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Discrete Distributions Week 5 Objectives • • • • On completion of this module you should be able to: find expected value, variance and standard deviation for a discrete random variable, calculate and interpret covariance, determine the appropriateness of the binomial distribution for certain situations, calculate probabilities of events for the binomial distribution, 2 Objectives On completion of this module you should be able to: • determine the appropriateness of the Poisson distribution for certain situations and • calculate probabilities of events for the Poisson distribution. 3 Example 5-1 The CEO of a large corporation is concerned that the corporations intranet has been suffering with a number of faults and is unavailable much of the time. He arranges for the collection of data which details the number of times where the intranet has been down, and for how long, over the last three hundred business days. The results are given on the following slide: 4 Example 5-1 Hours unavailable Frequency (per day) 0 210 1 24 2 30 3 24 4 6 5 0 6 6 a) Form the probability distribution for the number of hours per day the intranet was down. 5 Solution 5-1 Hours unavailable Frequency 0 210 1 24 2 30 3 24 4 6 5 0 6 6 Total 300 6 Solution 5-1 Hours unavailable Frequency Probability 0 210 1 24 2 30 3 24 4 6 5 0 6 6 Total 300 210 = 0.7 300 24 = = 0.08 300 30 = = 0.1 300 24 = = 0.08 300 6 = = 0.02 300 0 = =0 300 6 = = 0.02 300 = 7 Solution 5-1 Hours unavailable Probability (X) P(X) 0 0.7 1 0.08 2 0.1 3 0.08 4 0.02 5 0 6 0.02 This is the probability distribution. 8 P(X) 0.7 0.6 How can we tell what we might expect to happen “on average”? 0.5 0.4 0.3 0.2 0.1 0.0 0 1 2 3 4 Hours unavailable per day (X) 5 6 9 Expected value of a discrete random variable • The mean μ of a probability distribution is the expected value of its random variable. N E X XiP Xi i 1 where X i the ith outcome of X (the discrete random variable of interest) P X i probability of occurrence of the ith outcome of X 10 Example 5-1 (b) Compute the mean or expected number of hours that the intranet will be down per day. N E X XiP Xi i 1 0 0.7 1 0.08 6 0.02 0.72 Thus we can expect the intranet to be down for 0.72 of an hour, or about 43 minutes, each day. 11 Variance of a discrete random variable • The variance 2 of a probability distribution is: N X i E X P X i 2 2 i 1 • The standard deviation 2 of a probability distribution is: 2 N X i 1 E X P X i 2 i 12 Example 5-1 (c) Compute the standard deviation. N X i E X P X i 2 2 i 1 0 0.72 0.7 1 0.72 0.08 2 2 6 0.72 0.02 2 1.7216 2 1.7216 1.3121 (to 4 dec. pl.) 13 Example 5-1 (d) What is the probability that on any given day the intranet will be down for fewer than two hours? P fewer than 2 hours P X 2 P X 0 P X 1 0.7 0.08 0.78 So the probability that the intranet will be down for fewer than two hours is 0.78. 14 Example 5-1 (e) What is the probability that on any given day the intranet will not be down at all? P 0 hours P X 0 0.7 So the probability that the intranet will not be down at all is 0.7. 15 Example 5-1 (f) What is the probability that on any given day the intranet will be down for at least three hours? P at least 3 hours P X 3 P X 3 P X 4 P X 5 P X 6 0.08 0.02 0 0.02 0.12 So the probability that the intranet will be down at least three hours is 0.12. 16 Expected value of the sum of two random variables The expected value of the sum of two random variables is equal to the sum of the expected values. E X Y E X E Y 17 Variance of the sum of two random variables The variance of the sum of two random variables is equal to the sum of the variances plus twice the covariance. var X Y X2 Y X2 Y2 2 XY Standard deviation of the sum of two random variables X Y X2 Y 18 Example 5-2 One of your clients is deciding how they should invest a sum of money. They have obtained a report giving the predicted annual return for a $1000 investment in two different stocks. The following probability distribution was included in the