sol12 - CS 70 Discrete Mathematics and Probability Theory...

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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 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 + × 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.

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sol12 - CS 70 Discrete Mathematics and Probability Theory...

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