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Soa #320 exam p

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  • Soa #320 exam p

    Hi Guys,

    Does anyone know how to get to the joint probability Pr(x=0, Y=0) = 1/3 posted in the soa solutions?

    the problem is this

    # 320. Let X be a random variable that takes on the values –1, 0, and 1 with equal probabilities.

    Let Y= X^2

    Which of the following is true?


    (A) Cov(X, Y) > 0; the random variables X and Y are dependent.
    (B) Cov(X, Y) > 0; the random variables X and Y are independent.
    (C) Cov(X, Y) = 0; the random variables X and Y are dependent.
    (D) Cov(X, Y) = 0; the random variables X and Y are independent.
    (E) Cov(X, Y) < 0; the random variables X and Y are dependent.

    the solution (c) is this

    Cov(X,Y) = E(XY) – E(X)E(Y) = E(X^3) – E(X)E(X^2)
    E(X) = E(X^3)=(1/3)(-1+0+1) = 0
    E(X^2) = (1/3)(1+0+1) = 2/3
    Cov(X,Y) = 0 - 0(2/3)=0
    They are dependent, because
    Pr(X=0,Y=0) = Pr(X=0,X^2=0) = Pr(X=0) =1/3 ► here is my question..???????
    Pr(X=0)Pr(Y=0) = (1/3)(1/3) = 1/9 ≠ 1/3

    Thank you.



    I figured out the solution for my own question since I had no response.

    Y=X^2
    X = { -1, 0, 1} and P[x=1]= 1/3 ; P[x=2]= 1/3 P[x=3]= 1/3
    because all of them have the same probability and when you add them all , they are equal to one
    P[x=1] = P[x=2] = P[x=3], so 3P[x=i] =1 , ► P[x=i]=1/3
    Cov(x,y) = E(XY) - E(X)E(Y)

    E(X) = -1(1/3) + 0(1/3) + 1(1/3) = 0
    E(X^2) =(-1)^2*(1/3) + 0^2*(1/3) + 1^2*(1/3) = 2/3
    E(XY) = E(XX^2)=E(X^3) = (-1)^3*(1/3) + 0^3*(1/3) + 1^3*(1/3) = 0

    E(Y) = E(X^2) = 2/3 = 0*P(y=0) + 1*P(y=1)
    2/3 = 0 + P(y=1) ► P(y=1) = 2/3

    E(XY) = E(XX^2) = E(X^3) = 0 = XY*P(X,Y) ► E(x=1,y=1) = (x=1)(y=1)P(x=1,y=1)
    E(x=1,y=1) = 1*1*P(x=1,y=1)
    E(X=1)^3 = P(x=1,y=1)
    1/3 = P(x=1,y=1)

    P(x=1) = 1/3; P(y=1) = 2/3 ; P(x=1,y=1) =1/3; Cov(x,y) =0
    P(x=1)*P(y=1) = 1/3*2/3 = 2/9

    Therefore,

    P(x=1)*P(y=1) ≠ P(x=1,y=1)
    2/9 ≠ 1/3
    Last edited by Angelope25; September 20 2017, 02:20 AM. Reason: color fond
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