lec2 - CS648 Randomized Algorithms Lecture notes 2 1 Some...

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CS648 Randomized Algorithms Lecture notes : 2 1 Some probability problems Lemma 1.1 For a non-negative integer valued random variable X , E [ X ] = X t =0 Pr [ X > t ] Proof: Since X is a non-negative integer values random variable, the following equality holds. Pr [ X > t ] = X j = t +1 Pr [ X = j ] (1) If we plug Equation 1 in the expression t =0 Pr [ X > t ] , a term Pr [ X = j ] will appear j times in the sum for each integer j > 0 , hence X t =0 Pr [ X > t ] = X t =0 t Pr [ X = t ] = E [ X ] 2 Lemma 1.2 If we select k integers uniformly from set { 1 , ..., l } , the expected value of the smallest of them is ( l + 1) / ( k + 1) . We provide below a “calculative” way to prove the lemma above. We shall later present a simple and calculation-free proof. Let X be the random variable for the smallest of the k numbers selected from { 1 , ..., l } . Using Lemma 1.1, E [ X ] = X i =0 Pr [ X > i ] = X i =0 ( n - i k ) ( n k ) = 1 ( n k ) i = n - k X i =0 ± n - i k ! = 1 ( n k ) n k ! + ± n - 1 k ! + ... + ± k k !# = 1 ( n k ) "±± n + 1 k + 1 ! - ± n k + 1 !! + ±± n k + 1 ! - ± n - 1 k + 1 !! + · · · + ±± k + 1 k + 1 ! - ± k k + 1 !!# = 1 ( n k ) ± n + 1 k + 1 ! = n + 1 k + 1 1

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In the ﬁfth equality, we have used the well known binomial equality ( m - 1 r - 1 ) = ( m r ) - ( m - 1 r ) for each term. Calculation free proof for Lemma 1.2 : First we state a few simple observations. We say that two random variables X and Y have same probability distribution if for all t , Pr [ X = t ] = Pr [ Y = t ] . It is easy to see that if two random variables have the same probability distribution, then their expected values is also the same. Consider a circle with l + 1 nodes connected with edges of unit length each. All the nodes are colored white initially. If we select k + 1 nodes uniformly randomly from the set of nodes forming the circle, we get k + 1 segments. Each segment is the sequence of one of more edges with black nodes at the boundaries and all internal nodes colored white. Let us assign number 1 to the segment containing the top most node, and going in an anticlockwise direction, assign numbers to the all other segments from 2 to k + 1 . In case the topmost node is a black node, then the ﬁrst segment is the segment with topmost node as its right boundary node. Problem 1.1 Consider selecting k +1 nodes uniformly from l +1 nodes of a circle and coloring them black. What is expected length of i th segment ? Solution: The crucial point to be noted is that the nodes are selected uniformly and irrespective of their positions. Therefore, due to this uniform sampling and symmetry, lengths of each of the ( k + 1) -segments formed have identical probability distribution though they are not independent (since they all sum up to l + 1 ). This implies the following assertion : Let X i denote the length of i th segment, then E [ X 1 ] = E [ X 2 ] = E [ X 3 ] = ... = E [ X k +1 ] Also note that i X i = l + 1 . So ( k + 1) E [ X i ] = X j E [ X j ] = E [ X j X
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lec2 - CS648 Randomized Algorithms Lecture notes 2 1 Some...

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