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  • problem

    For a certain discrete random variable on the non-negative integers, the prob-
    ability function satisfies the relationships
    P(0) = P(1),
    P(k + 1) = 1/k P(k),
    k = 1,2,3 ...
    Find P(0)

  • #2
    p0=p1
    p2=p1 =PO
    p3=1/2 p2 =1/2 P0
    p4=1/3 p3 =(1/2)(1/3)PO
    P5=1/4 P4 =1/2*1/3*1/4 PO

    (P0+p1=2PO)
    1 = 2P0 + PO + PO/2! + PO/3!...........

    This is also 1=PO(2+1+1/2!+1/3!.......) and 1+1/2!+1/3!.......=e^x for x=1 so e^1

    e^x=Sigma(x^n/n!)

    So the solution should be 1=2PO+PO(e^1) unless I made an error but you can figure it out from here on!
    Puoya
    Actuary.com - Level III Poster
    Last edited by Puoya; May 1 2010, 03:33 AM.
    "As far as I'm concerned, I prefer silent vice to ostentatious virtue." Albert Einstein
    "It is hard to tell if a man is telling the truth when you know you would lie if you were in his place." H. L. Mencken

    Comment


    • #3
      Originally posted by Dabsy View Post
      For a certain discrete random variable on the non-negative integers, the prob-
      ability function satisfies the relationships
      P(0) = P(1),
      P(k + 1) = 1/k P(k),
      k = 1,2,3 ...
      Find P(0)
      Let P(0)=P(1)=p
      then P(2)=p/1, P(3)=p/2, P(4)=p/6, P(5)=p/24, ... , P(k+1)=p/k!
      P(0)+P(1)+P(2)+... = 1 = p+p/0!+p/1!+p/2!+... =p+p(1/0!+1/1!+1/2!+... )=p+pe
      p= 1/(1+e).

      Comment


      • #4
        Originally posted by Puoya View Post
        p0=p1
        p2=p1 =PO
        p3=1/2 p2 =1/2 P0
        p4=1/3 p3 =(1/2)(1/3)PO
        P5=1/4 P4 =1/2*1/3*1/4 PO

        (P0+p1=2PO)
        1 = 2P0 + PO + PO/2! + PO/3!...........

        This is also 1=PO(2+1+1/2!+1/3!.......) and 1+1/2!+1/3!.......=e^x for x=1 so e^1

        e^x=Sigma(x^n/n!)

        So the solution should be 1=2PO+PO(e^1) unless I made an error but you can figure it out from here on!
        We really need to know series expansion formulas and rules?

        Comment


        • #5
          NO, i don't think it's necessary. But the series for e is worth knowing.
          "As far as I'm concerned, I prefer silent vice to ostentatious virtue." Albert Einstein
          "It is hard to tell if a man is telling the truth when you know you would lie if you were in his place." H. L. Mencken

          Comment

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