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  • Is there a trick to this problem?

    An insurance company sells two types of auto insurance policies: Basic and Deluxe. The time until the next Basic Policy claim is an exponential random variable with mean 6 days. The time until the next Deluxe Policy claim is an independent exponential random variable with mean 7 days. What is the probability that the next claim will be a Deluxe Policy claim?

    a) 0.462
    b) 0.562
    c) 0.362
    d) 0.529
    e) 0.329

    I know how to do this but I can't get the double integral to work out...
    As I understand it:
    x - time until Basic claim
    y - time until Deluxe claim
    Find P(y < x)
    since the two functions are independent, we have:
    f_XY(x,y) = f_X(x) * f_Y(y) = (1/6)e^(-x/6) * (1/7)e^(-y/7)

    then you would just double integrate f(x,y) from
    int(0 to infinity) int(0 to x) of f_XY(x,y) dydx

    I don't know why I am having such a hard time integrating this... I came up with the answer .4487...
    Is there a trick to solving this other than just brute force. Any chance you can use the Survival function? :embarrassed:

  • #2
    Originally posted by Deltad View Post
    An insurance company sells two types of auto insurance policies: Basic and Deluxe. The time until the next Basic Policy claim is an exponential random variable with mean 6 days. The time until the next Deluxe Policy claim is an independent exponential random variable with mean 7 days. What is the probability that the next claim will be a Deluxe Policy claim?

    a) 0.462
    b) 0.562
    c) 0.362
    d) 0.529
    e) 0.329

    I know how to do this but I can't get the double integral to work out...
    As I understand it:
    x - time until Basic claim
    y - time until Deluxe claim
    Find P(y < x)
    since the two functions are independent, we have:
    f_XY(x,y) = f_X(x) * f_Y(y) = (1/6)e^(-x/6) * (1/7)e^(-y/7)

    then you would just double integrate f(x,y) from
    int(0 to infinity) int(0 to x) of f_XY(x,y) dydx

    I don't know why I am having such a hard time integrating this... I came up with the answer .4487...
    Is there a trick to solving this other than just brute force. Any chance you can use the Survival function? :embarrassed:
    I guess you should finish up your computation. The integral is not so difficult.

    Yes, there is a trick (better method). Please tell me if the answer is 6/13. Then I will show you.

    ctperng

    Comment


    • #3
      Originally posted by Deltad View Post
      An insurance company sells two types of auto insurance policies: Basic and Deluxe. The time until the next Basic Policy claim is an exponential random variable with mean 6 days. The time until the next Deluxe Policy claim is an independent exponential random variable with mean 7 days. What is the probability that the next claim will be a Deluxe Policy claim?

      a) 0.462
      b) 0.562
      c) 0.362
      d) 0.529
      e) 0.329

      I know how to do this but I can't get the double integral to work out...
      As I understand it:
      x - time until Basic claim
      y - time until Deluxe claim
      Find P(y < x)
      since the two functions are independent, we have:
      f_XY(x,y) = f_X(x) * f_Y(y) = (1/6)e^(-x/6) * (1/7)e^(-y/7)

      then you would just double integrate f(x,y) from
      int(0 to infinity) int(0 to x) of f_XY(x,y) dydx

      I don't know why I am having such a hard time integrating this... I came up with the answer .4487...
      Is there a trick to solving this other than just brute force. Any chance you can use the Survival function? :embarrassed:
      Hey Deltad, your on the right track. What you do is continue with the integration and you should get something like this:

      integral (0 to infinity) 1/6*e^(-x/6)-1/6*e^(-13*x/42) dx.

      The first part of the function integrates to 1 (original function) whereas in the second part you multiply by (13/42)*(42/13) and factor out the 1/6 to get

      1- (1/6) integral(0 to infinity) (42/13) * (13/42)e^(-13*x/42) dx

      Note the second part is an exponential function and evaluates to 1 when integrated from 0 to infinity. Consequently you have:

      1-(1/6)*(42/13)*(0-(-1))=1-7/13=6/13


      JGET

      Comment


      • #4
        well, I took this question off of saab website so I have no idea what the answer is. it was a random generated question so a good chance I wont see it again for some time.

