Topic 5: Discrete Random Variables & Expectations Reference Chapter 4

Transcription

1 Page 1 Topic 5: Discrete Random Variables & Epectations Reference Chapter 4 In Chapter 3 we studied rules for associating a probability value with a single event or with a subset of events in an eperiment. Now we will epand our analysis to consider all possible events in an eperiment. A random variable is a which assigns values to each possible outcome of an eperiment.

2 Page 2 Note: May need to distinguish between the of a random variable, and its value. Eample:{X=} means the variable X takes the value. Eample: Eperiment: choose student; ask if he/she is planning to major in business. Two discrete outcomes: Yes; No. Define random variables: = = 1; 0; Yes No = 0, 1 } indicator variable

3 Page 3 Eample: Midterm Test. Students may get a grade A to F Define: X=5; A X=4; B X=3; C X=2; D X=1; F Depending on the contet, sample space may be or continuous. If discrete, it may be finite or (countably).

4 Probability Distributions Page 4 Once an eperiment and its outcomes have been clearly defined and the random variable of interest has been defined, then the probability of the occurrence of any of the random variable can be specified. I.e. Can now specify the probability of each value of the random variable occurring.

5 Page 5 Eample: There are 5 projects to be done. Twenty-five government workers are to be allocated to teams (varying size) to work on them: Outcome Project # of Employees Value of X () Prob(X=) 1 1 5/ / / / / Probability Distribution

6 Can also depict graphically: Page 6 P(X) X The table or graph represent the probability distribution of X: Each value of X and its of occurrence. Similar to our earlier notion of frequency distribution, but now relates to all possible situations that can occur.

7 We can use probability distribution to determine probabilities of interesting events: Page 7 For Eample: P( 2) = P( assigned to project 2 or 3 or 4 or 5) = P( = 2) + P( = 3) + P( = 4) + P( = 5) = = =

8 Definitions: Page 8 If the individual values assigned to the random variable can be, it is called a discrete random variable. In this case, its probability distribution is called a probability mass function,. Note: Defining Characteristics of P.M.F.: () i 0 P( X = ) 1 ( ii) P( X = ) = 1

9 In the same manner, we can define the cumulative function (c.m.f.): Page 9 * * F( X ) = P( X ) = P( ) * The value of c.m.f. at any point is usually denoted F(*), where it is the of all values of the p.m.f. for all the values of the random variable that are *.

10 Page 10 Eample: Long-term eperience suggests the following p.m.f. for sales of flu medication in January: Type X # P(X) F(X) Aspirin for Flu Tylenol Flu Benadryl F(2) =P(=1 or =2)=P(Aspirin or Tylenol) If a customer buys flu medicine in January, the probability it is not Benadryl is 0.60.

11 P(>1)=1-P( 1)=1-P(=1)= =0.9 Page 11 P(1< 3)=P(=3)+P(=2)=0.9 =F(3)-F(1)=1-0.10=0.9 P(2< 3)=P(=3)=0.4 =F(3)-F(2)=1-0.60=0.4

12 Page 12 When we previously considered data distributions, we looked at summary measures such as the mean, variance, etc.. A distribution is just a special type of data distribution same motivation here to evaluate such measures. To do this, we define: The E Value: of the random variable X is: a weighted average of all values with μ = E( ) = P( ) probabilities as weights.

13 Page 13 The epected value of a random variable X is found by multiplying each value of the random variable by its probability and then summing all these products. The letter E usually denotes an e value. The epected value of X is the balancing point for the p.m.f..

14 Eample: Marks obtained by students: Marks X P() P() (midpoint) 0 < < < < < µ=e(x)=50 Page 14

15 Page 15 Sometimes we want to work out the epected value of some f of a random variable. For instance, instead of finding the mean of the random variable X, we might be interested in determining the epected of X 2, or Log X, or of e. If X is a random variable, then these functions of X are also variables, and their epected values can be determined. The epected value of g(): Eg ( ) = gp ( ) ( ) [ ] all

16 Eamples: Page 16 [ ] 2 2 E = ( ) P ( ) [ ] E log( ) = log( ) P( ) [ ] E ( + 2) = ( + 2) P( ) (Watch the units.)

17 Eample: Epected compounded return on stocks over 3 years: Stock 1-Yr. P() 3 P() Return () A B C E( 3 )= Page 17 Epected return over 3 years is 38.97%.

