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  • Problem from Actex manual

    Dr. Broverman,


    I have a question from your manual.
    Let X be a Poisson random variable with E(X)= Ln2. Calculate E[cos(22/7 X)]. How did you get Sum of x to infinity cos(22/7x)= Sum of x to infinity (-1)^x? Would you mind to explain? Ans (1/4)

    Thanks

    Rup

  • #2
    Rup,

    Suppose that X is a non-negative integer random variable,
    X = 0 , 1 , 2 , . . . (Poisson, for instance), and suppose that
    the probability function of X is P(X=x) = p(x) .
    Now suppose that g(X) is some function of X (for instance X^2 ,
    or cos(pi X) ). We want to find E[g(X)] .

    THe definition of E(X) is
    0*p(0) + 1*p(1) +2*p(2) + ... .

    THe definition of E[g(X)] is
    g(0)*p(0) + g(1)*p(1) + g(2)*p(2) + . . .

    In the example you mentioned, X has a Poisson dist with
    mean (lambda) of ln(2) , and g(X) is the function
    g(X) = cos(pi X) .
    Then E[cos(pi X)] = cos(0)*p(0) + cos(pi)*p(1) + cos(2 pi)*p(2) +...

    From the nature of the cosine function ,
    cos(0) = 1 , cos(pi) = -1 , cos(2 pi) = 1 , ... alternating 1 and -1 .
    Then E[cps(pi X)] becomes
    p(0) - p(1) + p(2) - . . .

    From the definition of the Poisson dist with mean lambda = ln(2) ,
    we have P(X=k) = p(k) = [e^(-lambda) * (lambda)^k]/k! .
    SInce e^(-ln 2) = 1/2 , we get p(k) = 1/2 * (ln 2)^k / k! .

    After factoring 1/2 from all terms, the series we are looking at becomes
    1/2 * [ 1 - (ln 2)/1! + (ln 2)^2 / 2! - (ln 2)^3 / 3! + . . . ]

    We then use the Taylor series
    e^(-t) = 1 - t + t^2 /2! - t^3 / 3! + . . . , so that
    1 - (ln 2)/1! + (ln 2)^2 / 2! - (ln 2)^3 / 3! + . . . = e^-(ln 2) = 1/2
    (the series above is the Taylor series with t = ln 2) .
    THe overall series sums to 1/2 * 1/2 = 1/4 .
    Sam Broverman

    [email protected]
    www.sambroverman.com

    Comment


    • #3
      Pi is commonly approximated as 22/7, so cos(22x/7) is approximately cos(pix). Also, notice that the summation from 0 to infinity of cos(pix)=1-1+1-1+1-1+. . . , forever continuing to oscillate between 1 and -1. Thus, in a summation, we can write cos(pix) as (-1)^x.

      So, E[cos(22/7 X)]=E[cos(pix)]=
      summation from x=0 to infinity of [cos(pix)*pdf of x] =
      summation from x=0 to infinity of [cos(pix)*(e^(-ln2)*(ln2)^x) / x!] =
      summation from x=0 to infinity of [1/2 * (-1)^x*(ln2)^x / x!].

      Now, for P, I believe you have to know certain Taylor series, and the Taylor series representation for the function e^x is:

      summation from n=0 to infinity of [x^n/n!]. Notice the we sum over all values of n, not x.

      Thus, we can write our expected value as:
      summation from x=0 to infinity of [1/2 * (-1*ln2)^x / x!]
      =1/2 * e^(-ln2)
      =1/2 * 1/2
      =1/4

      I think that's what Prof. Broverman intended when he wrote 22/7, but I'm just taking a guess! Hope this helps. Keep up the studying!

      Edit: Prof. Broverman posted as I posted, so, yeah, this is what he meant, of course explaining it how it should be =0
      Last edited by .Godspeed.; January 9 2006, 10:39 PM.
      act justly. walk humbly. .

      Comment


      • #4
        Re:

        Dr. Broverman,

        It was very clear. Thanks a million! You are great!

        Rup

        Comment


        • #5
          Re

          Thanks a lot Godspeed, It was perfect too.


          Rup

          Comment

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