Question

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Answer 1

X1 and X2 can take values of -1 and 1.

So, Y = X1 X2 can take values of -1 and 1

P(Y = -1) = P(X1 = -1, X2 = 1) + P(X1 = 1, X2 = -1) = P(X1 = -1) P(X2 = 1) + P(X1 = 1) P(X2 = -1) {X1 and X2 are independent

= (1/2) p + (1/2) (1-p) = 1/2

P(Y = 1) = P(X1 = -1, X2 = -1) + P(X1 = 1, X2 = 1) = P(X1 = -1) P(X2 = -1) + P(X1 = 1) P(X2 = 1) {X1 and X2 are independent

= (1/2) (1-p) + (1/2) p = 1/2

Now, if X2 and Y are independent, then

P(X2 = x, Y = y) = P(X2 = x) P(Y = y) for all values of x, y $\in$ (-1, 1)

P(X2 = -1, Y = -1) = P(X2 = -1) * P(Y = -1 | X2 = -1) = P(X2 = -1) * P(X1 = 1) { P(Y = -1 | X2 = -1) => P(X1 = 1) }

= (1-p) (1/2) = P(X2 = -1) P(Y = -1)

P(X2 = -1, Y = 1) = P(X2 = -1) * P(Y = 1 | X2 = -1) = P(X2 = -1) * P(X1 = -1) { P(Y = 1 | X2 = -1) => P(X1 = -1) }

= (1-p) (1/2) = P(X2 = -1) P(Y = 1)

P(X2 = 1, Y = -1) = P(X2 = 1) * P(Y = -1 | X2 = 1) = P(X2 = 1) * P(X1 = -1) { P(Y = -1 | X2 = 1) => P(X1 = -1) }

= p (1/2) = P(X2 = 1) P(Y = -1)

P(X2 = 1, Y = 1) = P(X2 = 1) * P(Y = 1 | X2 = 1) = P(X2 = 1) * P(X1 = 1) { P(Y = 1 | X2 = 1) => P(X1 = 1) }

= p (1/2) = P(X2 = 1) P(Y = 1)

Thus,

P(X2 = x, Y = y) = P(X2 = x) P(Y = y) for all values of x, y $\in$ (-1, 1)

and hence X2 and Y are independent.