E[X] and Var[X] of a Binomial with twist

Discussion in 'Probability' started by Rueful Rabbit, Oct 23, 2011.

  1. Hi All,

    x~Binomial
    p is 0.2
    N is 12
    E[X] = np = 2.4
    Var[X] = np(1-p) = 1.92

    Now suppose you add a charge of 75 for the first occurrence and 50 for
    subsequent occurrences
    what is a the E[X] and Var[X] of the cost?

    Thanks for reading.
     
    Rueful Rabbit, Oct 23, 2011
    #1
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  2. Rueful Rabbit

    Ray Koopman Guest

    Restate the "twist" as a charge of 50 for each occurrence,
    plus a "processing fee" of 25 if there are any occurrences.
    Then the expected cost is 50 np + 25 (1-(1-p)^n).

    The variance is a little sneakier. Use the alternate expression
    for variance: mean square minus squared mean. The expected square
    of the cost is 50^2 np + (75^2 - 50^2)(1-(1-p)^n). The variance is
    50^2 np + (75^2 - 50^2)(1-(1-p)^n) - (50 np + 25 (1-(1-p)^n))^2.
     
    Ray Koopman, Oct 23, 2011
    #2
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  3. Thanks again Ray, much appreciated.
    James
     
    Rueful Rabbit, Oct 23, 2011
    #3
  4. Rueful Rabbit

    Ray Koopman Guest

    Here's a simpler problem, that leads to a solution to the original
    problem:

    Suppose X is Binomial[n,p], except that its support is {z,1,2,...,n},
    where the arbitrary real value z occurs instead of 0, with the same
    probability as 0 would have. What are the mean and variance of X?
     
    Ray Koopman, Oct 24, 2011
    #4
  5. Rueful Rabbit

    Ray Koopman Guest

    Nice plan, but bad execution. The expected square of the cost is
    not as given above, but is much messier. So let me expand on the
    "simpler problem" that I mentioned in my second post yesterday.

    The costs are 0, 25+50, 25+100, 25+150, ... . The sequence is
    equally spaced, except the first space is 50% bigger than the others.
    That's a linear transformation of z, 1, 2, 3, ... , with z = -1/2.
    Then cost = 50(X + 1/2), which makes E[cost] = 50(E[X] + 1/2)
    and Var[cost] = 50^2 Var[X]; the mean and variance of X come from:
    E[X] = np + z(1-p)^n,
    E[X^2] = np(1-p) + (np)^2 + z^2 (1-p)^n,
    Var[X] = E[X^2] - (E[X])^2.
     
    Ray Koopman, Oct 24, 2011
    #5
  6. Thanks for the follow-up Ray

    James.
     
    Rueful Rabbit, Oct 26, 2011
    #6
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