report. Probability 0.2 Stock P -$50 Stock K -$105 0.2 0.4 0.2 $10 $30 $100 $2 $25 $200 19 Example 5-2 Let X = stock P and Y = stock K. (a) Find the expected return of stock P. N E X X XiP Xi i 1 50 0.2 10 0.2 30 0.4 100 0.2 $24 (b) Find the expected return of stock K. N E Y Y Yi P Yi i 1 105 0.2 2 0.2 25 0.4 200 0.2 $29.40 20 Notes on rounding answers • All final answers that are money amounts should be rounded to exactly 2 decimal places (for dollars and cents). • The only exception is if the value is whole dollars, when 0 decimal places are acceptable. • But, don’t round intermediate steps of working or you will introduce a rounding error (and be marked wrong in assessment items). • Correct: $4.57 $100.95 $5,984.56 $495 • Incorrect: $4.6 $100.9543 21 Example 5-2 (c) Find the standard deviation for stock P. N var X X i E X P X i 2 X 2 i 1 50 24 0.2 20 24 0.2 2 2 30 24 0.4 100 24 0.2 2 2 2304 X X2 2304 $48 22 Example 5-2 (d) Find the standard deviation for stock K. N var Y Yi E Y P Yi 2 Y 2 i 1 105 29.4 0.2 2 29.4 0.2 2 2 25 29.4 0.4 200 29.4 0.2 2 2 9591.44 Y Y2 9591.44 $97.94 23 Covariance The covariance between two discrete random variables X and Y is N XY X i E X Yi E Y P X iYi i 1 24 Example 5-2 (d) Find the covariance of stock P and stock K. N XY X i E X Yi E Y P X iYi i 1 50 24 105 29.4 0.2 10 24 2 29.4 0.2 30 24 25 29.4 0.4 100 24 200 29.4 0.2 4648.4 25 Example 5-2 • • • Based on your answers above, what recommendations might you make to the client. Explain. Stock K has higher expected return and a higher standard deviation – higher return but higher risk. Recommend stock K to client if they are willing to take a higher risk, but stock P if they want a safer investment. Covariance is positive – there is positive relationship between the stock – as one increases so does the other. 26 Portfolio expected return The portfolio expected return for a two-asset investment is: E P wE X 1 w E Y where E P portfolio expected return w portion assigned to asset X 1 w portion assigned to asset Y E X expected return of asset X E Y expected return of asset Y 27 Portfolio risk p w 1 w Y2 2w 1 w XY 2 2 X 2 28 Example 5-3 Returning to the scenario in Example 5-2, compute the portfolio expected return and the portfolio risk if 90% of the stock is invested in stock P. Repeat when 50% of the stock is invested in stock P. 29 Solution 5-3 • When 90% of the stock is invested in stock P, w = 0.9 and E P wE X 1 w E Y 0.9 24 0.1 29.4 $24.54 p w 1 w Y2 2w 1 w XY 2 2 X 2 0.9 2304 0.1 9591.44 2 0.9 0.1 4648.4 2 2 $52.90 30 Solution 5-3 • When 50% of the stock is invested in stock P, w = 0.5 and E P 0.5 24 0.5 29.4 $26.70 p 0.5 2304 0.5 9591.44 2 0.5 4648.4 2 2 2 $72.79 • Although both investment schemes are fairly risky, investing 50% in stock P offers a better return but much larger risk! 31 Binomial distribution The binomial distribution is appropriate if: • there are a fixed number of trials, n. • there are two possible outcomes for each trial: success (the desired outcome) and failure. • the probabilities of success (p) is constant over all observations (the probability of failure is therefore given by 1-p). • trials are independent. 32 Binomial distribution • The binomial distribution is given by: n! n X X P X p 1 p X ! n X ! where P(X) = probability of X successes n = number of trials (sample size) p = probability of success X = number of successes • We can use this formula, tables or Excel/PHStat2 (discussed in workshops) to find probabilities (make sure you can do all three!). 33 Example 5-4 An accountant works mainly with personal income tax cases. He knows from past experience that the probability of a customer being satisfied with the service he offers is 0.8. Given that he sees 8 clients today, determine the probability that: (a) all 8 customers are satisfied 34 Solution 5-4 We are told that n = 8 (he sees 8 clients), p = 0.8 and X = 8 (8 customers are satisfied – success). n! n X X P X x p 1 p X ! n X ! 8! 88 8 P X 8 0.8 1 0.8 8! 8 8 ! 