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Random Variables Discrete Random Variables For a discrete random variable X the probability distribution is described by the probability function p(x) = P[X = x], which has the following properties: 1. 0 p x 1 2. p x p x 1 x 3. i 1 P a x b i p x a x b Continuous random variables For a continuous random variable X the probability distribution is described by the probability density function f(x), which has the following properties : 1. f(x) ≥ 0 2. f x dx 1. 3. b P a X b f x dx. a The distribution function F(x) This is defined for any random variable, X. F(x) = P[X ≤ x] Properties 1. 2. 3. F(-∞) = 0 and F(∞) = 1. F(x) is non-decreasing (i. e. if x1 < x2 then F(x1) ≤ F(x2) ) F(b) – F(a) = P[a < X ≤ b]. 4. p(x) = P[X = x] =F(x) – F(x-) Here F x lim F u ux 5. If p(x) = 0 for all x (i.e. X is continuous) then F(x) is continuous. 6. For Discrete Random Variables F x P X x p u u x F(x) is a non-decreasing step function with F 0 and F 1 p x F x F x jump in F x at x. 1.2 F(x) 1 0.8 0.6 0.4 p(x) 0.2 0 -1 0 1 2 3 4 7. For Continuous Random Variables Variables x F x P X x f u du F(x) is a non-decreasing continuous function with F 0 and F 1 f x F x. f(x) slope F(x) 1 0 -1 0 1 x 2 To find the probability density function, f(x), one first finds F(x) then f x F x. Some Important Discrete distributions The Bernoulli distribution Suppose that we have a experiment that has two outcomes 1. Success (S) 2. Failure (F) These terms are used in reliability testing. Suppose that p is the probability of success (S) and q = 1 – p is the probability of failure (F) This experiment is sometimes called a Bernoulli Trial Let 0 if the outcome is F X 1 if the outcome is S q Then p x P X x p x0 x 1 The probability distribution with probability function q x 0 p x P X x p x 1 is called the Bernoulli distribution 1 0.8 0.6 p q = 1- p 0.4 0.2 0 0 1 The Binomial distribution We observe a Bernoulli trial (S,F) n times. Let X denote the number of successes in the n trials. Then X has a binomial distribution, i. e. n x n x p x P X x p q x 0,1, 2, x where 1. p = the probability of success (S), and 2. q = 1 – p = the probability of failure (F) ,n Example A coin is tossed n= 7 times. Let X denote the number of heads (H) in the n = 7 trials. Then X has a binomial distribution, with p = ½ and n = 7. Thus n x n x p x P X x p q x 0,1, 2, x 7 1 x 1 7 x 2 2 x 0,1, 2, , 7 x 7 1 7 2 x x 0,1, 2, ,7 ,n x 0 1/ p(x) p(x) 128 1 7/ 128 2 21/ 128 3 35/ 128 4 35/ 5 128 21/ 6 7/ 128 7 1/ 128 0.3 0.25 0.2 0.15 0.1 0.05 0 0 1 2 3 4 x 5 6 7 128 Example If a surgeon performs “eye surgery” the chance of “success” is 85%. Suppose that the surgery is perfomed n = 20 times Let X denote the number of successful surgeries in the n = 20 trials. Then X has a binomial distribution, with p = 0.85 and n = 20. Thus n x n x p x P X x p q x 0,1, 2, , n x 20 x 7 x .85 .15 x 0,1, 2, , 20 x x p (x ) x p (x ) x p (x ) x p (x ) 0 0.0000 6 0.0000 12 0.0046 18 0.2293 1 0.0000 7 0.0000 13 0.0160 19 0.1368 2 0.0000 8 0.0000 14 0.0454 20 0.0388 3 0.0000 9 0.0000 15 0.1028 4 0.0000 10 0.0002 16 0.1821 5 0.0000 11 0.0011 17 0.2428 0.3000 0.2500 p(x) 0.2000 0.1500 0.1000 0.0500 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x The probability that at least sixteen operations are successful = P[X ≥ 16] = p(16) + p(17) + p(18) + p(19) + p(20) = 0.1821 + 0.2428 + 0.2293 + 0.1368 + 0.0388 = 0.8298 Other discrete distributions • • • • Poisson distribution Geometric distribution Negative Binomial distribution Hypergeometric distribution The Poisson distribution • Suppose events are occurring randomly and uniformly in time. • Let X be the number of events occuring in a fixed period of time. Then X will have a Poisson distribution with parameter l. p x lx x! e l x 0,1, 2,3, 4, Some properties of the probability function for the Poisson distribution with parameter l. x 0 x 0 p x 1. l x x! e x 0 l l x x! e l 1 2 3 4 l l l l e 1 l 2! 3! 4! e l el 1 2 3 4 u u u u using e 1 u 2! 3! 