=+Now w E Hn C implies that P[H||], P[{}||5] =1 and @ EH'nC implies that P[H|],

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=+Now w E Hn C implies that P[H||], ≥ P[{}||5] =1 and @ EH'nC implies that P[H|], < P[] -[w]]] =0. Thus @ ECimplies that P[H ]=IH(w). But since H and are independent, P[H|] =

P(H) =; with probability 1, a contradiction.

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Probability And Measure Wiley Series In Probability And Mathematical Statistics

ISBN: 9788126517718

3rd Edition

Authors: Patrick Billingsley

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=+Now w E Hn C implies that P[H||], P[{}||5] =1 and @ EH'nC implies that P[H|],

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Probability And Measure Wiley Series In Probability And Mathematical Statistics

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