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Independence and Bernoulli Trials

Independence and Bernoulli Trials. Independence. A, B independent implies: are also independent. Proof for independence of :. Events A and B are independent if. Application. let : Then Also Hence it follows that Ap and Aq are independent events!.

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Independence and Bernoulli Trials

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  1. Independence and Bernoulli Trials

  2. Independence • A, B independent implies: are also independent. Proof for independence of : Events A and B are independent if 2

  3. Application • let: • Then • Also • Hence it follows that Ap and Aq are independent events! Example 3

  4. Some Properties of Independence • Let P(A)=0, • Then: • Thus: P(AB) = P(A) P(B) = 0 The event of zero probability is independent of every other event Other independent events cannot be ME • Since: 4

  5. Application • A family of events {Ai} are said to be independent, if for every finite sub collection Remark Application • If and Ais are independent, • Application in solving number theory problems. 5

  6. Example Each switch remains closed with probability p. (a) Find the probability of receiving an input signal at the output. Three independent parallel switches Solution • Let Ai = “Switch Siis closed” and R = “input signal is received at the output”. Then 6

  7. Example – continued Another Solution • Since any event and its complement form a trivial partition, • Thus: Note:The events A1, A2, A3 do not form a partition, since they are not ME. Moreover, 7

  8. Example – continued (b) Find the probability that switch S1 is open given that an input signal is received at the output. Solution • From Bayes’ theorem • Also, because of the symmetry : 8

  9. Repeated Trials • A joint performance of the two experiments with probability models (1, F1, P1) and (2, F2, P2). • Two models’ elementary events: 1, 2 • How to define the combined trio (, F, P)? •  = 1 2 • Every elementary event in  is of the form  = (, ). • Events: Any subset A  B of  such that AF1 and B  F2 • F : all such subsets A  B together with their unions and complements. 9

  10. Repeated Trials • In this model, The events A  2 and 1  B are such that • Since the events A  2 and 1  B are independent for any A  F1 and B  F2. • So, for all A  F1 and B  F2 • Now, we’re done with definition of the combined model. 10

  11. Generalization • Experiments with • let • Elementary events • Events in are of the form and their unions and intersections. • If s are independent, and is the probability of the event in then 11

  12. Example • A has probability p of occurring in a single trial. • Find the probability that A occurs exactly k times, k  n in n trials. • The probability model for a single trial : (, F, P) • Outcome of n experiments is : where and • A occurs at trial # i , if • Suppose A occurs exactly k times in . Solution 12

  13. Example - continued • If belong to A, • Ignoring the order, this can happen in N disjoint equiprobable ways, so: • Where 13

  14. Bernoulli Trials • Independent repeated experiments • Outcome is either a “success” or a “failure” • The probability of k successes in n trials is given by the above formula, where p represents the probability of “success” in any one trial. 14

  15. Example Example • Toss a coin n times. Obtain the probability of getting k heads in n trials. • “head” is “success” (A) and let • Use the mentioned formula. • In rolling a fair dice for eight times, find the probability that either 3 or 4 shows up five times ? • Thus Example 15

  16. Bernoulli Trials • Consider a Bernoulli trial with success A, where • Let: • As are mutually exclusive, • so, 16

  17. Fig. 2.2 Bernoulli Trials • For a given n and p what is the most likely value of k ? • Thus if or • Thus as a function of k increases until • The is the most likely number of successes in n trials. 17

  18. Effects Of p On Binomial Distribution 18

  19. Effects Of p On Binomial Distribution 19

  20. Effects Of n On Binomial Distribution 20

  21. Example • In a Bernoulli experiment with n trials, find the probability that the number of occurrences of A is between k1 and k2. Solution 21

  22. Example • Suppose 5,000 components are ordered. The probability that a part is defective equals 0.1. What is the probability that the total number of defective parts does not exceed 400 ? Solution 22

  23. If kmax is the most likely number of successes in n trials, • or • So, • This connects the results of an actual experiment to the axiomatic definition of p. 23

