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  • 2 problems

    1. The number of eggs laid on a tree leaf by an insect of a certain type is a Poisson random variable with parameter λ. However, such a random variable can only be observed if it is positive, since if it is 0 then we cannot know that such an insect was on the leaf. If we let Y denote the observed number of eggs, then
    P{Y = i} = P{X = i|X > 0}
    where X is Poisson with parameter λ. Find E[Y].

    2. The amount of time that a certain type of component functions before failing is a random variable with probability density function
    f(x) = 2x, 0 < x < 1
    Once the component fails it is immediately replaced by another one of the same type. If we let Xi denote the lifetime of the i-th component to be put in use, then Sn = Σ(i=1, n) Xi represents the time of the n-th failure. The long-term rate at which failures occur, call it r, is defined by
    r = lim (n/Sn)
    Assuming that the random variables Xi, i >or= 1 are independent, determine r.



    Any help will be appreciated.

  • #2
    I just got 1/λ for #1, but I have no idea if that's right.
    I did E[Y] = Σ(k=1 to infinity) (k(e^-λ)(λ^k))/k!
    = (e^-λ)*Σ(k=1 to infinity) (λ^k)/(k-1)!
    = (e^-λ)*(λ + λ^2 + (λ^3)/2 + (λ^4)/3.....)
    = (e^-λ)*(1/λ)*(1 + λ + (λ^2)/2 + (λ^3)/3.....)
    = (e^-λ)*(1/λ)*(e^λ)
    = 1/λ
    Is this right?
    Also, if somebody could help out on #2 that would be great.
    Thanks.

    Comment


    • #3
      Originally posted by tdl5730 View Post
      1. The number of eggs laid on a tree leaf by an insect of a certain type is a Poisson random variable with parameter λ. However, such a random variable can only be observed if it is positive, since if it is 0 then we cannot know that such an insect was on the leaf. If we let Y denote the observed number of eggs, then
      P{Y = i} = P{X = i|X > 0}
      where X is Poisson with parameter λ. Find E[Y].

      2. The amount of time that a certain type of component functions before failing is a random variable with probability density function
      f(x) = 2x, 0 < x < 1
      Once the component fails it is immediately replaced by another one of the same type. If we let Xi denote the lifetime of the i-th component to be put in use, then Sn = Σ(i=1, n) Xi represents the time of the n-th failure. The long-term rate at which failures occur, call it r, is defined by
      r = lim (n/Sn)
      Assuming that the random variables Xi, i >or= 1 are independent, determine r.



      Any help will be appreciated.
      #1. P{Y = i} = P{X = i|X > 0} = P{(X = i)and(X>0)}/P(X>0)
      P(X>0) = 1-P(0) = 1 - e^(-λ), thus P(Y= i) = (λ^x)/{x![(e^λ)-1]}
      Now, E(Y) = Σ(k=1 to infinity) ((λ^k)/{(k-1)![(e^λ)-1]}= Σ(k=0 to infinity) ((λ^[k+1])/{k![(e^λ)-1]}= λ*e^λ/[(e^λ) - 1]

      #2. I would find lim(n/E[Sn]) which is 3/2, but it's probably wrong.

      Comment


      • #4
        P(Y=i) = P(X=i and X > 0) / P(X >0)

        where P(X=i) and X > 0 intersect is at P(X = i)

        =P(X=i) / P(X >0)



        = 0 *((exp(-λ) * λ^0) / ((1- exp(-λ))*0!) + 1 *((exp(-λ) * λ^1) / ((1- exp(-λ))*1! + 2*......


        (common factor... also common factor out a lambda to get...)

        = (exp(-λ) / ((1- exp(-λ)) * { λ + 2λ/2! + 3λ^3/3! + 4λ^4/4! + .....}
        ^cancel out leading coefficients and factor out a λ

        = λ (exp(-λ) / ((1- exp(-λ)) * { 1+ λ/1! + λ^2/2! + λ^3/3! + .....}

        ^this is the series sum (n=1 to inf) x^n/n! = exp(x)



        = λ / ((1- exp(-λ))
        Last edited by intoronto; May 9 2010, 10:54 PM.

        Comment


        • #5
          P(Y=i) = P(X=i and X > 0) / P(X >0)

          where P(X=i) and X > 0 intersect is at P(X = i)

          =P(X=i) / P(X >0)



          = 0 *((exp(-λ) * λ^0) / ((1- exp(-λ))*0!) + 1 *((exp(-λ) * λ^1) / ((1- exp(-λ))*1! + 2*......


