Statistics for Business and Economics

Transcription

1 Statistics for Business and Economics Chapter 5 Continuous Random Variables and Probability Distributions Ch. 5-1

2 Probability Distributions Probability Distributions Ch. 4 Discrete Continuous Ch. 5 Probability Probability Distributions Distributions Binomial Uniform Normal Ch. 5-2

3 5.1 Continuous Probability Distributions A continuous random variable is a variable that can assume any value in an interval thickness of an item time required to complete a task temperature of a solution height, in inches These can potentially take on any value, depending only on the ability to measure accurately. Ch. 5-3

4 Cumulative Distribution Function The cumulative distribution function, F(x), for a continuous random variable X expresses the probability that X does not exceed the value of x F(x) = P(X x) Let a and b be two possible values of X, with a < b. The probability that X lies between a and b is P(a < X < b) = F(b) F(a) Ch. 5-4

5 Probability Density Function The probability density function, f(x), of random variable X has the following properties: 1. f(x) > 0 for all values of x 2. The area under the probability density function f(x) over all values of the random variable X is equal to The probability that X lies between two values is the area under the density function graph between the two values Ch. 5-5

6 Probability Density Function (continued) The probability density function, f(x), of random variable X has the following properties: 4. The cumulative density function F(x 0 ) is the area under the probability density function f(x) from the minimum x value up to x 0 x f(x0 ) = f(x)dx x 0 m where x m is the minimum value of the random variable x Ch. 5-6

7 Probability as an Area Shaded area under the curve is the probability that X is between a and b f(x) P ( a X b ) = P ( a < X < b ) a b x (Note that the probability of any individual value is zero) Ch. 5-7

8 The Uniform Distribution Probability Distributions Continuous Probability Distributions Uniform Normal Ch. 5-8

9 The Uniform Distribution The uniform distribution is a probability distribution that has equal probabilities for all possible outcomes of the random variable f(x) Total area under the uniform probability density function is 1.0 a b x Ch. 5-9

10 The Uniform Distribution (continued) The Continuous Uniform Distribution: f(x) = 1 b a if a x b 0 otherwise where f(x) = value of the density function at any x value a = minimum value of x b = maximum value of x Ch. 5-10

11 Properties of the Uniform Distribution The mean of a uniform distribution is µ = a + 2 b The variance is σ 2 = (b - a) 12 2 Ch. 5-11

12 Uniform Distribution Example Example: Uniform probability distribution over the range 2 x 6: 1 f(x) = 6-2 =.25 for 2 x 6 f(x).25 a + b µ = = = x σ 2 = (b - a) 12 2 = (6-2) 12 2 = Ch. 5-12

13 5.2 Expectations for Continuous Random Variables The mean of X, denoted µ X, is defined as the expected value of X µ X = E(X) The variance of X, denoted σ X 2, is defined as the expectation of the squared deviation, (X - µ X ) 2, of a random variable from its mean σ 2 X = E[(X µ X ) 2 ] Ch. 5-13

14 Linear Functions of Variables Let W = a + bx, where X has mean µ X and variance σ X 2, and a and b are constants Then the mean of W is µ = E(a+ bx) = a+ bµ W X the variance is 2 W σ = Var(a + bx) = the standard deviation of W is σ = bσ W X b 2 σ 2 X Ch. 5-14

15 Linear Functions of Variables (continued) An important special case of the previous results is the standardized random variable Z = X µ σ X X which has a mean 0 and variance 1 Ch. 5-15

16 5.3 The Normal Distribution Probability Distributions Continuous Probability Distributions Uniform Normal Ch. 5-16

17 The Normal Distribution (continued) Bell Shaped Symmetrical Mean, Median and Mode are Equal Location is determined by the mean, µ Spread is determined by the standard deviation, σ The random variable has an infinite theoretical range: + to - f(x) µ σ Mean = Median = Mode x Ch. 5-17

18 The Normal Distribution (continued) The normal distribution closely approximates the probability distributions of a wide range of random variables Distributions of sample means approach a normal distribution given a large sample size Computations of probabilities are direct and elegant The normal probability distribution has led to good business decisions for a number of applications Ch. 5-18

19 Many Normal Distributions By varying the parameters µ and σ, we obtain different normal distributions Ch. 5-19

20 The Normal Distribution Shape f(x) Changing µ shifts the distribution left or right. σ Changing σ increases or decreases the spread. µ x Given the mean µ and variance σ we define the normal distribution using the notation 2 X ~ N(µ, σ ) Ch. 5-20

21 The Normal Probability Density Function The formula for the normal probability density function is f(x) = 1 2πσ e (x µ) 2 /2σ 2 2 Where e = the mathematical constant approximated by π = the mathematical constant approximated by µ = the population mean σ = the population standard deviation x = any value of the continuous variable, - < x < Ch. 5-21

22 Cumulative Normal Distribution For a normal random variable X with mean µ and variance σ 2, i.e., X~N(µ, σ 2 ), the cumulative distribution function is F(x0 ) = P(X x0) f(x) P(X x0 ) 0 x 0 x Ch. 5-22

