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6 b from the definition of expectation we have e x 4

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6. (b) From the definition of expectation, we have E [ X ] = 4 X k =0 kp X ( k ) = 0 · 1 2 + 1 · 1 4 + 2 · 1 8 + 3 · 1 16 + 4 · 1 16 = 15 16 . 6. (c) From the definition of variance, we have Var ( X ) = E [ X 2 ] - ( E [ X ]) 2 = 4 X k =0 k 2 p X ( k ) - ( E [ X ]) 2 = 0 2 · 1 2 + 1 2 · 1 4 + 2 2 · 1 8 + 3 2 · 1 16 + 4 2 · 1 16 - 15 16 2 = 367 256 . 6. (d) Note that Y = 0 when X = 0 or 3, Y = 1 when X = 1 or 4, and Y = 2 when X = 2 . The PMF of Y is thus P ( Y = 0) = P ( X = 0) + P ( X = 3) = 1 2 + 1 16 = 9 16 P ( Y = 1) = P ( X = 1) + P ( X = 4) = 1 4 + 1 16 = 5 16 P ( Y = 2) = P ( X = 2) = 1 8 . 6. (e) E [ Y ] = 2 k =0 k P ( Y = k ) = 0 · 9 16 + 1 · 5 16 + 2 · 1 8 = 9 16 . 4
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