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Unformatted text preview: 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 wellshuffled 52card 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 + × 1 3 + 2 × 1 3 = E ( Y 2 ) = ( 2 ) 2 × 1 3 + 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 ) 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 + × 1 3 + 2 × 1 3 + 5 × 1 6 = 5 6 E ( Z 2 ) = ( 4 ) 2 × 1 6 + 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 (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 ) ....
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This note was uploaded on 12/08/2010 for the course CS 70 taught by Professor Papadimitrou during the Fall '08 term at Berkeley.
 Fall '08
 PAPADIMITROU

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