sol12 - CS 70 Fall 2010 Discrete Mathematics and...

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CS 70 Discrete Mathematics and Probability Theory Fall 2010 Tse/Wagner Soln 12 1. (24 pts.) Practice with variance Here are some calculations that we are hoping you will have down solid. Please make sure you are comfort- able with these calculations. (a) Let X be an indicator random variable for the event that the top card of a well-shuffled 52-card deck is the Ace of Spades. Calculate E ( X ) and Var ( X ) . Answer: In general, when X is an indicator random variable with Pr [ X = 1 ] = p , E ( X ) = p and Var ( X ) = p ( 1 - p ) . Here, Pr [ X = 1 ] = 1 52 . So E ( X ) = 1 52 and Var ( X ) = 51 52 2 . (b) Let Y be a random variable with the following distribution: Y = 2 with probability 1 3 , Y = 0 with probability 1 3 , and Y = - 2 with probability 1 3 . Calculate Var ( Y ) . Answer: E ( Y ) = - 2 × 1 3 + 0 × 1 3 + 2 × 1 3 = 0 E ( Y 2 ) = ( - 2 ) 2 × 1 3 + 0 2 × 1 3 + 2 2 × 1 3 = 8 3 Var ( Y ) = E ( Y 2 ) - E ( Y ) 2 = 8 3 . Alternative answer: As calculated above, E ( Y ) = 0 (which could alternatively be derived by noticing that the distribution fo Y is symmetrical around 0). Therefore, Var ( Y ) = E (( Y - 0 ) 2 ) = E ( Y 2 ) . The calculation of E ( Y 2 ) can be done as above. (c) Let Z be a random variable with the following distribution: Z = 5 with probability 1 6 , Z = 2 with probability 1 3 , Z = 0 with probability 1 3 , and Z = - 4 with probability 1 6 . Calculate Var ( Z ) . Answer: E ( Z ) = - 4 × 1 6 + 0 × 1 3 + 2 × 1 3 + 5 × 1 6 = 5 6 E ( Z 2 ) = ( - 4 ) 2 × 1 6 + 0 2 × 1 3 + 2 2 × 1 3 + 5 2 × 1 6 = 49 6 Var ( Z ) = E ( Z 2 ) - E ( Z ) 2 = 269 36 . Alternative answer: As calculated above, E ( Z ) = 5 6 . Also Var ( Z ) = E (( Z - 5 6 ) 2 ) . The random vari- able ( Z - 5 6 ) 2 has the following distribution: it is ( 25 6 ) 2 with probability 1 6 , ( 7 6 ) 2 with probability 1 3 , ( - 5 6 ) 2 with probability 1 3 , and ( - 29 6 ) 2 with probability 1 6 . Therefore, Var ( Z ) = E (( Z - 5 6 ) 2 ) = 25 6 2 × 1 6 + 7 6 2 × 1 3 + - 5 6 2 × 1 3 + - 29 6 2 × 1 6 = 1614 216 = 269 36 . CS 70, Fall 2010, Soln 12 1
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(d) With Z as defined in part (c), calculate Var ( Z + 10 ) . Answer: The variance is invariant to shifts in the random variable, i.e., for any random variable X and any constant c , Var ( X + c ) = Var ( X ) . So Var ( Z + 10 ) = Var ( Z ) = 269 36 . Let R , S , T be independent r.v.’s with R Binomial ( 100 , 1 2 ) , S Binomial ( 20 , 1 2 ) , and T Binomial ( 90 , 1 3 ) . (e) Find Var ( R ) , Var ( S ) , and Var ( T ) . Answer: In general, if X Binomial ( n , p ) , Var ( X ) = np ( 1 - p ) . Var ( R ) = 100 × 1 2 × 1 - 1 2 = 25 Var ( S ) = 20 × 1 2 × 1 - 1 2 = 5 Var ( T ) = 90 × 1 3 × 1 - 1 3 = 20 (f) Let U = R + S . Calculate Var ( U ) . Answer: Var ( U ) = Var ( R + S ) = Var ( R )+ Var ( S ) since R and S are independent. Therefore, Var ( U ) = 25 + 5 = 30. Alternative answer: R counts the number of Heads among 100 flips of a fair coin, and S the number of Heads among another 20 flips of the fair coin. This means that U = R + S counts the number of Heads among 120 flips of a fair coin, i.e., U Binomial ( 120 , 1 2 ) . By the formula mentioned above, this means Var ( U ) = 120 × 1 2 × 1 - 1 2 = 30 . (g) Let V = R + T . Calculate Var ( V ) . Answer: Var ( V ) = Var ( R + T ) = Var ( R ) + Var ( T ) since R and T are independent. So Var ( U ) = 25 + 20 = 45. (h) Let W = 2 R + T . Calculate Var ( W ) . Answer: Var ( W ) = Var ( 2 R + T ) = Var ( 2 R )+ Var ( T ) since R and T are independent and therefore the random variables R 0 = 2 R and T are also independent. Var ( 2 R ) = 2 2 Var ( R ) = 4Var ( R ) since for any random variable X and any constant c , Var ( cX ) = c 2 Var ( X ) . Therefore, Var ( W ) = 4Var ( R )+ Var ( T ) = 4 × 25 + 20 = 120.
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