Claims in a portfolio are believed to arise as an Exp(lambda) distribution. There is a retention limit of 1000 in force and claims in excess of 1000 are paid by the reinsurer. The insurer, wishing to estimate lambda, observe a random sample of 100 claims, and finds that the average amount of the 90 claims that do not exceed 1000 is 82.9. There are 10 claims that do exceed the retention limit. Find the MLE for lambda. I proceeded doing the following : L = P(X<=1000)^90 . P(X>1000)^10 = P(X<=1000)^90 . (e^-1000lambda)^10 How to solve the first part? I tried to integrate 82.9e^-82.9x.dx from (-infinity) to 1000 but could not solve it, because e^infinity is coming!
Hi Chandrima, The MLE is equal to - (lambda.*e^-(lambda *x)) multiplied for n= 1 to 90 * (e^-lambda*1000) multiplied for n=91 to 100. Here since we know each of the claim amount below 1000 exactly, we need to use the actual PDF instead of the CDF you have used. This will reduce to - ( lambda^90*e^-(82.9*90*lambda)* (e^- 10000*lambda) = Lambda^90*e^-17461*lambda. If you take logs and differentiate with respect to lambda, you will get the estimate of lambda. Please let me know if any you have any questions
