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1. 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

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