An insurance policy is written to cover a loss X where X has density function (3x^2)/8 over 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.
I found f(x,y) but when they find the marginal density of Y and begin to solve P[Y>=3], how do they jump to the min (y,2)^2 being 4? If the lower limit is 3, then how do they deem 2 to be the minimum? If anyone can break this down to me "Barney Style" I would appreciate the help.
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.
I found f(x,y) but when they find the marginal density of Y and begin to solve P[Y>=3], how do they jump to the min (y,2)^2 being 4? If the lower limit is 3, then how do they deem 2 to be the minimum? If anyone can break this down to me "Barney Style" I would appreciate the help.
SOA Problem 120