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  • Question 46 of 123

    All my steps are perfect and correct until the final integration of the last line... Can someone run through the propper steps of the last integration of e^-(t/3)dt from 2 to inf.
    i keep getting -3+3e^-2/3 (just for the last part and it doesnt add up with the first part of the equation...

  • #2
    Originally posted by faibs
    All my steps are perfect and correct until the final integration of the last line... Can someone run through the propper steps of the last integration of e^-(t/3)dt from 2 to inf.
    i keep getting -3+3e^-2/3 (just for the last part and it doesnt add up with the first part of the equation...

    U can try this way: Let u = t/3, then du = dt(1/3) -> 3du=dt, ok then sub into the equation, it gonna be..........integration of e^-u 3(du) from 2 to inf.

    after integration, it gonna be, 3[-e^-u] from 2 to inf -> 3[-e^-(t/3)] from 2 to inf (because u=t/3)

    Therefore the answer is 3e^-(2/3)

    of course there are a lot of different way to do this integration, but this one is more easy to understand. I hope it helps

    T
    Tears stream down your face
    I promise you I will learn from my mistakes

    Comment


    • #3
      Hey I had a question on this problem....

      how do you deal with the X=max(T,2). I dont know what that means.

      Chris

      Comment


      • #4
        Originally posted by EXAMER
        Hey I had a question on this problem....

        how do you deal with the X=max(T,2). I dont know what that means.

        Chris

        Hi Chris,

        X=max(T,2) has two conditions. First, if 2 is the max, then t<2. Second, if T is the max, then t>2. Then, u use these two conditions to find E[X]. Like this, integration of 2 * f(t) dt from 0 to 2 + integration of t*f(t) dt from 2 to inf.
        Then you should be able to find the answer.
        Does it helps??

        T
        Tears stream down your face
        I promise you I will learn from my mistakes

        Comment


        • #5
          I think so, so Max(T,2) means...

          If less than 2, T=2
          If more than 2, T = t

          so when you find the E(T) on the interval of (0,2) then youre integrating the (pdf)(2) where as if youre on the interval (2,inifniti) then its itegrating (t)(pdf)?

          If thats correct then i got it. Is that always what it means? Max(x,y) would mean that if less than y, you would use y in the expected value, and then if greater than y you use x in your expected value?

          Thanks,

          Chris

          Comment


          • #6
            Let me tell u this way:
            X=max(T,2), that means X is the max of (T,2). There has 2 possibilities.
            1) If X=T, which is T is the max. It also means T is bigger than 2, that's why T is the max. So, T>2
            2) If X=2, which is 2 is the Max. So, T<2

            Originally posted by EXAMER
            I think so, so Max(T,2) means...

            If less than 2, T=2
            If more than 2, T = t

            so when you find the E(T) on the interval of (0,2) then youre integrating the (pdf)(2) where as if youre on the interval (2,inifniti) then its itegrating (t)(pdf)?

            If thats correct then i got it. Is that always what it means? Max(x,y) would mean that if less than y, you would use y in the expected value, and then if greater than y you use x in your expected value?

            Thanks,

            Chris
            No!!
            The expected value for a continuous random variable is "integration x*f(x) dx
            from (-inf to inf)".
            In this questions, pdf is f(t) which 0<t<inf.

            Part I: if x=2, then interval of t is 0<t<2 (because 2 is max.)
            So, expected value for this one is "integration 2 * f(t) dt from 0 to 2.

            Part II: if x=t, then interval of t is 2<t<inf (because t is max, of course t is larger than 2).
            So, expected value for this one is "integration of t*f(t) dt from 2 to inf.

            the answer is Part I + Part II

            sorry, maybe I confused you last time. I hope this time is more detail and clear.

            T
            Tears stream down your face
            I promise you I will learn from my mistakes

            Comment


            • #7
              Thanks for the help! I appreciate it.

              Hmm, I get the expected values,

              I guess i just dont get the Max(2,t), though I understand that its describining time for the expected life time,

              so its liek E(2) and E(T), the first being 2 becuase they dont check anything prior to 2 years.

              They used the Max() denotation in another problem, problem 103...incidentally i dont know how to do that one either.

              Chris

              Comment


              • #8
                Originally posted by EXAMER
                Thanks for the help! I appreciate it.

                Hmm, I get the expected values,

                I guess i just dont get the Max(2,t), though I understand that its describining time for the expected life time,

                so its liek E(2) and E(T), the first being 2 becuase they dont check anything prior to 2 years.

                They used the Max() denotation in another problem, problem 103...incidentally i dont know how to do that one either.

                Chris
                If you haven't got 103, then here is an explanation of how to handle these type of problems. Let Y = Max {X_1,X_2,X_3} be the maximum of these three random variables. Then F(y) = P(Y <= y) . Note that if Y, the maximum of the three variables, is less than y, then that means each one of those variables is <= y. So, F(y) = P(X_1 <= y, X_2 <= y, X_3 <= y). If the joint pdf is known, the you can calculate cdf of Y. If, we assume that the variables are independent, then


                F(y) = P(X_1 <= y, X_2 <= y, X_3 <= y) = P(X_1 <= y) *P(X_2 <= y)*P(X_3 <= y) = F_1(y)*F_2(y)*F_3(y), F_1, F_2 and F_3 are the respective cdfs.

                Once you get F(y), you can take the derivative to find the pdf.

                So, in problem no. 103, You need to find P(Y > 3).
                Now, P(Y>3) = 1 - P(Y<= 3) = 1 - P(S <= 3)*P(F <= 3)*P(T<= 3)

                = 1 - (1-e^-(3))*(1-e^-(3/1.5))*(1-e^-(3/2.4))

                Does that help?

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