Negative Binomial Lecture

Negative Binomial Lecture - Lecture Sep 24 Review Geometric...

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Unformatted text preview: Lecture Sep 24 Review Geometric Distribution. Background: Generalizing binomial coefficients: First note that for any positive integer n, we can write n ( n − 1)… ( n − k + 1) ⎛n⎞ n! . Therefore, for any real number α , it seems reasonable to = ⎜ ⎟= k! ⎝ k ⎠ k !( n − k ) ! define the “binomial coefficient” ⎜ ⎟ = Calculus Binomial Theorem. Let α be a nonzero real number. For all real numbers x such that x < 1, we have (1 + x ) = α ⎛α ⎞ ⎝k⎠ α (α − 1)… (α − k + 1) k! . ∑ ⎜ k ⎟ x . (Proof uses Taylor’s Theorem with remainder) k k =0 ∞ ⎛α ⎞ ⎝⎠ Now, onto the negative binomial distribution, which we cover in better detail than our text. Model: A coin comes up heads with probability p. We toss the coin repeatedly until we get r heads. We let X be the toss number of the r th head and Y be the number of tails preceding the r th head. Then according to what book you are reading either X or Y has the negative binomial distribution. Notes: 1. Here r is a fixed positive number of our choosing—if we have r = 1 then we are dealing with the geometric distribution. 2. X = Y + r. Probability mass function of X : Obviously the possible values of X are k , k + 1, k + 2,… . Also, it is clear that for each such k we will have X = k exactly when we have had r − 1 heads in the first k − 1 tosses and then one more head on the k th toss. Thus we have ⎛ k − 1⎞ r −1 k − r ⎛ k − 1⎞ r k − r P( X = k) = ⎜ ⎟p q p=⎜ ⎟ p q for k = r , r + 1,… . ⎝ r −1⎠ ⎝ r −1⎠ Probability mass function of Y , take 1: The possible values of Y are 0,1,2, … , and for k = 0,1, 2,… we have P (Y = k ) = P ( X = k + r ) = ⎜ ⎛ k + r − 1⎞ k + r k ⎟ p q . ⎝ r −1 ⎠ Intermezzo: ⎜ ( −1) k ( −r )( −r − 1)… ( −r − k + 1) = k! ⎛ k + r − 1⎞ ⎛ k + r − 1⎞ ( r − 1 + k ) ( r − 1 + k − 1) ⎟=⎜ ⎟= k! ⎝ r −1 ⎠ ⎝ k ⎠ ( r − 1 + 1) = ( −1) k ⎛ −r ⎞ ⎜ ⎟ . Therefore ⎝k⎠ ⎛ −r ⎞ k ⎟ ( −q ) , k = 0,1, 2,…. ⎝k⎠ Probability mass function of Y , take 2: P (Y = k ) = p r ⎜ Verify that P ( X < ∞ ) = P (Y < ∞ ) = 1. Proof: ∞ ∞ ⎛ −r ⎞ −r k P (Y = k ) = p r ∑ ⎜ ⎟ ( −q ) = p r (1 − q ) = 1. ∑ k =0 k =0 ⎝ k ⎠ Final note: The pmf of Y is what gives this distribution its name. Hypergeometric, Poissson distributions. ...
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This note was uploaded on 06/12/2010 for the course MATH 307 taught by Professor Luikonnen during the Fall '08 term at Tulane.

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