18 The epectation operator, E( ), can be used to help us define the v of a random variable. Page 18 Recall, the variance of a frequency distribution was: 1 N ( ) 2 μ f i i (Average of ( i -µ) 2 values.) So, the of a random variable, X, is: [ ] V( ) = σ = ( μ) P( ) = P( ) μ

19 Eample: Volatility of Stock returns over one year: µ =E()=11.5% Page 19 Stock Return () μ 2 A 15% B 10% C 5% P() ( ) P( )

20 Rules of Epectation : E ( ) = P ( ) Page 20 1) The E(c)=c, where c is a constant. E( c) = cp( c) = c 1 = c eg. E(10)= 2) V(c)=0; The variance of a constant is zero. Recall: V()=E(-µ) 2 ; V(c)=E(c-µ) 2 =E(c-c) 2 =E(0) 2 =0 C X

21 3) E(aX)=aE(X): Multiply X by a constant: = = = E( a) ap( ) a P( ) ae( ) eg. E(10)=10E() Page 21 4) E(+b)=E()+b: Add a constant E( + b) = ( + b) P( ) = ( b) P( ) + ( b) P( ) = E( ) + b P( ) = E( ) + b since P( ) = 1.

22 5) V(a)=a 2 V(): Recall V()=E(-µ) 2 ; Page 22 V(a)=E(a-E(a)) 2 =E(a-aE()) 2 =E(a(-E())) 2 =E(a 2 (-E()) 2 ) V(a)=a 2 E(-E()) 2 =a 2 V() Eample: V(120)=

23 6) V(+b)=V() Page 23 V(+b)=E((+b)-E(+b)) 2 =E(+b-E()-b) 2 =E(-E()) 2 =V() E(a±b)=aE() ±b V(a±b)= Eamples: V(+10)= V(-20)=V() ***************************

24 One advantage of knowing these rules is to speed up calculations. Page 24 Eample: Data in U.S. $: epected value=u.s. $10 and standard deviation =U.S. $2. What about the results in $ CAN? Suppose C$1 =U.S. $.996 (U.S. $1 =C$1.004) E(a)=aE(); a =1.004 So, epected value C$ ( )= C$10.04 V(a)= a 2 V(); a= Std. deviation (a)= a S.D. (X) =C$( )=C$2.008

25 Eample: F=random variable of temperature (ºF) E(F)= 72.4ºF; V(F)=64(ºF) 2. Page 25 What is E(C) and V(C), where C is temperature (ºC)? F = C; C = F 32 = F 9 5 And since E(a+b)=aE()+b: E( C) = E( F) = b 2 V( C) = ( ) V( F) = a a

26 Page 26 Note: these epectation and variance rules also enable us to standardize random variables for comparison purposes: E()= μ ; V()=σ 2 ; S.D.()=σ. X μ Z = X Then let: = 1 { μ σ σ { σ 1 EZ ( ) = EX ( ) μ σ σ = 1 μ = σ μ σ 0 a So: E(Z)=0; V(Z)=1 b V( Z) = 2 V( X) 1 σ = = = 2 2 σ σ σ σ 1

27 Bivariate Probability Distributions: Page 27 In many situations, an eperiment may involve outcomes that are related to or more random variables This section considers probability functions that involve more than one variable: such functions are called multivariate probability functions We will only deal with the case of bivariate (two variables) probability functions. When a sample space involves two or more random variables, the function describing their combined probability is called a j probability function.

28 Eample: Behaviour of Households: Page 28 X=Income of the oldest member of the household Y= Years work eperience of this member By looking at P(X= and Y=y) for all, y values, we get the JOINT PROBABILITY DISTRIBUTION. X Income ($) Eperience 25,000 30,000 50,000 (Years) Y Entries in the table are j probabilities!

29 Two properties for Joint Probability Functions: Page 29 () i 0 P(, y) 1 ( ii) P(, y) where P(,y)=P(X= and Y=y). y = 1 This joint distribution also helps us with calculation of conditional probabilities.

