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Question about evaluating E(T(X))

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  • Question about evaluating E(T(X))

    My study manual defines "e-circle sub x" as E(T(X)) and defines that as the integral over the defined interval of the Survival function of T(X) ie integral of tpx. This varies from the usual expected value equation which is the integral of t times f(t) dt. I notice that it can be evaluated both ways. That implies that x times "force of mortality" of t+x is 1. Is this true? If so can anyone explain why it is true?

  • #2
    Why do you think that implies x force of mortality is 1?

    An easy counter example would be constant force of mortality. As x increases, the product would increase.
    Whether you are the lion or the gazelle, when the sun comes up, you better be running.

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    • #3
      I agree it doesn't sound like it makes sense. Yet the expected value can be calculated in two different way yielding the exact same result. One way is the integral of the Survival function[S(X)]. The other way is by integrating x times the density function[f(X)].

      Now I know that f(X) is equal to S(X) times the force of mortality. So on the one hand, we have the integral of S(X), and on the other we have the integral of X times S(X) times Force of Mortality. So tell me how those can both give the same result without being equal to one another?

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      • #4
        Originally posted by yos9694
        My study manual defines "e-circle sub x" as E(T(X)) and defines that as the integral over the defined interval of the Survival function of T(X) ie integral of tpx. This varies from the usual expected value equation which is the integral of t times f(t) dt. I notice that it can be evaluated both ways. That implies that x times "force of mortality" of t+x is 1. Is this true? If so can anyone explain why it is true?
        The integrand of E(T(X)) is not tpx*x*mu(x+t), but tpx*t*"mu(x+t), which is I think what you meant.

        E(T(X)) equaling both 1) the integral of t*tpx*mu(x+t) from 0 to infinity AND 2) the integral of tpx from 0 to infinity is a result from integration by parts. As you state, tpx*mu(x+t)=fx(t), so we can write E(T(X)) as int(t*fx(t)dt). Now for the integration by parts, take u=t, du=dt, dv=fx(t) dt, v=tqx.
        Thus int(t*fx(t))=t*tqx (evaluated from 0 to inf.) - int(tqx dt) from 0 to inf.

        But t*tqx (evaluated from 0 to inf.) really just equals t evaluated at "infinity". And the derivative of t with respect to t is just 1. So, with some grouping, we see that int(t*fx(t))=int(1-tqx dt) from 0 to inf. = int(tpx dt) from 0 to inf.

        Again, the product t*mu(x+t) does not necessarily equal 1; the two integrals' being equal follows from integration by parts. Does this help any?
        act justly. walk humbly. .

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        • #5
          I can't figure out where you are gettin f(X) = s(X) times Force of mortality?

          Is this true?

          The relationship I know if f(X)=-s'(X).

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          • #6
            Originally posted by PhillyP
            I can't figure out where you are gettin f(X) = s(X) times Force of mortality?

            Is this true?

            The relationship I know if f(X)=-s'(X).
            Yeah, that's definitely true, PhillyP. Pages 20-21 of Loss Models confirms this.

            And this is the crux of the question. We can write the pdf of t for age (x), fx(t), in terms of the force of mortality of (x) at age t, mu(x+t), and the probaility of (x) surviving at least t years, tpx: fx(t)=mu(x+t)*tpx. This is analagous to the base relationship mu(x)=f(x)/S(x).
            Last edited by .Godspeed.; February 23 2006, 10:55 PM.
            act justly. walk humbly. .

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            • #7
              Originally posted by .Godspeed.
              The integrand of E(T(X)) is not tpx*x*mu(x+t), but tpx*t*"mu(x+t), which is I think what you meant.
              Absolutely right, I did mean that, it's just a question of naming the variable.

              Originally posted by .Godspeed.
              Again, the product t*mu(x+t) does not necessarily equal 1; the two integrals' being equal follows from integration by parts. Does this help any?
              Yes, it really does help. Thank you very much. I was thinking along those lines-that two integrals can be equal even though the integrands themselves are not identical or even equal. But I wanted to see if the conclusion was true or not, and you have shown that it is not.

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              • #8
                Integral of the survival function

                Originally posted by yos9694
                My study manual defines "e-circle sub x" as E(T(X)) and defines that as the integral over the defined interval of the Survival function of T(X) ie integral of tpx. This varies from the usual expected value equation which is the integral of t times f(t) dt. I notice that it can be evaluated both ways. That implies that x times "force of mortality" of t+x is 1. Is this true? If so can anyone explain why it is true?
                The fact that the expected value of any non-negative random variable is the integral of its survival function is actually something you should have learned in your study of probability. This is proved by integration by parts. It is actually done in Bowers et al., as well. And this begs some questions: aren't you reading Bowers et al.? And, what textbook did you use for exam P, as you really should have learned it then. My ASM manual for exam P has a complete proof of this, and I have posted several exercises illustrating this property (The Darth Vader Rule, as I named it) in the exam P forum.

                Yours,
                Krzys' Ostaszewski

                P.S. Maybe you should consider the exam M seminar that I will run with Dr. Kubicka starting March 10.
                Want to know how to pass actuarial exams? Go to: smartURL.it/pass

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                • #9
                  humm , i think

                  t * f(t) approximately equals s ( t ) which is tpx


                  thats why you intregate the survival function to its domain to find E( t(x)

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