SOA 120

[SPratt]

An insurance policy is written to cover a loss X where X has density function

f(x)= (3/8)x^2 for 0<x<2

The time (in hours) to process a claim of size x, where 0 ≤ x ≤ 2, is uniformly distributed
on the interval from x to 2x.

Calculate the probability that a randomly chosen claim on this policy is processed in three hours or more.

So given this f(t|X=x) = 1/x

f(x,t)= (1/x)(3x^2/8) =3x/8

When I graph the function I see why you would integrate from t/2 to 2 for dx and from 3 to 4 for dt, but why can't you integrate from 1.5 to 2 for dx and from 3 to 2x for dt?

[Actuary2011]

I tried it both ways and they both give the same answer. So I guess the answer to your question is that there is no reason why you can't do it the other way. Were you getting different answers?

[SPratt]

Thanks, got the same answer the second time around. The calc part of the exam is definitely where I make the most mistakes.

[Shaun2357]

An insurance policy is written to cover a loss X where X has density function

f(x)= (3/8)x^2 for 0<x<2


The time (in hours) to process a claim of size x, where 0 ≤ x ≤ 2, is uniformly distributed
on the interval from x to 2x.

Calculate the probability that a randomly chosen claim on this policy is processed in three hours or more.

So given this f(t|X=x) = 1/x

f(x,t)= (1/x)(3x^2/8) =3x/8

When I graph the function I see why you would integrate from t/2 to 2 for dx and from 3 to 4 for dt, but why can't you integrate from 1.5 to 2 for dx and from 3 to 2x for dt?

I think this may be to much math. But look up on Complex Analysis. It is essential one step up from Real Analysis ( Advanced Calculus ) . More specifically, look up integration. But not improper integral. In Complex Analysis you have some pretty good examples on how to double integrate function as your following a particular path. It you are very interested tgen check out how they have formulas for transferring any "real" function into a complex function then use the residue formula to find its integral.