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Math Help - what is the appropriate approach this time!?

  1. #1
    alv
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    Question what is the appropriate approach this time!?

    hi all and especially mr_fantastic! remember the post "what is the...". that one was the case of Bayesian theorem. now what about this one?

    Consider two players A and B who are accompanying in a sequence of independent games. Probability of A win is p^2, probability of B win is q^2, and probability of equality (neither A nor B wins) is 2pq (p+q=1 and 0<p<1). At each run, the winner gets 2 points and in the situation of equality each player gets 1 point. Suppose X and Y are the total points of relatively A and B after the end of n runs of game. What is COV(X, Y)?
    ps. power is denoted by ^.
    {hence my absolute unfamiliarity with posting rules!!}
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  2. #2
    Flow Master
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    Quote Originally Posted by alv View Post
    hi all and especially mr_fantastic! remember the post "what is the...". that one was the case of Bayesian theorem. now what about this one?


    Consider two players A and B who are accompanying in a sequence of independent games. Probability of A win is p^2, probability of B win is q^2, and probability of equality (neither A nor B wins) is 2pq (p+q=1 and 0<p<1). At each run, the winner gets 2 points and in the situation of equality each player gets 1 point. Suppose X and Y are the total points of relatively A and B after the end of n runs of game. What is COV(X, Y)?

    ps. power is denoted by ^.
    {hence my absolute unfamiliarity with posting rules!!}
    Cov(X, Y) = E(XY) - E(X) E(Y).


    Define the following random variables:

    W_A is the number of games won by player A. Then W_A ~ Binomial (n, p^2).

    W_B is the number of games won by player B. Then W_B ~ Binomial (n, q^2).

    D_A = D_B is the number of games drawn by both players.

    Note that:

    1. W_A + W_B + D_A = n.

    2. The joint distribution of W_A, W_B and D_A is multinomial.


    From 1:

    X = 2W_A + D_A = W_A - W_B + n.

    Y = 2W_B + D_B = W_B - W_A + n.

    Note that:

    3. X + Y = 2n \Rightarrow (X + Y)^2 = 4n^2 \Rightarrow X^2 + 2XY + Y^2 = 4n^2.


    E(X) = E(W_A - W_B + n) = E(W_A) - E(W_B) + n = np^2 - nq^2 + n = 2np.

    E(Y) = E(W_B - W_A + n) = E(W_B) - E(W_A) + n = 2nq.

    From 3: E(XY) = 2n^2 - \frac{1}{2} E(X^2) - \frac{1}{2} E(Y^2).

    But E(X^2) = Var(X) + [E(X)]^2 and:

    * Var(X) = Var(W_A - W_B + n) = Var(W_A) + Var(W_B) + 2 Cov(W_A, W_B).

    * Cov(W_A, W_B) = -np^2 q^2 (this is a standard result following from 2).

    etc. (the details are left left for you).
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  3. #3
    alv
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    Exclamation thanks but a general question?

    by the way where can i find help on how to use math tags like what you guys do? it's really hard to explain the problems in plain text...those math tags are necessary. i have no information on how to use those tags...where can i look?
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by alv View Post
    by the way where can i find help on how to use math tags like what you guys do? it's really hard to explain the problems in plain text...those math tags are necessary. i have no information on how to use those tags...where can i look?
    Take a look at this thread in the Latex Help forum.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Cov(X, Y) = E(XY) - E(X) E(Y).


    Define the following random variables:

    W_A is the number of games won by player A. Then W_A ~ Binomial (n, p^2).

    W_B is the number of games won by player B. Then W_B ~ Binomial (n, q^2).

    D_A = D_B is the number of games drawn by both players.

    Note that:

    1. W_A + W_B + D_A = n.

    2. The joint distribution of W_A, W_B and D_A is multinomial.


    From 1:

    X = 2W_A + D_A = W_A - W_B + n.

    Y = 2W_B + D_B = W_B - W_A + n.

    Note that:

    3. X + Y = 2n

    [snip]
    A shorter approach might now be to note that X + Y = 2n \Rightarrow Var(X + Y) = Var(2n) = 0.

    But Var(X + Y) = Var(X) + Var(Y) + 2 Cov(X, Y).

    Therefore:

    Var(X) + Var(Y) + 2 Cov(X, Y) = 0

    \Rightarrow Cov(X, Y) = -\frac{1}{2} (Var(X) + Var(Y)).


    Now note that:

    (a) Var(X) = Var(W_A - W_B + n) = Var(W_A) + Var(W_B) + 2 Cov(W_A, W_B)

    = n p^2 (1 - p^2) + n q^2 (1 - q^2) + 2 Cov(W_A, W_B)

    = {\color{red}2npq(1-pq)} + 2 Cov(W_A, W_B).

    (You might want to check the simplification in red).


    (b) Var(Y) = Var(W_B - W_A + n) = Var(W_A) + Var(W_B) + 2 Cov(W_A, W_B) = Var(X).


    (c) Cov(W_A, W_B) = -np^2 q^2 (this is a standard result following from the fact that the joint distribution of W_A, W_B and D_A is multinomial).


    Therefore Cov(X, Y) = .........
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Cov(X, Y) = E(XY) - E(X) E(Y).


    Define the following random variables:

    W_A is the number of games won by player A. Then W_A ~ Binomial (n, p^2).

    W_B is the number of games won by player B. Then W_B ~ Binomial (n, q^2).

    D_A = D_B is the number of games drawn by both players.

    Note that:

    1. W_A + W_B + D_A = n.

    2. The joint distribution of W_A, W_B and D_A is multinomial.


    From 1:

    X = 2W_A + D_A = W_A - W_B + n.

    Y = 2W_B + D_B = W_B - W_A + n.

    Note that:

    3. X + Y = 2n
    [snip]
    One more thought:

    Therefore Cov(X, Y) = Cov(X, 2n - X) = .......
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