introduction-probability.pdf

# The measure n mσ 2 is called gaussian distribution

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The measure N m,σ 2 is called Gaussian distribution 10 ( normal distribu- tion ) with mean m and variance σ 2 . Given A ∈ B ( R ) we write N m,σ 2 ( A ) = A p m,σ 2 ( x ) dx with p m,σ 2 ( x ) := 1 2 πσ 2 e - ( x - m ) 2 2 σ 2 . The function p m,σ 2 ( x ) is called Gaussian density . 1.3.6 Exponential distribution on R with parameter λ > 0 (1) Ω := R . (2) F := B ( R ) Borel σ -algebra. (3) For A and G as in Subsection 1.3.5 we define, via the Riemann -integral, P 0 ( A ) := n i =1 b i a i p λ ( x ) dx with p λ ( x ) := 1I [0 , ) ( x ) λe - λx Again, P 0 satisfies the assumptions of Proposition 1.2.17, so that we can extend P 0 to the exponential distribution μ λ with parameter λ and density p λ ( x ) on B ( R ). 9 Georg Friedrich Bernhard Riemann, 17/09/1826 (Germany) - 20/07/1866 (Italy), Ph.D. thesis under Gauss. 10 Johann Carl Friedrich Gauss, 30/04/1777 (Brunswick, Germany) - 23/02/1855 (G¨ ottingen, Hannover, Germany).

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1.3. EXAMPLES OF DISTRIBUTIONS 29 Given A ∈ B ( R ) we write μ λ ( A ) = A p λ ( x ) dx. The exponential distribution can be considered as a continuous time version of the geometric distribution. In particular, we see that the distribution does not have a memory in the sense that for a, b 0 we have μ λ ([ a + b, ) | [ a, )) = μ λ ([ b, )) with the conditional probability on the left-hand side. In words: the proba- bility of a realization larger or equal to a + b under the condition that one has already a value larger or equal a is the same as having a realization larger or equal b . Indeed, it holds μ λ ([ a + b, ) | [ a, )) = μ λ ([ a + b, ) [ a, )) μ λ ([ a, )) = λ a + b e - λx dx λ a e - λx dx = e - λ ( a + b ) e - λa = μ λ ([ b, )) . Example 1.3.4 Suppose that the amount of time one spends in a post office is exponential distributed with λ = 1 10 . (a) What is the probability, that a customer will spend more than 15 min- utes? (b) What is the probability, that a customer will spend more than 15 min- utes from the beginning in the post office, given that the customer already spent at least 10 minutes? The answer for (a) is μ λ ([15 , )) = e - 15 1 10 0 . 220. For (b) we get μ λ ([15 , ) | [10 , )) = μ λ ([5 , )) = e - 5 1 10 0 . 604. 1.3.7 Poisson’s Theorem For large n and small p the Poisson distribution provides a good approxima- tion for the binomial distribution. Proposition 1.3.5 [Poisson’s Theorem] Let λ > 0 , p n (0 , 1) , n = 1 , 2 , ... , and assume that np n λ as n → ∞ . Then, for all k = 0 , 1 , . . . , μ n,p n ( { k } ) π λ ( { k } ) , n → ∞ .
30 CHAPTER 1. PROBABILITY SPACES Proof . Fix an integer k 0. Then μ n,p n ( { k } ) = n k p k n (1 - p n ) n - k = n ( n - 1) . . . ( n - k + 1) k ! p k n (1 - p n ) n - k = 1 k ! n ( n - 1) . . . ( n - k + 1) n k ( np n ) k (1 - p n ) n - k . Of course, lim n →∞ ( np n ) k = λ k and lim n →∞ n ( n - 1) ... ( n - k +1) n k = 1 . So we have to show that lim n →∞ (1 - p n ) n - k = e - λ . By np n λ we get that there exists a sequence ε n such that np n = λ + ε n with lim n →∞ ε n = 0 . Choose ε 0 > 0 and n 0 1 such that | ε n | ≤ ε 0 for all n n 0 . Then 1 - λ + ε 0 n n - k 1 - λ + ε n n n - k 1 - λ - ε 0 n n - k .

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