Unformatted text preview: CSE 21: Solutions  Problem Set 5
1. Let X =the number of changeovers in n flips. We want E(X). For i = 1, 2, , n  1 (not n), define Xi = 1, if the (i + 1)th flip differs from the ith 0, if the (i + 1)th flip is the same as the ith Then X = X1 + X2 + + Xn1 , thus E(X) = E(X1 ) + E(X2 ) + + E(Xn1 ). For each i, E(Xi ) = 0 P r(Xi = 0) + 1 P r(Xi = 1) = P r(Xi = 1). P r(Xi = 1) means the (i + 1)th flip differs from the ith flip. There are only two outcomes for the 1 1 1 (i + 1)th flip, each of which has probability . Therefore, P r(Xi = 1) = . Then, E(X) = (n  1). 2 2 2 2. (a) Since Var(X) = E[X 2 ]  E[X]2 , we can use E[X 2 ] = Var(X) + E[X]2 to find E[X 2 ]. (b) By the linearity of expected values, E[(a + bX)2 ] = E[a2 + 2ab X + b2 X 2 ] = a2 + 2ab E[X] + b2 E[X 2 ], where E[X] is given, and E[X 2 ] is known from part (a). (c) Similar to part (a), we have Var(a + bX) = E[(a + bX)2 ]  E[a + bX]2 , where E([(a + bX)2 ] is known from part (b), and E[a + bX] can be computed as E[a + bX] = a + bE[X]. 3. (a) Each number can be 2, 4 or 6. By rule of product, there are 3 3 = 9 such results. The total number of results is 6 6 = 36. Then P r(E) = 9/36 = 1/4. (b) List all the possible pairs where the first one is smaller: (1, 4), (1, 5), (1, 6), (2, 5), (2, 6), (3, 6). In each pair, switching the two numbers gives us a new pair. Then there are 12 possible pairs. P r(F ) = 12/36 = 1/3. (c) E F = the two numbers are both even and differ by at least 3. There are only two such pairs: (2, 6), (6, 2). Then P r(E F ) = 2/36 = 1/18. (d) P r(E)P r(F ) = (1/4)(1/3) = 1/12 = P r(E F ) = 1/18. Then E and F are not independent. 1 ...
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This note was uploaded on 04/23/2008 for the course CSE 21 taught by Professor Graham during the Spring '07 term at UCSD.
 Spring '07
 Graham

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