6. Exam-like question Assume the random variable X has distribution X Bin(9,0.5) and let Y =...

Question

Question

Answer 1

Answer #1

X can take values 0,1,2,3,4,5,6,7,8,9

So, Y can take values (-1)o = (-1)-= (-1) = (-1)6 = (-1)8-1

and (-1)^1 = (-1)^3 = (-1)^5= (-1)^7= (-1)^9= -1

1) The probability mass function of Y is

P( Y = -1) = rac{1}{2} since P(X = 0) + P(X = 2) + P(X = 4) +P(X = 6) + P(X = 8) = 1/2

and 2 P(Y = 1) = since P(X = 1) + P(X = 3) + P(X = 5) +P(X = 7) + P(X = 9) = 1/2

2) The mean of Y is

$E(Y) = sum_{y}y*P(Y=y) = -1*rac{1}{2} + 1*rac{1}{2} = 0$

3) Moreover,

$E(Y^2) = sum_{y}y^2*P(Y=y) = (-1)^2*rac{1}{2} + 1^2*rac{1}{2} = 1$

Hence,

The variance of Y = Var(Y) = E(Y^2) - [E(Y)]^2 = 1^2 -0 = 1

Answer 2

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