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Unformatted text preview: 4.4 The Exponential and Gamma Distribution . A continuous variable X is said to have exponential distribution with parameter λ , if its density function is ) ( )  ( x f x f = λ = λ eλ x , if x ≥ 0 0, otherwise. The graphs of exponential distributions are given below. 1 Remarks : (i) Each value of λ denotes an exponential distribution. So, better denote it by ) ( λ x f ; λ is called the “parameter”. (ii) { } : )  ( λ λ x f denotes a family of exponential distributions. (iii) Proportion of t X : . ) ( ∫ ∞ = t t e dx x f λ Example. The time (sec) it takes a librarian to locate an entry in a file of records on checkedout books has an exponential distribution with λ = 0.05. (a ) What is the probability of all location times that are less than 20 seconds? At least 25 seconds? Between 10 and 25 seconds? (b) Does a location of 1 sec qualify as one of the fastest 5% of all times? 2 Solution . (a) Using the integration shortcut mentioned, we have P (X < 20) = 1 P (X ≥ 20) = 1  e λ (20) = 1  e(.05)(20) = .632. P (X ≥ 25) = e λ (25) = e(.05)(25) = .287. P (10 ≤ X ≤ 25) = P (X >10) – P (X>25) = e.5 e1.25 = .320. (b) Let x denote the time that exceeds the fastest 5% of times. Then, P (X < x ) = .05, or equivalently, P(X ≥ x ) = .95. Therefore, x e λ = .95, so  λ x = ln(.95) and x =ln(.95)/ λ = .05129329/.05 = 1.026. Yes, x = 1 qualifies as one of the fastest 5%, since x < x . Example: If the variable X follows exponential distributions with parameter λ: Find the variance, sd and proportion of X beyond μ σ + Solution . We need μ and σ for an exponential random variable. 3 ∫ ∞ = d x e x x λ λ μ Using integration by parts let x x e v dx du dx e dv x u λ λ λ = = = = , , , So, ] =  + = = ∞ ∞ ∞ ∞ ∫ ∫ λ λ λ λ λ λ λ 1 1 x x x x e dx e e x dx e x That is, = λ μ 1 . Now, 2 2 2 μ λ σ λ = ∫ ∞ d x e x x Using integration by parts twice: Fist we obtain: x x e v xdx du dx e dv x u λ λ λ = = = = , 2 , , 2 [ ] ∫ ∫ ∫ ∞ ∞ + = = 2 2 2 2 d x x e d x x e e x d x e x x x x x λ λ λ λ λ Then, using integration by parts again: x x e v d x d u d x e d v x u λ λ λ = = = = 1 , , , 2 2 2 2 1 2 2 1 2 2 λ λ λ λ λ λ λ λ λ = =  =  = ∞ ∞ ∞ ∫ ∫ x x x x e dx e xe dx xe 4 Thus, =  = = 2 2 2 2 2 2 1 1 2 2 λ λ λ μ λ σ That is, = 2 2 1 λ σ and = λ σ 1 Finally: ( 29 135 ....
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This note was uploaded on 07/25/2008 for the course STT 430 taught by Professor Nane during the Spring '08 term at Michigan State University.
 Spring '08
 NANE

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