E=P[X>100]*E[N]*E[X-100|X>100] I agree with this. However E[X-100|X>100] term is made up of another term which has a P[X>100] on the denominator, I agree with this. So the P[X>100] terms cancel leaving just the integral in the numerator of the E[X-100|X>100] term times by E[N]. However the ASET solutions the P[X>100] seems to come back again (bottom P28). I thought it cancelled? Maybe it's getting late but im going mad trying to see how it came back.
Hello I don't think ASET has added it back in. If L=lamda and A=alpha.... Note that, in order to derive the final expression for E at the foot of p28, we have taken the expression for E from half way down the page and: 1) taken L^A outside the intergal 2) multiplied inside integral by (L+100)^A and divided by the same amount outside the integral. We then use the expression for the mean of a Pareto (but with parameters A and (L+100) to evaluate the expression inside integral. I think this is OK. Does this help? I hope so.