23 Finding Normal Probabilities The probability for a range of values is measured by the area under the curve P(a < X < b) = F(b) F(a) a µ b x Ch. 5-23

24 Finding Normal Probabilities F(b) = P(X < b) (continued) a µ b x F(a) = P(X < a) a µ b x P(a < X < b) = F(b) F(a) a µ b x Ch. 5-24

25 The Standardized Normal Any normal distribution (with any mean and variance combination) can be transformed into the standardized normal distribution (Z), with mean 0 and variance 1 f(z) Z ~ N(0, 1) Need to transform X units into Z units by subtracting the mean of X and dividing by its standard deviation Z = X µ σ 0 1 Z Ch. 5-25

26 Example If X is distributed normally with mean of 100 and standard deviation of 50, the Z value for X = 200 is X µ Z = = = σ This says that X = 200 is two standard deviations (2 increments of 50 units) above the mean of 100. Ch. 5-26

27 Comparing X and Z units X 2.0 Z (µ = 100, σ = 50) ( µ = 0, σ = 1) Note that the distribution is the same, only the scale has changed. We can express the problem in original units (X) or in standardized units (Z) Ch. 5-27

28 Finding Normal Probabilities Φ = Φ < < = < < σ µ a σ µ b σ µ b Z σ µ a P b) X P(a a b x f(x) σ b µ σ a µ Z µ 0 Ch. 5-28

29 Probability as Area Under the Curve The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below f(x) P( < X < µ) = 0.5 P(µ < X < ) = µ P( < X < ) = 1.0 X Ch. 5-29

30 Appendix Table 1 The Standardized Normal table in the textbook (Appendix Table 1) shows values of the cumulative normal distribution function For a given Z-value a, the table shows Φ(a) (the area under the curve from negative infinity to a ) Φ(a) = P(Z < a) 0 a Z Ch. 5-30

31 The Standardized Normal Table Appendix Table 1 gives the probability Φ(a) for any value a Example: P(Z < 2.00) = Z Ch. 5-31

32 The Standardized Normal Table For negative Z-values, use the fact that the distribution is symmetric to find the needed probability:.9772 (continued) Example: P(Z < -2.00) = = Z Z Ch. 5-32

33 General Procedure for Finding Probabilities To find P(a < X < b) when X is distributed normally: Draw the normal curve for the problem in terms of X Translate X-values to Z-values Use the Cumulative Normal Table Ch. 5-33

34 Finding Normal Probabilities Suppose X is normal with mean 8.0 and standard deviation 5.0 Find P(X < 8.6) 8.0 X 8.6 Ch. 5-34

35 Finding Normal Probabilities Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(X < 8.6) X µ Z = = = σ (continued) µ = 8 σ = 10 µ = 0 σ = X Z P(X < 8.6) P(Z < 0.12) Ch. 5-35

36 Solution: Finding P(Z < 0.12) Standardized Normal Probability Table (Portion) z Φ(z) P(X < 8.6) = P(Z < 0.12) Φ(0.12) = Z Ch. 5-36

37 Upper Tail Probabilities Suppose X is normal with mean 8.0 and standard deviation 5.0. Now Find P(X > 8.6) 8.0 X 8.6 Ch. 5-37

38 Upper Tail Probabilities Now Find P(X > 8.6) P(X > 8.6) = P(Z > 0.12) = P(Z 0.12) = = (continued) = Z 0 Z Ch. 5-38

39 Finding the X value for a Known Probability Steps to find the X value for a known probability: 1. Find the Z value for the known probability 2. Convert to X units using the formula: X = µ + Zσ Ch. 5-39

40 Finding the X value for a Known Probability (continued) Example: Suppose X is normal with mean 8.0 and standard deviation 5.0. Now find the X value so that only 20% of all values are below this X.2000? 8.0? 0 X Z Ch. 5-40

41 Find the Z value for 20% in the Lower Tail 1. Find the Z value for the known probability Standardized Normal Probability Table (Portion) z Φ(z) 20% area in the lower tail is consistent with a Z value of ? X Z Ch. 5-41

42 Finding the X value 2. Convert to X units using the formula: X = µ + Zσ = ( 0.84)5.0 = 3.80 So 20% of the values from a distribution with mean 8.0 and standard deviation 5.0 are less than 3.80 Ch. 5-42

43 5.4 Normal Approximation for the average of Bernoulli random var. Bernoulli random variable X i : By Central Limit Theorem, X i =1 with probability p X i =0 with probability 1-p E(X i ) = p By Central Limit Theorem, Var(X i ) = p(1-p) 1 n n (X i E(X i )) N 0,Var(X i ) i=1 ( ) Ch. 5-43

44 Normal Approximation for the average of Bernoulli random var. (continued) The shape of the average of independent Bernoulli is approximately normal if n is large X p = 1 n n $ ( X i p) N & 0, % i=1 p(1 p) n ' ) ( Standardize to Z from the average of Bernoulli random variable: Z = X E(X) Var(X) = X p p(1 p)/n Ch. 5-44

45 Normal Distribution Approximation for Binomial Distribution Let Y be the number of successes from n independent trials, each with probability of success p. Y = n X i i=1 Y E(Y ) Var(Y ) = Y ne(x) Var(nX) = nx = Y np np(1 p) N 0,1 ( ) Ch. 5-45

