hw4sol - Stat 310A/Math 230A Theory of Probability Homework...

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Stat 310A/Math 230A Theory of Probability Homework 4 Solutions Andrea Montanari Due on October 25, 2010 Exercises on independent random variables and product measures Exercise [1.4.18] We will use the following fact repeatedly: the probability that X is divisible by j is P ( D j ) = X k =1 ( kj ) - s ( s ) = j - s . 1. Let p i be an enumeration of distinct primes. By deﬁnition, it suﬃces to show that for any ﬁnite sub-collection { p j } n j =1 and any n < , P ( n \ j =1 D p j ) = n Y j =1 P ( D p j ) . Indeed, all the primes p j in the sub-collection divide k if and only if k is divisible by their product. So the LHS is just ( Q n i =1 p j ) - s which by the fact we derived before, equals the RHS. 2. Euler’s formula is the statement that { X = 1 } if and only if X is not divisible by any prime number. Indeed, as the latter event is T p D c p , by continuity from below of P ( · ) we have that 1 ζ ( s ) = P ( X = 1) = lim n →∞ P ( n \ j =1 D c p j ) . In part (a) we veriﬁed the mutual independence of the collections { D p } , p prime, each of which is trivially a π -system. This implies the mutual independence of σ ( D p ), p prime (see Corollary 1.4.7), hence that of the events D c p . With P ( D p ) = p - s , we get that P ( T n j =1 D c p j ) = Q n j =1 (1 - p - s j ) leading to Euler’s formula when taking n → ∞ . 3. The event that no perfect square other than 1 divides X , is precisely the event that p 2 does not divide X for every prime p , which is T p D c p 2 . Similarly to part (a), it is not hard to verify that

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hw4sol - Stat 310A/Math 230A Theory of Probability Homework...

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