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Unformatted text preview: EE 278 Handout #4 Statistical Signal Processing Monday, July 13, 2009 Homework #1 Solutions 1. (15 points) (a) The sample space consists of triplets of the form (curtain with gold behind it, curtain chosen by player, curtain that Monty opens). If we denote the curtains by A , B , and C , then we can write our sample space as Ω = { ( A,A,B ) , ( A,A,C ) , ( A,B,C ) , ( A,C,B ) , ( B,B,A ) , ( B,B,C ) , ( B,A,C ) , ( B,C,A ) , ( C,C,A ) , ( C,C,B ) , ( C,B,A ) , ( C,A,B ) } . (b) As dicussed in lecture notes 1, for a discrete sample space, the probability measure is completely specified by the probabilities of the single outcome events. For this problem we can calculate the probabilities as follows: P( A,A,B ) = P { Monty opens B  gold behind A , player’s first choice is A } · P { gold behind A , player’s first choice is A } = 1 2 P { gold behind A } · P { player’s first choice is A } = 1 2 × 1 3 × 1 3 = 1 18 . Note that Monty Hall’s choice of curtain to open is not independent of where the gold is placed and what the player’s initial choice was. That is why we used conditional probability in the above derivation. Using the same approach we find P( A,A,C ) = P( B,B,A ) = P( B,B,C ) = P( C,C,A ) = P( C,C,B ) = 1 18 . This covers all the cases where the gold placement and the initial choice of con testant coincide. For the other cases, P { ( A,B,C ) } = P { Monty opens C  Gold behind A , player’s first choice is B } × P { Gold is behind A , player’s first choice is B } = 1 × P { Gold is behind A } × P { player’s first choice is B } = 1 × 1 3 × 1 3 = 1 9 . Using similar arguments, P( A,C,B ) = P( B,A,C ) = P( B,C,A ) = P( C,A,B ) = P( C,B,A ) = 1 9 and we now have a complete description of the probability space for this random experiment. To find the probability of winning if the player decides to switch, we 1 need to find the subset W of the sample space corresponding to this event, which is W = { ( A,B,C ) , ( A,C,B ) , ( B,A,C ) , ( B,C,A ) , ( C,A,B ) , ( C,B,A ) } . The probability of W is the sum of the probabilities of its members, and therefore the probability of winning is P( W ) = 6 × 1 9 = 2 3 . Somewhat surprisingly, the player is better off switching to the other curtain. 2. (15 points) (a) P( { ( k,m ) : k ≥ m } ) = summationdisplaysummationdisplay { ( k,m ): k ≥ m } p ( k,m ) = ∞ summationdisplay m =1 ∞ summationdisplay k = m p 2 (1 p ) k + m 2 = ∞ summationdisplay m =1 ∞ summationdisplay k =0 p 2 (1 p ) k +2 m 2 = p 2 (1 p ) 2 ∞ summationdisplay m =1 (1...
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This note was uploaded on 10/26/2011 for the course ECE 180 taught by Professor Eggers during the Spring '11 term at Aarhus Universitet.
 Spring '11
 Eggers
 Signal Processing

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