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851lecture04

# 851lecture04 - Statistics 851(Fall 2013 Prof Michael...

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Statistics 851 (Fall 2013) September 11, 2013 Prof. Michael Kozdron Lecture #4: There is no uniform probability on ([0 , 1] , 2 [0 , 1] ) Our goal for today is to prove the first of the claims made last lecture, namely that there does not exist a uniform probability on the sample space [0 , 1] with the σ -algebra 2 [0 , 1] . Suppose that P is our candidate for the uniform probability on ([0 , 1] , 2 [0 , 1] ). Motivated by our experience with elementary probability, it is desirable for such a uniform probability to satisfy P { [ a, b ] } = b a for any interval [ a, b ] [0 , 1]. In other words, the probability of any interval is just its length. In fact, if 0 a < b 1, then the uniform probability should satisfy P { [ a, b ] } = P { ( a, b ) } = P { [ a, b ) } = P { ( a, b ] } = b a. In particular, P { a } = 0 for every 0 a 1 . Furthermore, the uniform probability should also satisfy countable additivity since this is one of the axioms for probability. That is, if 0 a 1 < b 1 < · · · < a n < b n < · · · 1, then the uniform probability should also satisfy P i =1 [ a i , b i ] = i =1 P { [ a i , b i ] } = i =1 ( b i a i ) . For instance, the probability that the outcome is in the interval [0 , 1 / 4] is 1 / 4, the probability the outcome is in the interval [1 / 3 , 1 / 2] is 1/6, and the probability that the outcome is in either the interval [0 , 1 / 4] or [1 / 3 , 1 / 2] should be 1 / 4 + 1 / 6 = 5 / 12. That is, P { [0 , 1 / 4] [1 / 3 , 1 / 2] } = P { [0 , 1 / 4] } + P { [1 / 3 , 1 / 2] } = 1 4 + 1 6 = 5 12 .

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