hw04soln - ECE 310 Spring 2005 HW 4 solutions Problem E3.5...

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ECE 310, Spring 2005, HW 4 solutions Problem E3.5 a) Using Definition 3.4.1 , (1 - 1 n ) r n - 1 ( A ) + 1 n I A ( ω n ) = n - 1 n × 1 n - 1 n - 1 X i =1 I A ( ω i ) + 1 n I A ( ω n ) = 1 n n X i =1 I A ( ω i ) = r n ( A ) / b) From part a), r n ( A ) - r n - 1 ( A ) = 1 n ( I A ( ω n ) - r n - 1 ( A )). ⇒ | r n ( A ) - r n - 1 ( A ) | = 1 n | I A ( ω n ) - r n - 1 ( A ) | 6 1 n c) No. Consider the following construction of a binary sequence as a counter- example: Let x 1 = 1 and r i = 1 i j = i j =1 x j For i > 2 if r i > 2 / 3, then x i +1 = 0 else if r i 6 1 / 3, then x i +1 = 1 else if 1 / 3 < r i < 2 / 3 and x i = 1, then x i +1 = 1 else if 1 / 3 < r i < 2 / 3 and x i = 0, then x i +1 = 0 The sequence of relative frequencies will oscillate between 1/3 and 2/3. Problem E3.8 The polynomial has the following roots: p 1 = 2, p 2 = 2 - 1, p 3 = - 2 - 1, p 4 = - 0 . 5, and p 5 = 0 . 5 . The fifth root, p 5 = 0 . 5, is the only that belongs to the interval [0 , 1], therefore p = 0 . 5 is the answer. Problem E3.9 P ( { a } ) = P ( { a, b }∩{ a, c } ) = P ( { a, b } )+ P ( { a, c } ) - P ( { a, b, c } ) = . 5+ . 7 - 1 = . 2 Similarly, we find P ( { b } ) = 0 . 3 and P ( { c } ) = 0 . 5 Problem E3.13 a) P ( A c ) = 1 - P ( A ) = . 3 b) P ( A B ) = P ( A ) + P ( B ) - P ( A B ) P ( A B ) + P ( A B ) = P ( A ) + P ( B ) = . 3 c) By the Union Bound, P ( A B ) 6 P ( A ) + P ( B ) = . 6 and as P ( A B ) > max( P ( A ) , P ( B )) = . 4, we have . 4 6 P ( A B ) 6 . 6. Problem E3.14 a) The answer is P ( B ) = . 5 and occurs for the case A B with P ( A ) = . 1.
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