        Comment


        • #5
          oh... my... gosh...
          this is the stupidest mistake ever made... I got to the second to last step and instead of taking down (-13/42) I did (-13/43)!!! ARGGGGH. how the '''' did that even happen...???
          1 - (1/6)(43/13) = .4487

          I did this problem TWICE and did the same darn mistake!!!
          OK ctperng, can you tell me a shortcut/trick to this so I wont make this stupid mistake again

          Comment


          • #6
            See next post.
            Last edited by ctperng; August 1 2007, 01:13 PM. Reason: deleting because of repetition

            Comment


            • #7
              Originally posted by Deltad View Post
              oh... my... gosh...
              this is the stupidest mistake ever made... I got to the second to last step and instead of taking down (-13/42) I did (-13/43)!!! ARGGGGH. how the '''' did that even happen...???
              1 - (1/6)(43/13) = .4487

              I did this problem TWICE and did the same darn mistake!!!
              OK ctperng, can you tell me a shortcut/trick to this so I wont make this stupid mistake again
              All right, there is no trick, but a simple observation: we just need to integrate the density function of the first order statistic, f_Y(y)*S_X(y) (here I am keeping your notations and S_X is survival function for X) , from 0 to infinity. Therefore you get
              int(0 to infinity) of (1/7)e^(-y/7)*e^(-y/6)
              = int(0 to infinity) of (1/7)e^(-13/42y) = (1/7)*(42/13) = 6/13.

              By the way, instead of doing the integral

              int(0 to infinity) int(0 to x) of f_XY(x,y) dydx,

              you can do

              int(0 to infinity) int(y to infinity) of f_XY(x,y) dxdy (by Fubini's theorem: changing order of integration doesn't matter!).

              Then you get

              int(0 to infinity)(1/7)e^(-y/7)*{int(y to infinity}(1/6)e^(-x/6)dx}dy.

              Note that inside the braces you have the survival function S_X(y), therefore this shows that the beginning observation is true.

              ctperng

              PS. One has to be careful, for general context of order statistics involving more than 2 random variables, the assumption is that each pdf looks like the same. But in our case here, the two pdf's are different. Since we have only two orders to compare, the general result applies to this case with slight modifcation only.

              Comment


              • #8
                is it just 6/(6+7)? if the question asked 5 days and 7 days, would the answer be 5/12? i don't know for sure, but i think it would be.

                Comment


                • #9
                  Originally posted by brothertupelo View Post
                  is it just 6/(6+7)? if the question asked 5 days and 7 days, would the answer be 5/12? i don't know for sure, but i think it would be.
                  Yes, that would be a trick. Now you got it. What I provided is honest work, but you got the trick!

                  ctperng

                  Comment


                  • #10
                    Originally posted by brothertupelo View Post
                    is it just 6/(6+7)? if the question asked 5 days and 7 days, would the answer be 5/12? i don't know for sure, but i think it would be.
                    Oh man... are you kidding me? lol... will this work for all exponential function? can anyone prove this?
                    anyway, the "trick" doesn't make sense to me... it would make more sense if it was 7/(6+7) since you want the probability that the Deluxe happens first.

                    I didn't get to order statistic since my friend who passed the test already said it wont be on exam P, I can see that it'll help though.

                    Thank you ctperng!

                    Comment


                    • #11
                      Originally posted by Deltad View Post
                      Oh man... are you kidding me? lol... will this work for all exponential function? can anyone prove this?
                      anyway, the "trick" doesn't make sense to me... it would make more sense if it was 7/(6+7) since you want the probability that the Deluxe happens first.

                      I didn't get to order statistic since my friend who passed the test already said it wont be on exam P, I can see that it'll help though.

                      Thank you ctperng!
                      You are welcome!

                      But the information from your friend is incorrect. Order statistic in on the syllabus: See

                      http://www.soa.org/files/pdf/edu-catfall07-exam-p.pdf
                      (part 3. item transformations and order statistics)


                      The previous trick works for expenential distributions as can be seen from my two proofs, either using order statistics or by direct integrations.

                      This makes sense, because if the mean is higher, the probability of first occurence should be lower. Of course, if I have time I can also show a different way of looking at this phenomenon.

                      Anyway, good luck on your study!

                      ctperng

                      Comment


                      • #12
                        The first time I tried this problem I tried doing P(D<B), or P(D-B>0). I then calculated E(D-B)=3-2=1. The variance is sqrt(3^2+2^2)=sqrt(13).
                        I then did (0-1)/sqrt(13)=-.277, which returns a probability of .3936. What is the reason that normal approximation doesn't work here? Is it because there is nothing that says that there are at least 30 independent claims?

                        Comment

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