30 Page 30 X Income ($) Eperience 25,000 30,000 50,000 P(y) (Years) Y P() P( ) = P( X = ) = P( X = and Y = y) y = "marginal probability of X" Py ( ) = PY ( = y) = P( X = and Y = y) = "marginal probability of Y"

31 Page 31 A marginal probability for any value of Y may be found by summing all the probabilities involving that value of Y. The table depicts joint probability distributions of X and Y.

32 When we discussed events, we had: Page 32 P(A I B) P(A B) =. P(B) Conditional Joint Marginal or P(X= Y=y) = P(X = and Y = y) P(Y = y). P( y) = P(,y) P(y).

33 From the last eample: Page 33 (i) P($25,000 and 5 years eperience)= (ii) P($25,000)= ; P(5Years)=0.26 (iii) P($25,000 5 years eperience)= =

34 Independence Page 34 As with events, two random variables are in if: P( y) = P() for all and y. P(,y)=P()P(y); for all and y If the knowledge of a certain on one variable does not affect the probability of the occurrence of values of the other random variable, then the two variables are independent. I.e. X and Y are independent if the conditional probability of X given Y is the same as the marginal probability of X.

35 If X and Y are independent, the P(,y)=. Page 35 An eample of not independent from the last eample: P(=$25,000 and y=5 years)= P(=$25,000)= P(y=5 years) =0.26 However: (0.35)(0.26) Years of eperience and income are dependent random variables.

36 Epectations of Combined Random Variables: Page 36 When dealing with the joint probability distribution of X and Y, the rule for finding epectations is similar to the rule for finding the epected value of a function. Basic rule: [ ] Eg ( ) = gp ( ) ( ) all If we combine two random variables together into h(x,y), then: [ ] = EhXY (, ) hypy (, ) (, ) y

37 Eample: Selling items of different value: EXY [ ] = ( ypy ) (, ) y X Quantity Price P(y) ($) Y P() EXY [ ] = ypy ( ) (, ) y = [( 1 4 0) + ( ) + ( ) + ( ) + ( ) + ( ) + Page 37 ( ) + ( ) + ( ) = $

38 Covariance of X and Y Page 38 When we have 2 random variables it is interesting to measure the etent to which their random behaviours are related. The covariance, C(X,Y), measures how much the two random variables with each other, (how they co-vary). Look at the variation of each random variable relative to its respective. Let μ = μ = y E( X ) EY ( ): ( μ )( μ ) y [ ] [( μ )( μ )] y Cov( X, Y) = E y = y y P(, y) Cov( X, Y) = E( X, Y) E( X ) E( Y)

39 Note: if X and Y are independent: P(X,Y)=P(X) * P(Y) Page 39 [( )( )] μ μy Cov( X, Y) = y P(, y) = = y P( ) y P( y) y ( μ ) ( ) μy P ( ) P ( ) ypy ( ) μy Py ( ) y y ( ) μ ( ) [ μ ] [ ] μy = E() E(y) - = 0 = ( μ μ = 0)( μ μ = 0) y y

40 Page 40 If high values of X (relative to μ ) tend to be associated with high values of Y (relative to μ y ) then C(X,Y) will be a large number. If low values of one variable tend to be associated with high values of the other, then C(X,Y) will be a large number. Eample continued: X Quantity Price P(y) ($) Y P()

41 [ ] E Quantity( X ) = X P( X ) = [( ) + ( ) + ( ) = Page 41 [ ] E Price( Y) = Price P( price) y = [( ) + ( ) + ( ) = Cov[ X, Y] = ( μ)( y μy) P(, y) Y = [( )( )( 0)] + [( )( )( 0. 05) + L+ [( )( )( 0. 3)] = or: COV(, y) = E( XY) E( X ) E( Y) = [ (2.28)(5.55) = 0.016

42 Page 42 One limitation of the covariance measure is that it depends on u of the data, and it can take any positive or negative value. To compensate for both of these failings, especially when wanting to across different data sets, consider the following measure: Cov X Y ρ = y.(, ) = σσ y where σ σ correlation is the standard deviation of X, = variance() = E( μ ) 2

