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Question 40 from 123 problems

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  • Question 40 from 123 problems

    if you graph out the pdf of X and then factor in the deductible, then why dont you just integrate the pdf of X from C to C + .5.

    How did they arrive at integrating the pdf of X from 0 to C + .5?

    Chris

  • #2
    Originally posted by EXAMER
    if you graph out the pdf of X and then factor in the deductible, then why dont you just integrate the pdf of X from C to C + .5.

    How did they arrive at integrating the pdf of X from 0 to C + .5?

    Chris
    Because in figuring out the probability of the payment being less than 0.5, you can not ignore the cases where there is no payment due to losses being under the deductible. Obviously, a no payment is less than a payment of 0.5 right?

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    • #3
      ah, greg, right you are. Thanks for the help (again).

      Chris

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      • #4
        Originally posted by EXAMER
        if you graph out the pdf of X and then factor in the deductible, then why dont you just integrate the pdf of X from C to C + .5.

        How did they arrive at integrating the pdf of X from 0 to C + .5?

        Chris
        Another way to look at it is to write the payment as a function of X. The insurance payment in the presence of an ordinary deductible is not the loss X but X-deductible=X-c. We are asked to find c such that P(payment<.5)=.64. We can rewrite this probability as P(X-c<.5)=P(X<c+.5). This is just the cdf of X evaluated at c+.5, and we know that X ranges from 0 to 1. Thus, our bounds of the integral must be 0 to c+.5.

        Just a different way of looking at it; hopefully this helps.
        act justly. walk humbly. .

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        • #5
          godpseed,

          i thought of it that way, except, I was thinking P(0<X-C<.5) and then it became P(C<X<C+.5). Nothing like Brain Farts in the last few days.

          Thanks for the help.

          Comment


          • #6
            Originally posted by EXAMER
            godpseed,

            i thought of it that way, except, I was thinking P(0<X-C<.5) and then it became P(C<X<C+.5). Nothing like Brain Farts in the last few days.

            Thanks for the help.
            I made the same mistake while doing that problem for the first time.
            Good luck.

            Comment

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