Bernoulli Processes

Stochastic Processes A stochastic process is a collection of random variables: –{X(t), t  T} t may be either discrete: –E.g., number of trials or continuous: –E.g., time, location

Bernoulli Processes A Bernoulli process is a collection of independent, identically distributed Bernoulli random variables: –P(X i = 1) = p –P(X i = 0) = 1 – p –E(X) = p –Variance(X) = p – p 2

Number of Successes in n Trials Number of successes in n trials: –Binomial(n, p) –, k = 0, 1, …, n –E(K) = n p –Variance(K) = n p (1 – p)

Number of Trials until a Success Number of trials between 2 arrivals (inter-arrival time): –Geometric(p) –P(N = 1) = p –P(N = 2) = p (1 – p), … –P(N = n) = p (1 – p) n-1, n = 1, 2, … –E(N) = 1/p –Variance(N) = (1 – p)/p 2

Number of Trials until a Success Number of remaining trials until next arrival is also geometric(p): –Regardless of how many trials have already elapsed! The process is “memory-less”

Number of Trials until r th Success Number of trials until the r th success (r th -order inter-arrival time): –Pascal distribution (Drake), or “negative binomial” distribution (Ross) –P(r th success on n th trial) = P(r–1 successes in n–1 trials) P(success on n th )=, n = r, r + 1, … –E(N r ) = r/p –Variance(N r ) = r (1 – p)/p 2

# of Failures until r th Success Number of failures until the r th success: –Negative Binomial distribution (Drake) –P(k failures before r th success) = P(k failures in k+r–1 trials) P(success on k+r) =, k = 0, 1, 2, … –E(K r ) = r (1/p – 1) = r (1 – p)/p

Sample Applications Number of customers wanting a particular service (Binomial) Number of defects out of n products (Binomial) Number of wasted items to complete an order (Negative Binomial) Number of customers until we run out (Pascal) Number of visits until a success call (Geometric) Number of uses until an item fails (Geometric): –If it is memory-less!

Random Erasures Consider a Bernoulli process with success probability p 1 Now assume that each success is randomly “erased” (turned into a failure) with probability p 2 The resulting sequence of successes and failures is a Bernoulli process with success probability p 1 (1 - p 2 )

Systematic Erasures Consider a Bernoulli process with success probability p 1 Now assume that every 2 nd success is “erased” (turned into a failure) Is the resulting sequence of successes and failures a Bernoulli process? –Think about the assumptions of a Bernoulli process!

Example Fred is giving out samples of dog food: –He makes calls door to door –He leaves a sample when the door is answered and a dog is there –P(door answered) = 3/4 –P(dog there) = 2/3 –Assume that “door answered” and “dog there” are independent (?) –Assume also that successive calls are independent (?)

Example By independence: –P(answered) = 3/4 –P(dog) = 2/3 –P(door answered and dog there) = 3/4 (2/3) = 1/2

Example Probability Fred gives out 1 st sample on 3 rd call: –P(no samples on 1 st 2 calls) P(sample on 3 rd )= (1 – 1/2) 2 1/2 = (1/4) (1/2) = 1/8 Probability Fred gives out 5 th sample on 11 th call: –Given that he gave out 4 samples in 1 st 8 calls –P[Fred gives out (5-4) th sample on (11-8) th call] = (1 – 1/2) 2 1/2 = (1/4) (1/2) = 1/8 Geometric distribution!

Example Probability Fred gives out 2 nd sample on 5 th call: –P(1 success in 1 st 4 trials) P(success on 5 th ) = – = 4 (1/2) 5 = 1/8 –Pascal distribution Probability Fred gives out 2 nd sample on 5 th call: –Given that he did not give out 2 nd sample on 2 nd call –P(2 nd sample on 5 th call)/P(2 nd sample not on 2 nd call) = [1/8]/[1 – (1/2) 2 ] = (1/8)/(1 – 1/4) = (1/8)/(3/4) = 1/6 –Conditional probability!

Example Fred needs new supply after he gives out last can: –Starts with two cans Probability that Fred makes  5 calls before he needs new supply: –Probability that Fred gives 2 nd sample  5 th call = –1 – P(Fred gives 2 nd sample on 2 nd, 3 rd, or 4 th call) = – = 1 – 1/4 – 1/4 - 3/16 = 5/16 –Each individual term is a Pascal distribution!