4! n x n x 2. If pBin x p, n p 1 p x is the probability function for the Binomial distribution with parameters n and p, and we allow n → ∞ and p → 0 such that np = a constant (= l say) then l x l lim pBin x p, n pPoisson x l e n , p 0 x! n x n x Proof: pBin x p, n p 1 p x l Suppose np l or p n x n! l pBin x p, n pBin x l , n x ! n x ! n l x l 1 n n x n! l l 1 1 x x ! n n x ! n n x n x n n 1 n x 1 l l l 1 1 x! nn n n n x n x l 1 x 1 l l 1 1 1 1 1 x! n n n n x n Now lim pBin x l , n n l 1 lim 1 x ! n n x l l lim 1 x ! n n x x 1 l l 1 1 1 n n n x n n n Now using the classic limit lim pBin x l , n n l u lim 1 eu n n l n lx lim 1 e l pPoisson x l x ! n n x! x Graphical Illustration Suppose a time interval is divided into n equal parts and that one event may or may not occur in each subinterval. n subintervals time interval - Event occurs - Event does not occur X = # of events is Bin(n,p) As n→∞ , events can occur over the continuous time interval. X = # of events is Poisson(l) Example The number of Hurricanes over a period of a year in the Caribbean is known to have a Poisson distribution with l = 13.1 Determine the probability function of X. Compute the probability that X is at most 8. Compute the probability that X is at least 10. Given that at least 10 hurricanes occur, what is the probability that X is at most 15? Solution • X will have a Poisson distribution with parameter l = 13.1, i.e. p x l x x! e l x 13.1 13.1 e x! x 0,1, 2,3, 4, x 0,1, 2,3, 4, Table of p(x) x 0 1 2 3 4 5 6 7 8 9 p (x ) 0.000002 0.000027 0.000175 0.000766 0.002510 0.006575 0.014356 0.026866 0.043994 0.064036 x 10 11 12 13 14 15 16 17 18 19 p (x ) 0.083887 0.099901 0.109059 0.109898 0.102833 0.089807 0.073530 0.056661 0.041237 0.028432 P at most 8 P X 8 p 0 p 1 p 8 .09527 P at least 10 P X 10 1 P X 9 1 p 0 p 1 p 9 .8400 P at most 15 at least 10 P X 15 X 10 P X 15 X 10 P X 10 p 10 p 11 .8400 p 15 P 10 X 15 P X 10 0.708 The Geometric distribution Suppose a Bernoulli trial (S,F) is repeated until a success occurs. Let X = the trial on which the first success (S) occurs. Find the probability distribution of X. Note: the possible values of X are {1, 2, 3, 4, 5, … } The sample space for the experiment (repeating a Bernoulli trial until a success occurs is: S = {S, FS, FFS, FFFS, FFFFS, … , FFF…FFFS, …} (x – 1) F’s p(x) =P[X = x] = P[{FFF…FFFS}] = (1 – p)x – 1p Thus the probability function of X is: P[X = x] = p(x) = p(1 – p)x – 1 = pqx – 1 A random variable X that has this distribution is said to have the Geometric distribution. Reason p(1) = p, p(2) = pq, p(3) = pq2 , p(4) = pq3 , … forms a geometric series p x p 1 +p 2 +p 3 + x 1 p pq pq pq 2 3 p p 1 1- q p The Negative Binomial distribution Suppose a Bernoulli trial (S,F) is repeated until k successes occur. Let X = the trial on which the kth success (S) occurs. Find the probability distribution of X. Note: the possible values of X are {k, k + 1, k + 2, k + 3, 4, 5, … } The sample space for the experiment (repeating a Bernoulli trial until k successes occurs) consists of sequences of S’s and F’s having the following properties: 1. each sequence will contain k S’s 2. The last outcome in the sequence will be an S. A sequence of length x containing exactly k S’s SFSFSFFFFS FFFSF … FFFFFFS The # of S’s in the first x – 1 trials is k – 1. The last outcome is an S x 1 k x k p x P X x p q k 1 x k , k 1, k 2, The # of ways of choosing from the first x – 1 trials, the positions for the first k – 1 S’s. The probability of a sequence containing k S’s and x – k F’s. The Hypergeometric distribution Suppose we have a population containing N objects. Suppose the elements of the population are partitioned into two groups. Let a = the number of elements in group A and let b = the number of elements in the other group (group B). Note N = a + b. Now suppose that n elements are selected from the population at random. Let X denote the elements from group A. (n – X will be the number of elements from group B.) Find the probability distribution of X.\ Population GroupB (b elements) Group A (a elements) x sample (n elements) N-x Thus the probability function of X is: The number of ways x elements can be chosen Group A . a b x n x p x P X x N n The number of ways n - x elements can be chosen Group B . The total number of ways n elements can be chosen from N = a + b elements A random variable X that has this distribution is said to have the Hypergeometric distribution. The possible values of X are integer values that range from max(0,n – b) to min(n,a)