02_Probability_part5

# 02_Probability_part5 - explanations 1 if n large the pmf of...

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NTHU MATH 2820, 2008, Lecture Notes Ch1~6, p.73 explanations. 1. if n large, the pmf of B ( n, p ) is not easily calculated. Then, we can approximate them by pmf of P ( λ ), where λ = np . 2. Let X be the number of times some event occurs in a given time interval I . Divide the interval into many small subintervals I k , k =1 ,...,n ,o f equal length. Let N k be the number of events occurring in I k .Wh en we can assume N 1 ,...,N n are independent and approximately B ( p ), X has a distribution near P ( λ ), where λ = . pmf: p ( x )= λ x x ! e λ ,x =0 , 1 , 2 ,... 0 , otherwise . mgf: e λ ( e t 1) ,t R . mean: λ variance: λ parameter: λ > 0 example: number of phone calls coming into an exchange during a unit of time made by Shao-Wei Cheng (NTHU, Taiwan) Ch1~6, p.74 Note: Let X i P ( λ i ), i ,...,k ,and X 1 ,...,X k are independent. Then, Y = X 1 + ··· + X k P ( λ 1 + + λ k ). Definition 4.8 (Hypergeometric distribution HG(r, n, m) , sec 2.1.4) Suppose that an urn contains n black balls and m white balls. Let X denote the number of black balls drawn when taking r balls without replacement. Then, X follows hypergeometric distribution. pmf: p ( x n x m r x n + m r , x , 1 ,..., min( r, n ) , r x m 0 , otherwise

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NTHU MATH 2820, 2008, Lecture Notes Ch1~6, p.75 mgf: No explicit form. mean: rn n + m variance: rnm ( n + m r ) ( n + m ) 2 ( n + m 1) parameter: r, n, m, =1 , 2 ,..., r n + m example: sampling industrial products for defect inspection Notes. relationship between hypergeometric and binomial dis- tributions: Let m , n →∞ in such a way that p m,n n m + n p , where 0 <p< 1. Then, n x m r x n + m r r x p x (1 p ) r x . made by Shao-Wei Cheng (NTHU, Taiwan) Ch1~6, p.76 • continuous distributions Definition 4.9 (Uniform distribution U(a, b) , sec 2.2) Chooseanumberatrandombetween a and b . Note: U (0 , 1) is useful for pseudo-random number generation pdf: f ( x )= 1 b a ,a x b 0 , otherwise . cdf: F ( x 0 ,x < a x a b a x b 1 > b . mgf: e bt e at t ( b a ) ,t R . mean: a + b 2 variance: ( b a ) 2 12 parameter: a,b R ,a< b
NTHU MATH 2820, 2008, Lecture Notes Ch1~6, p.77 Definition 4.10 (Exponential distribution E( λ ) , sec 2.2.1) pdf: f ( x )= λ e λ x ,x 0 0 < 0 . cdf: F ( x 1 e λ x 0 0 < 0 . mgf: λ λ t ,t< λ . mean: 1 λ variance: 1 λ 2 parameter: λ > 0 example: lifetime or waiting time made by Shao-Wei Cheng (NTHU, Taiwan) Ch1~6, p.78 Notes: 1. memoryless (future independent of past): Let T E ( λ ), then P ( T>t + s | T>s P ( + s and ) P ( ) = P ( + s ) P ( ) = e λ ( t + s ) e λ s = e λ t = P ( ) ( ) If a continuous distribution is memoryless, it is exponential. It does not mean the two events and + s are independent. 2. relationship between exponential (or Gamma) and Poisson distribution Let T 1 ,T 2 3 ,... be i.i.d. E ( λ )and S k = T 1 + ··· + T k , k =1 , 2 . Let X i be the number of S k ’s that falls in [ t i 1 ,t i ], i ,...,n ,then X 1 ,...,X n are independent, and X i P ( λ ( t i t i 1 )). The reverse statement is also ture.

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02_Probability_part5 - explanations 1 if n large the pmf of...

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