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August 23rd, 2017, 06:11 PM  #1 
Newbie Joined: Aug 2017 From: Cincinnati, OH Posts: 1 Thanks: 0  Expected value of a function of a discrete random variable
I'm new here, so perhaps this is beyond the intended purpose of the board. But, that doesn't mean it's beyond its members. Here is the question: An insurance company sells a oneyear automobile policy with a deductible of 2. The probability that the insured will incur a loss is 0.05. If there is a loss, the probability of a loss of amount N is K/N for N = 1; ... ; 5 and K a constant. These are the only possible loss amounts and no more than one loss can occur. Determine the net premium for this policy. I have determined that K=60/137 Here is a chart of the probability distribution that I found: Pr(N=1)= p(1) = 60/137 Pr(N=2) = 30/137 Pr(N=3) = 20/137 Pr(N=4) = 15/137 Pr(N=5) = 12/137 Here's where I start to get unsure, but think it's still on the right track: E(K/N) = .05K*E(1/N) = K*(1(p(1)) + 1/2(p(2)) + 1/3(p(3)) + ... 1/5(p(5)) I'm not sure what to do with the deductible of 2. The answer is .0314 but I haven't found a way to get to it. Any help? Last edited by skipjack; August 24th, 2017 at 12:00 AM. 
August 24th, 2017, 12:12 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038  

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discrete, expected, function, random, variable 
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