46 Example 40% of all voters support ballot proposition A. What is the probability that between 0.36 and 0.40 fraction of voters indicate support in a sample of n = 100? E(X) = p = 0.40 Var(X) = p(1-p)/n = (0.40)(1-0.40)/100 = # P(0.36 < X < 0.40) = P% $ (0.4)(1 0.4)/100 Z (0.4)(1 0.4)/100 = P( 0.82 < Z < 0) = Φ(0) Φ( 0.82) = = & ( ' Ch. 5-46

47 5.6 Joint Cumulative Distribution Functions Let X 1, X 2,...X k be continuous random variables Their joint cumulative distribution function, F(x 1, x 2,...x k ) defines the probability that simultaneously X 1 is less than x 1, X 2 is less than x 2, and so on; that is ({ X < x } { X < x } { X }) F(x < 1, x 2,, x k ) = P ! k x k Ch. 5-47

48 Joint Cumulative Distribution Functions (continued) The cumulative distribution functions F(x 1 ), F(x 2 ),...,F(x k ) of the individual random variables are called their marginal distribution functions The random variables are independent if and only if F(x1,x2,,x k) = F(x1)F(x2)! F(x k ) Ch. 5-48

49 Covariance Let X and Y be continuous random variables, with means µ x and µ y The expected value of (X - µ x )(Y - µ y ) is called the covariance between X and Y Cov(X, Y) )(Y µ An alternative but equivalent expression is Cov(X, Y) = E[(X µ = E(XY) µ x µ y If the random variables X and Y are independent, then the covariance between them is 0. However, the converse is not true. x y )] Ch. 5-49

50 Correlation Let X and Y be jointly distributed random variables. The correlation between X and Y is ρ = Corr(X,Y) = Cov(X, Y) σ X σ Y Ch. 5-50

51 Sums of Random Variables Let X 1, X 2,...X k be k random variables with means µ 1, µ 2,... µ k and variances σ 12, σ 22,..., σ k2. Then: The mean of their sum is the sum of their means E(X X2 +! + Xk ) = µ 1 + µ 2 +! µ k Ch. 5-51

52 Sums of Random Variables (continued) Let X 1, X 2,...X k be k random variables with means µ 1, µ 2,... µ k and variances σ 12, σ 22,..., σ k2. Then: If the covariance between every pair of these random variables is 0, then the variance of their sum is the sum of their variances Var(X X2 +! + Xk ) = σ1 + σ2 +! σk However, if the covariances between pairs of random variables are not 0, the variance of their sum is Var(X1 + X2 +! + Xk ) = σ1 + σ2 +! + σk + 2 Cov(X i,x j) K 1 K i= 1 j= i+ 1 Ch. 5-52

53 Differences Between Two Random Variables For two random variables, X and Y The mean of their difference is the difference of their means; that is E(X Y) = µ X µ Y If the covariance between X and Y is 0, then the variance of their difference is Var(X Y) = σ X σy If the covariance between X and Y is not 0, then the variance of their difference is Var(X Y) = σ 2 X + σ 2 Y 2Cov(X,Y) Ch. 5-53

54 Linear Combinations of Random Variables A linear combination of two random variables, X and Y, (where a and b are constants) is W = ax + by The mean of W is µ + W = E[W] = E[aX + by] = aµ X bµ Y Ch. 5-54

55 Linear Combinations of Random Variables (continued) The variance of W is σw = a σx + b σy + 2abCov(X, Y) Or using the correlation, σ W = a σx + b σy + 2abCorr(X,Y)σ X σ Y If both X and Y are joint normally distributed random variables then the linear combination, W, is also normally distributed Ch. 5-55

56 Portfolio Analysis A financial portfolio can be viewed as a linear combination of separate financial instruments Return on = portfolio Proportion of Stock 1 portfolio value return in stock1 + Proportion of Stock portfolio value return in stock2 2! + Proportion of Stock N portfolio value return instock N Ch. 5-56

57 Portfolio Analysis Example Consider two stocks, A and B The price of Stock A is normally distributed with mean 12 and standard deviation 4 The price of Stock B is normally distributed with mean 20 and standard deviation 16 The stock prices have a positive correlation, ρ AB =.50 Suppose you own 10 shares of Stock A 30 shares of Stock B Ch. 5-57

58 Portfolio Analysis Example The mean and variance of this stock portfolio are: (Let W denote the distribution of portfolio value) (continued) µ B W = 10µ A + 20µ = (10)(12) + (30)(20) = 720 σ 2 W = 10 = σ 2 A (4) = 251, σ 2 2 B + (2)(10)(30)Corr(A,B)σ (16) 2 σ + (2)(10)(30)(.50)(4)(16) A B Ch. 5-58

59 Portfolio Analysis Example (continued) What is the probability that your portfolio value is less than $500? µ W = 720 σ W = 251,200 = The Z value for 500 is Z = = 0.44 P(Z < -0.44) = So the probability is 0.33 that your portfolio value is less than $500. Ch. 5-59