1 0.8 0.2 8 0 0.1678 (to 4 dec. pl.) So the probability that the accountant will have exactly 8 satisfied customers is 0.1678. 35 Solution 5-4 (b) at least 6 customers are satisfied. n = 8 and p = 0.8 P X 6 P X 6 P X 7 P X 8 Using the binomial tables, find n and p, and then the relevant probabilities are: 0.2936 0.3355 0.1678 0.7969 The probability that the accountant will have at least 6 satisfied customers in the day is 0.7969. 36 Solution 5-4 OR calculate each probability using the binomial formula as below (we do not need to recalculate the probability that 8 customers are satisfied): 8! 8 6 6 P X 6 0.8 1 0.8 0.2936 6! 8 6 ! 8! 8 7 7 P X 7 0.8 1 0.8 0.3355 7! 8 7 ! 37 Solution 5-4 (c) fewer than 5 customers are satisfied. P X 5 P X 0 P X 1 P X 2 P X 3 P X 4 OR 1 P X 8 P X 7 P X 6 P X 5 1 0.1678 0.3355 0.2936 0.1468 0.0563 So the probability that the accountant will have less than 5 satisfied customers is 0.0563. 38 Solution 5-4 (d) What assumptions are necessary in (a) to (c)? • Trials are independent – the outcome for one customer does not affect other customers. – Note: Is this actually true? Do customers make recommendations to others? Can we prove that the assumption has been violated? • There are only two possible outcomes: satisfied (success) and not satisfied (failure). Normally we would check the assumptions prior to doing any calculations!! 39 Mean and standard deviation of the binomial distribution E X np np 1 p 2 40 Solution 5-4 (e) What are the mean and standard deviation of the probability distribution? E X np 8 0.8 6.4 np 1 p 8 0.8 0.2 1.1314 (to 4 dec. pl.) 41 Poisson distribution The Poisson distribution is appropriate if: • you are interested in the rate at which some event occurs (in a particular time, area etc) • events are rare • events are random and independent (The textbook goes in to more detail – make sure you read this…) 42 Poisson distribution • The Poisson distribution is given by: X e P X X! where P(X) = probability of X successes given knowledge of = expected number of successes e = constant (2.71828…) X = number of successes per unit • We can use this formula, tables or Excel/PHStat2 (discussed in workshops) to find probabilities (make sure you can do all three!). 43 Example 5-5 A new call centre is being set up for a large telephone company. Based on experience at their other similar centres, they expect that calls will come in at an average of 10 per minute. In order to plan their staffing requirements, the company would like to answer the following questions: (a) What is the probability that less than five calls will be received in any given minute? 44 Solution 5-5 • We know that = 10. P X 5 P X 0 P X 1 P X 2 P X 3 P X 4 • Since this would be a lot of values to calculate, we’ll look this up in tables (Table E.7 in the text)… P X 5 0.000 0.0005 0.0023 0.0076 0.0189 0.0293 45 Solution 5-5 • So the probability that the company receives less than five calls in a given minute is 0.0293. • We’ll look at two examples of using the formula to do the calculations: e X P X X! e 10102 P X 2 0.0023 (4 dec. pl.) 2! 10 4 e 10 P X 4 0.0189 (4 dec. pl.) 4! 46 Example 5-5 (b) What is the probability that at least twenty calls will be received in any given minute? P X 20 P X 20 P X 21 P X 22 0.0019 0.0009 0.0004 0.0002 0.0001 0.0035 Note: probabilities for X=25, 26, … exist but are effectively zero to 4 decimal places. So the probability that the company will receive at least twenty calls in a given minute is 0.0035. 47 Example 5-5 (c) What is the probability that between seven and thirteen calls will be received in any given minute? P 7 X 13 P X 8 P X 9 P X 10 P X 11 P X 12 0.1126 0.1251 0.1251 0.1137 0.0948 0.5713 So the probability that the company will receive between seven and thirteen calls in a given minute is 0.5713. 48 After the lecture each week… • Review the lecture material • Complete all readings • Complete all of recommended problems (listed in SG) from the textbook • Complete at least some of additional problems • Consider (briefly) the discussion points prior to tutorials 49