lecture4 - Lecture-4 Binomial r.v Approximations and...

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1 Lecture-4 Binomial r.v Approximations and Conditional Probability Density Functions Let X represent a Binomial r.v as in (3.42). Then from (2.32) Since the binomial coefficient grows quite rapidly with n , it is difficult to compute (1) for large n . In this context, two approximations are extremely useful. a. The Normal Approximation (Demoivre-Laplace Theorem) Suppose with p held fixed. Then for k in the neighborhood of np , we can approximate (29 = - = = = 2 1 2 1 . ) ( 2 1 k k k k n k k k k n q p k n k P k X k P (1) ! )! ( ! k k n n k n - = n npq
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2 (2) . 2 1 2 / ) ( 2 npq np k k n k e npq q p k n - - - π Thus if and in (1) are within or around the neighborhood of the interval we can approximate the summation in (1) by an integration. In that case (1) reduces to where We can express (3) in terms of the normalized integral that has been tabulated extensively (See Table 4.1). 1 k 2 k ( 29 , , npq np npq np + - (29 , 2 1 2 1 2 / 2 / ) ( 2 1 2 2 1 2 2 1 dy e dx e npq k X k P y x x npq np x k k - - - = = (3) ) ( 2 1 ) ( 0 2 / 2 x erf dy e x erf x y - = = - (4) . , 2 2 1 1 npq np k x npq np k x - = - =
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3 For example, if and are both positive ,we obtain Example 4.1: A fair coin is tossed 5000 times. Find the probability that the number of heads is between 2475 to 2525. Solution: We need Here n is large so that we can use the normal approximation. In this case so that and Since and the approximation is valid for and Thus Here (29 ). ( ) ( 1 2 2 1 x erf x erf k X k P - = 1 x 2 x ). 2525 2475 ( X P (5) , 2 1 = p 2500 = np . 35 npq , 2465 = - npq np , 2535 = + npq np 2475 1 = k . 2525 2 = k - = 2 1 2 . 2 1 2 / 2 1 x x y dy e k X k P π . 7 5 , 7 5 2 2 1 1 = - = - = - = npq np k x npq np k x
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4 x erf( x ) x erf( x ) x erf( x ) x erf( x ) 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.01994 0.03983 0.05962 0.07926 0.09871 0.11791 0.13683 0.15542 0.17364 0.19146 0.20884 0.22575 0.24215 0.25804 0.27337 0.80 0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50 0.28814 0.30234 0.31594 0.32894 0.34134 0.35314 0.36433 0.37493 0.38493 0.39435 0.40320 0.41149 0.41924 0.42647 0.43319 1.55 1.60 1.65 1.70 1.75 1.80 1.85 1.90 1.95 2.00 2.05 2.10 2.15 2.20 2.25 0.43943 0.44520 0.45053 0.45543 0.45994 0.46407 0.46784 0.47128 0.47441 0.47726 0.47982 0.48214 0.48422 0.48610 0.48778 2.30 2.35 2.40 2.45 2.50 2.55 2.60 2.65 2.70 2.75 2.80 2.85 2.90 2.95 3.00 0.48928 0.49061 0.49180 0.49286 0.49379 0.49461 0.49534 0.49597 0.49653 0.49702 0.49744 0.49781 0.49813 0.49841 0.49865 2 1 ) ( 2 1 ) ( 0 2 / 2 - = = - x G dy e x erf x y π Table 4.1
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5 Since from Fig. 4.1(b), the above probability is given by where we have used Table 4.1 b. The Poisson Approximation As we have mentioned earlier, for large n , the Gaussian approximation of a binomial r.v is valid only if p is fixed, i.e., only if and what if np is small, or if it does not increase with n ? , 0 1 < x (29 , 516 . 0 7 5 2 |) (| ) ( ) ( ) ( 2525 2475 1 2 1 2 = = + = - = erf x erf x erf x erf x erf X P 1 np . 1 npq . 258 . 0 ) 7 . 0 ( = erf Fig. 4.1 x (a) 1 x 2 x 2 / 2 2 1 x e - π 0 , 0 2 1 x x x (b) 1 x 2 x 2 / 2 2 1 x e - 0 , 0 2 1 < x x
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6 Obviously that is the case if, for example, as such that is a fixed number. Many random phenomena in nature in fact follow this pattern. Total number of calls on a telephone line, claims in an insurance company etc. tend to follow this type of behavior. Consider random arrivals such as telephone calls over a line. Let n represent the total number of calls in the interval From our experience, as we have so that we may assume Consider a small interval of duration as in Fig. 4.2. Had there been only a single call coming in, the probability
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lecture4 - Lecture-4 Binomial r.v Approximations and...

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