  24. Bernoulli’s Theorem • A: an event whose probability of occurrence in a single trial is p. • k: the number of occurrences of A in n independent trials • Then , • i.e. the frequency definition of probability of an event and its axiomatic definition ( p) can be made compatible to any degree of accuracy. 24

  25. Proof of Bernoulli’s Theorem • Direct computation gives: • Similarly, 25

  26. Proof of Bernoulli’s Theorem - continued • Equivalently, • Using equations derived in the last slide, 26

  27. Proof of Bernoulli’s Theorem - continued • Using last two equations, • For a given can be made arbitrarily small by letting n become large. • Relative frequency can be made arbitrarily close to the actual probability of the event A in a single trial by making the number of experiments large enough. • As the plots of tends to concentrate more and more around Conclusion 27

  28. Example: Day-trading strategy • A box contains nrandomly numbered balls (not necessarily 1 through n) • m = np(p<1)of them are drawn one by one by replacement • The drawing continues untila ball is drawn with a number larger than the first mnumbers. • Determine the fraction pto be initially drawn, so as to maximize the probability of drawing the largest among the nnumbers using this strategy. 28

  29. Day-trading strategy: Solution • Let drawn ball has the largest number among: alln balls, the first kballs is in the group of first mballs, k > m. • = where A = “largest among the first k balls is in the group of first m balls drawn” B = (k+1)st ball has the largest number among all n balls”. • Since A and B are independent , • So, P (“selected ball has the largest number among all balls”) 29

  30. Day-trading strategy: Solution • To maximize with respect to p, let: • So, p = e-1≈ 0.3679 • This strategy can be used to “play the stock market”. • Suppose one gets into the market and decides to stay up to 100 days. • When to get out? • According to the solution, when the stock value exceeds the maximum among the first 37 days. • In that case the probability of hitting the top value over 100 days for the stock is also about 37%. 30

  31. Example: Game of Craps • A pair of dice is rolled on every play, • Win: sum of the first throw is 7 or 11 • Lose: sum of the first throw is 2, 3, 12 • Any other throw is a “carry-over” • If the first throw is a carry-over, then the player throws the dice repeatedly until: • Win: throw the same carry-over again • Lose: throw 7 Find the probability of winning the game. 31

  32. Game of Craps - Solution P1: the probability of a win on the first throw Q1: the probability of loss on the first throw B: winning the game by throwing the carry-over C: throwing the carry-over 32

  33. Game of Craps - Solution The probability of win by throwing the number of Plays that do not count is: The probability that the player throws the carry-over k on the j-th throw is: 33

  34. Game of Craps - Solution the probability of winning the game of craps is 0.492929 for the player. Thus the game is slightly advantageous to the house. 34

  35. Game Of Craps Using Biased Dice • Now assume that for each dice, • Faces 1, 2 and 3 appear with probability • Faces 4, 5 and 6 appear with probability • If T represents the combined total for the two dice, we get 35

  36. Game Of Craps Using Biased Dice • This gives the probability of win on the first throw to be • and the probability of win by throwing a carry-over to be • Thus • Although perfect dice gives rise to an unfavorable game, 36

  37. Game Of Craps Using Biased Dice • Even if we let the two dice to have different loading factors and (for the situation described above), similar conclusions do follow. • For example, • gives: 37

  38. Euler, Ramanujan and Bernoulli Numbers

  39. Euler’s Identity • S. Ramanujan in (J. of IndianMath Soc; V, 1913): • if are numbers less than unity • where the subscripts are the series of prime numbers, then • The terms above are arranged so that the product obtained by multiplying the subscripts are the series of all natural numbers • This follows by observing that the natural numbersare formed by multiplying primes and their powers. 39

  40. Euler’s Identity • This is used to derive a variety of interesting identities including the Euler’s identity. • By letting • This gives the Euler identity • The right side can be related to the Bernoulli numbers (for s even). • Bernoulli numbers are positive rational numbers defined through the power series expansion of the even function • Thus if we write • Then 40

  41. Euler’s Identity • Alse we can obtain : • This is another way to define Bernoulli numbers • Further • which gives • Thus 41

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