hmwk3_prob4_soln.pdf - Math 464 Fall 18 Homework 3 4(Exposition Let X be a discrete RV whose range is 0 1 2 3 � � � Prove that E[X = ∞ X P(X > k k=0

hmwk3_prob4_soln.pdf - Math 464 Fall 18 Homework 3...

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Math 464 - Fall 18 - Homework 3 4. ( Exposition ) Let X be a discrete RV whose range is 0 , 1 , 2 , 3 , · · · . Prove that E [ X ] = X k =0 P ( X > k ) Proof. Consider the discrete random variable X with image { 0 , 1 , 2 , 3 , · · ·} . Recognizing the definition P ( X > k ) = P ( X = k + 1) + P ( X = k + 2) + · · · = X l = k +1 P ( X = l ) , the right side of the given equality follows as X k =0 P ( X > k ) = X k =0 X l = k +1 P ( X = l ) . Conditions are suitable to exchange the series, since P ( X = l ) 0. The indexing changes as (see the figure on page 3) X k =0 X l = k +1 P ( X = l ) = X l =1 l - 1 X k =0 P ( X = l ) . This becomes X k =0 P ( X > k ) = X k =0 X l = k +1 P ( X = l ) = X l =1 l - 1 X k =0 P ( X = l ) = X l =1 P ( X = l ) l - 1 X k =0 1 = X l =1 lP ( X = l ) = X l =0 lP ( X = l ) = E [ X ] . This verifies the given equality. 1
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If you want a more challenging problem you can instead prove that E [ X 2 ] = X k =0 P ( X > k )(2 k + 1) Proof. Again consider the discrete random variable X with image { 0 , 1 , 2 , 3 , · · ·} . We will apply the same definition for P ( X > k ) as in the previous proof, so that X k =0 P ( X > k )(2 k + 1) = X
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Unformatted text preview: k =0 ∞ X l = k +1 P ( X = l )(2 k + 1) . We can exchange the series, since P ( X = l )(2 k + 1) ≥ 0, and it follows that ∞ X k =0 P ( X > k )(2 k + 1) = ∞ X k =0 ∞ X l = k +1 P ( X = l )(2 k + 1) = ∞ X l =1 l-1 X k =0 P ( X = l )(2 k + 1) = ∞ X l =1 P ( X = l ) l-1 X k =0 (2 k + 1) = ∞ X l =1 P ( X = l ) " 2 l-1 X k =0 k + l-1 X k =0 1 # = ∞ X l =1 P ( X = l ) " 2 l-1 X k =1 k + l-1 X k =0 1 # = ∞ X l =1 P ( X = l ) 2 l ( l-1) 2 + l = ∞ X l =1 P ( X = l ) l 2 = ∞ X l =1 l 2 P ( X = l ) = ∞ X l =0 l 2 P ( X = l ) = E [ X 2 ] . The fifth line was obtained using the expression for the sum of n integers, n X k =1 k = n ( n + 1) 2 . The given equality is confirmed. 2 3...
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