          (common factor... also common factor out a lambda to get...)

          = (exp(-λ) / ((1- exp(-λ)) * { λ + 2λ/2! + 3λ^3/3! + 4λ^4/4! + .....}
          ^cancel out leading coefficients and factor out a λ

          = λ (exp(-λ) / ((1- exp(-λ)) * { 1+ λ/1! + λ^2/2! + λ^3/3! + .....}

          ^this is the series sum (n=1 to inf) x^n/n! = exp(x)



          = λ / ((1- exp(-λ))
          Last edited by intoronto; May 9 2010, 10:54 PM.

          Comment


          • #6
            Originally posted by intoronto View Post
            P(Y=i) = P(X=i and X > 0) / P(X >0)

            where P(X=i) and X > 0 intersect is at P(X = i)

            =P(X=i) / P(X >0)



            = 0 *((exp(-λ) * λ^0) / ((1- exp(-λ))*0!) + 1 *((exp(-λ) * λ^1) / ((1- exp(-λ))*1! + 2*......


            (common factor... also common factor out a lambda to get...)

            = (exp(-λ) / ((1- exp(-λ)) * { λ + 2λ/2! + 3λ^3/3! + 4λ^4/4! + .....}
            ^cancel out leading coefficients and factor out a λ

            = λ (exp(-λ) / ((1- exp(-λ)) * { 1+ λ/1! + λ^2/2! + λ^3/3! + .....}

            ^this is the series sum (n=1 to inf) x^n/n! = exp(x)



            = λ / ((1- exp(-λ))
            Isn't sum (n=0 to inf) x^n/n! = exp(x) ?

            Comment


            • #7
              Originally posted by AsyaK View Post
              Isn't sum (n=0 to inf) x^n/n! = exp(x) ?
              Thank you both. Both answers are the same.
              Yes. It is from 0 to inf. I think when he pulls out the λ it becomes the summation(n=1 to inf) of (λ^(n-1))/(n-1)! which is the same as the summation(n=0 to inf) of (λ^n/n!)
              Thanks again.

              AsyaK, would you mind showing how to get get 3/2 from taking that limit? (For #2) I'm having trouble. Thanks.

              Comment


              • #8
                Originally posted by tdl5730 View Post

                2. The amount of time that a certain type of component functions before failing is a random variable with probability density function
                f(x) = 2x, 0 < x < 1
                Once the component fails it is immediately replaced by another one of the same type. If we let Xi denote the lifetime of the i-th component to be put in use, then Sn = Σ(i=1, n) Xi represents the time of the n-th failure. The long-term rate at which failures occur, call it r, is defined by
                r = lim (n/Sn)
                Assuming that the random variables Xi, i >or= 1 are independent, determine r.
                Is 3/2 the right answer? If so, then
                E(Xi) = integral from 0 to 1 of (x* 2x) = 2/3.
                Since implied in the problem that variables are independent, E[Sn] = n*2/3.
                Then r = n/(n*2/3) = 3/2... I really think this is not the right solution, we don't even need the lim.... so we should wait for some expert to solve the problem correctly.

                Comment


                • #9
                  Originally posted by AsyaK View Post
                  Is 3/2 the right answer? If so, then
                  E(Xi) = integral from 0 to 1 of (x* 2x) = 2/3.
                  Since implied in the problem that variables are independent, E[Sn] = n*2/3.
                  Then r = n/(n*2/3) = 3/2... I really think this is not the right solution, we don't even need the lim.... so we should wait for some expert to solve the problem correctly.
                  I don't know the correct answer right now. I might be able to get it tomorrow. And that's better than anything I've tried, so thank you.

                  Comment


                  • #10
                    Originally posted by AsyaK View Post
                    Isn't sum (n=0 to inf) x^n/n! = exp(x) ?
                    ya so you get exp(λ)

                    when you multiply this by the exp(-λ) the exponents are added together and you get exp(0) which is 1

                    mine is the same as yours if you factor out an exp(-λ) from the denominator and make it positive in the numerator

                    Comment


                    • #11
                      Originally posted by intoronto View Post
                      ya so you get exp(λ)

                      when you multiply this by the exp(-λ) the exponents are added together and you get exp(0) which is 1

                      mine is the same as yours if you factor out an exp(-λ) from the denominator and make it positive in the numerator
                      You are right! And your answer is much cleaner then mine.

                      Comment

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