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#1




p exam problem continuous random variables
Hi guys,
I'm having trouble resolving the following problem. Please help me. A group insurance policy covers the medical claims of the employees of a small company. The value, V, of the claims made in one year is described by V = 100000Y where Y is a random variable with density function f(x) =k(1−y)4 0 < y < 1 0 otherwise where k is a constant. What is the conditional probability that V exceeds 40,000, given that V exceeds 10,000? 
#3




First, you have to find the value of k and this is done by equating the density function i.e. f(x) to 1. You will get k=5
Then you find P(100,000Y>40,000)100,000Y>10,000) which is same as P(Y>0.4Y>0.1) I hope you can go on from here. Your final answer should be 0.132 
#4




Quote:

#6




I'm making a mistake but I don't know where.
I find k by saying that the integral of k*(1y)^4 from 0 to 1 is equal to 1. That means that the limit of k/5*(1y)^5 ( y goes towards 1) minus the limit of k/5*(1y)^5 ( y goes towards 0) is equal to 1. Since for y goes towards 1 the limit of f(x) is 0, minus the limit of k/5( 1y)^5 with y goes towards 0 is equal to 1. That gives me k=5. Where did I make a mistake? thank you for the help. Last edited by efast; 11162017 at 11:03 PM.. 
#8




You might find this thread helpful.

#9




Thank you Gandalf and Z3ta.
what is the rule to integrate k*(1y)^4? Last edited by efast; 11172017 at 11:11 AM.. 
#10




Thank God, I got it using the substitution rule.

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