HW1_soln

# HW1_soln - EE 568 Homework Solution 1 EE568 Homework...

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EE 568 Homework Solution 1 EE568 Homework Solution 1 Problem 1.1 The probability that w ( u ) = j is simply the number of ways to have j 1’s in a length n vector - ( n j ) - multiplied by the probability of obtaining a speciﬁc vector with j 1’s - ± j (1 - ± ) n - j : Pr { w ( u ) = j } = ± n j ² ± j (1 - ± ) n - j You may recognize this as the binomial distribution from your probability class. Finally, Pr { w ( u ) > d } is obtained via summation: Pr { w ( u ) > d } = n X j = d +1 Pr { w ( u ) = j } = 1 - d X j =0 Pr { w ( u ) = j } Problem 1.2 (a) First we decompose A B , A and B as unions of disjoint events: P ( A B ) = P ( A B 0 ) + P ( B A 0 ) + P ( A B ) P ( A ) = P ( A B 0 ) + P ( A B ) P ( B ) = P ( B A 0 ) + P ( A B ) By substituting P ( A B 0 ) and P ( B A 0 ) from the lower two equations into the top equation, we obtain P ( A B ) = P ( A ) + P ( B ) - P ( A B ) . From the axioms of probability, P ( A B ) is non-negative, and hence, P ( A B ) P ( A ) + P ( B ) . (b) Let event D = A B . By a repetitive use of the previous subdivision, we have that P ( A B C ) = P ( D C ) P ( D ) + P ( C ) P ( A ) + P ( B ) + P ( C ) (c) We prove this result by induction. Let Prop( n ) denote the proposition that P ³ n [ i =1 A i ! n X i =1 P ( A i ) is true.

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EE 568 Homework Solution 2 We know that Prop(2) is true from subdivision (a) of this problem (note that Prop(1) is trivially true). Assume that Prop( k ) is true for some k 2, i.e., P ± k [ i =1 A i ! k
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## This note was uploaded on 02/27/2008 for the course EE 568 taught by Professor Chugg during the Fall '07 term at USC.

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HW1_soln - EE 568 Homework Solution 1 EE568 Homework...

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