a +0.3 $$ I leave the case of $a > 0.2$, where only the first integral is non-zero, to you.