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Lecture 03-2007

# Lecture 03-2007 - Lecture III Uniform Probability Measure I...

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Unformatted text preview: Lecture III Uniform Probability Measure I think that Bieren’s discussion of the uniform probability measure provides a firm basis for the concept of probability measure. First, we follow the conceptual discussion of placing ten balls numbered 0 through 9 into a container. Next, we draw out an infinite sequence of balls out of the container, replacing the ball each time. Table 1. Random Draws of Single Digits Ball Drawn Draw 1 Draw 2 Draw 3 1 7 3 2 4 2 3 1 9 2 4 4 6 2 5 8 4 6 3 5 4 Taking each column, we can generate three random numbers { 0.741483, 0.029645, 0.302204} . Note that each of these sequences are contained in the unit interval Ω =[0,1]. The primary point of the demonstration is that the number draw ( x Ω = [0,1] ) is a probability measure. Taking x =0.741483 as the example, we want to prove that P ([0,x=0.741483]) = 0.741483. To do this we want to work out the probability of drawing a number less than 0.741483. As a starting point, what is the probability of drawing the first number in Table 1 less than 7, it is 7 ~{ 0,1,2,3,4,5,6} . Thus, without consider the second number, the probability of drawing a number less than 0.741483 is somewhat greater than 7/10. Next, we consider drawing a second number given that the first number drawn is greater than or equal to 7. Now, we are interested in the scenario were the number drawn is equal to...
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Lecture 03-2007 - Lecture III Uniform Probability Measure I...

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