note 4 - MA2216/ST2131 Probability Notes 4 § 1. Discrete...

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Unformatted text preview: MA2216/ST2131 Probability Notes 4 § 1. Discrete Uniform Distribution. 1. If X assumes the values x 1 < x 2 < ... < x n , with equal probabilities, then X is called a discrete uniform r.v. The probability distribution of X is called the discrete uniform distribution . The probability density function of X is given by f X ( x ) = 1 n , x = x 1 ,x 2 ,...,x n . 2. The mean and variance of X are, respectively, EE ( X ) = 1 n n X i =1 x i , Var( X ) = 1 n n X i =1 x 2 i- ˆ 1 n n X i =1 x i ! 2 . 3. Distribution Function. The c.d.f. F X ( x ) is given by F X ( x ) = IP { X ≤ x } = X i : x i ≤ x f X ( x i ) = # of x i ’s with x i ≤ x n (to be more explicit) = , for x < x 1 , j n , for x j ≤ x < x j +1 for 1 ≤ j < n , 1 , for x n ≤ x . (It is obvious to see that F X ( · ) is a step function which is right-continuous with left limits.) 1 § 2. Geometric Distribution. 1. The model which generates a geometric distribution can be described as follows: Conduct independent Bernoulli trials, IP(success) = p, IP(failure) = q = 1- p, < p < 1 , until a success is obtained. Let X be the number of trials required. Then, f X ( n ) = IP( X = n ) = q n- 1 p, n = 1 , 2 ,..., and X is called a geometric r.v. with parameter p , denoted by X ∼ Geom( p ) . X is also said to have a geometric distribution . Note that f X ( n ) is indeed a genuine p.d.f., for f X ( n ) > 0 and ∞ X n =1 q n- 1 p = p ∞ X n =0 q n = p 1- q = 1 . Obviously, the above infinite sum has to do with a geometric series, which explains why it is called a geometric distribution. Remark. Recall Ex.1 in Tut. 1, which can be re-phrased as a geometric distribution model. 2. Example: During busy hours, a telephone exchange is very near capac- ity, so people cannot find a line to use. If the probability of a connection during busy time is p = 0 . 1, what is the probability that at least 5 attempts are needed for a successful call? Sol. Let X denote the number of attempts so as to get the first success. It is obvious to see that X has a geometric distribution with parameter p = 0 . 1. The desired probability is given by IP { X ≥ 5 } = ∞ X x =5 p (1- p ) x- 1 = (1- p ) 5- 1 = (0 . 9) 4 . 2 Note. Here, we actually derive a useful formula for geometric r.v.’s. Namely, IP { X ≥ k } = ∞ X x = k p (1- p ) x- 1 = (1- p ) k- 1 = q k- 1 , (2 . 1) for k = 1 , 2 ,.... It is quite straightforward to figure out a “ probabilistic ” derivation. The event { X ≥ k } simply means that each of the first k- 1 trials results in a failure, etc. 3. For a geometric r.v. X with parameter p , EEX = 1 p , Var( X ) = EE [ X 2 ]- [ EEX ] 2 = 1- p p 2 . (2 . 2) It is well-known that ∞ X n =1 q n- 1 = ∞ X n =0 q n = 1 1- q ....
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note 4 - MA2216/ST2131 Probability Notes 4 § 1. Discrete...

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