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Probability: Coinflips & Variance

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data
    You are flipping a fair coin n times. Let X be the number of times you get a head followed by a tail (HT) and Y is the number of times you get tail and then a head (TH).

    Find Var(X) and Var(Y).

    2. Relevant equations
    Cov(X,X) = Var(X)

    3. The attempt at a solution
    My approach is the break up the variables X and Y into sums of Bernoulli random variables. For instance, X = X1 + X2 +...+ Xn-1
    (because there are n-1 slots for a HT in n flips).
    So Xi = 1 if the ith slot of two flips contains a HT and 0 otherwise
    I am using the fact that Cov(X, X) = Var(X) and that
    Cov(Xi, Xj) = E(Xi Xj) - E(Xi)E(Xj) and then summing this up over all i and j.

    For the case where i=j I got 3/16, which seems reasonable. I am have trouble with the case where i≠j, my answers keeps coming out negative and therefore making my whole variance negative, which is not right.

    What I've done is:
    Cov(Xi, Xj) = E(Xi Xj) - E(Xi)E(Xj)
    = P(Xi Xj =1) - (1/4)(1/4)
    = P(Xi = 1, Xj = 1) - (1/16)
    Here is seems to me that P(Xi = 1, Xj = 1) = 0 since if you have HT in the ith slot then the jth slot begins with a T and thus cannot be HT so Xj = 0. But then this gives you that Cov(Xi, Xj) for i≠j = -1/16. Then summing up you get

    (n-1)(3/16) + (n2 - n)(-1/16)
    =(1/16)(-n2 + 4n - 3) and this is not positive for most legitimate values of n.

    Can anyone help explain what I am missing? Thanks.
  2. jcsd
  3. Nov 17, 2009 #2
    Never mind. I think I figured it out.
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