PostLecture03EN(3).pdf - Lecture 3 Chapters 2.5(to...

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Lecture 3 Chapters 2.5 (to factorial moments), 3.3 (the distributions will be discussed in a different order than in the book) Plan for lecture 3 1. Example for Chebyshev’s inequality (Lecture 2) 2. Moment generating functions 3. The uniform distribution 4. The Normal distribution-part 1
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The Chebychev inequality Theorem (2.4.7) For every random variable X with expected value μ and variance σ 2 > 0 , it holds that P ( | X - μ | ≥ k σ ) 1 k 2 and P ( | X - μ | ≤ k σ ) 1 - 1 k 2 I Interpretation: for every random variable X with expected value μ and variance σ 2 > 0, I at least 75% of the realisations lies within 2 σ of the expected value I at least 91% of the realisations lies within 3 σ of the expected value Note For Chebyshev’s inequality we only need the expected value and standard deviation. Example for the Markov and Chebychev inequality Assume tthat he number of products manufactured during a week is a random variable X with expected value 500. a. Give a bound on the probability that the next week’s production will be at least 1000 products. b. If the variance of the production equals 100, what can one say about the probability of the next week’s production being between 300 and 600? J (1) Solution a. From the Markov inequality it follows that: P ( X 1000 ) E [ X ] 1000 = 1 2 .
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b. We want to calculate the probability P ( 300 < X < 600 ) or find a lower bound/upper bound for this probability using the Chebychev inequality. I Chebyshev’s inequality gives: P ( | X - 500 | < 10 k ) 1 - 1 k 2 . I which is equivalent to P ( 500 - 10 k X 500 + 10 k ) 1 - 1 k 2 . I We can make two choices: 500 - 10 k = 300 or 500 + 10 k = 600. I The first one gives k = 20 and P ( 300 X 700 ) 1 - 1 400 . Since P ( 300 X 600 ) P ( 300 X 700 ) , we can’t say much about P ( 300 X 600 ) I The second choice gives 500 + 10 k = 600 and k = 10. Chebyshev’s inequality gives now P ( 400 X 600 ) 0 . 99 . Furthermore, P ( 300 X 600 ) P ( 400 X 600 ) 0 . 99 . Moment generating functions I The moment generating function M X : R 7→ R of a random variable X is defined as M X ( t ) = E ( e t X ) . I If X is a discrete random variable with probability mass function P , then M X ( t ) = X x e t x P ( X = x ) .
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