43 We call ρ y the simple correlation between the random variables X and Y. Page 43 () i ( ii) correlation is unitless. 1 ρ 1 X Quantity Price P(y) ($) Y P()

44 Page 44 [ ] E Quantity( X ) = X P( X ) = 228. [ Price ] E ( Y) = Y P( Y) = 555. y COV(, y) = E( XY) E( X) E( Y) =0.016 EXY [ ] = ( ypy ) (, ) = $ y

45 [ ] 2 2 E Quantity( X ) = X P( X ) = [( ) + ( ) + ( ) = 5. 7 Page 45 [ 2 ] E Price( Y ) = Y 2 P( Y) y = [( ) + ( ) + ( ) = VAR( ) = E( X ) [ E( )] σ = = [2.28] 2 =

46 VAR( Y) = E( y ) [ E( y)] 2 2 = [5.55] 2 = Page 46 σ y = ρ, y COV(, y) = = σσ y = = positive correlation between and y.

47 Linear Combinations of Random Variables Page 47 In general, we had: EhXY [ (, )] = hypy (, ) (, ) y Now consider the epectation of the weighted sum of: h(x,y)=ax +by where a and b are constants:

48 i) E(aX + by) = ( ax + by) P(,y) = ( ax) P(,y) + (by) P(,y) = a y [ ] P(,y) + b y P(,y) y y = a P() + b yp(y) = ae(x) + be(y) y y Page 48

49 More directly: E(aX +by) =E(aX) +E(bY) =a E(X) + b E(Y) (ii) Var(aX + by) [( ) ( )] = E ax + by E ax + by [( )] = E ax ae(x) + by be(y) = Ea(X E(X)) + b(y E(Y)) [ ] [ 2 2] [ 2 2] = E a (X E(X)) + E b (Y E(Y)) + E 2ab(X E(X))(Y E(Y)) [ ] 2 2 = a V(X) + b V(Y) + 2abCov.(X,Y) Page 49

50 Page 50 X Quantity Price P(y) ($) Y P() E[ Quantity( X )] = 228. E [ Y ] Price( ) = 555. COV(, y) =0.016 EXY [ ] = $

51 Page 51 [ 2 ] [ 2 Y ] = E Quantity( X ) = 57. E Price( ) VAR( ) = σ = VAR( Y) = σ = y ρ, y =

52 Page 52 Let R=3X+7Y. What is the epected value and the standard deviation of R? E(R)=3E() + 7E(y) 3(2.28)+7(5.55)=45.69 V(R)=V(3+7y) =9V()+49V(y)+2(3)(7)Cov(,y) =(9)(0.5016)+(49)(0.3475)+(2)(3)(7)(0.016) = = Std. Dev.=

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54 Discrete Probability Distributions Page 54 While it is often useful to determine probabilities for a specific discrete random variable or combined random variables, there are many situations in statistical inference and decision-making that involve the s type of probability function. I.e. Certain probability distributions have general characteristics which can be eploited. Recognize the similarities between certain types of eperiments and match a given case to the general formulas for mean, variance, independence and other characteristics of the random variables.

55 Page 55 The eamples we have seen so far have been different, specific, probability distributions. We are given information in each case. Let s consider one specific, common, discrete random variable. In Topic 6 we will consider one specific continuous random variable.

56 The Binomial Distribution: Page 56 Many eperiments share the common trait that their can be classified into one of two events. For Eample: 1. Select a person: male or 2. Apply for a job: Success or 3. Your credit card bill: paid or due It is often possible to describe the outcomes of events as either a success or failure. Use general terminology of success or failure to distinguish.

57 Page 57 Eperiments involving repeated independent trials, each with just two possible outcomes, form the basis of the distribution ( probability distribution). Definition: Bernoulli Trial: Each repetition of an eperiment involving only outcomes is called a Bernoulli trial. Interested in a series of independent, repeated Bernoulli trials. Independent results mean any one trial cannot influence the results of any other trial.

58 Note: Name the trials: t 1 t 2 t 3..., then,: P( t 2 =success t 1 = success)=p( t 2 =success); etc. Page 58 By repeated trials we mean that each time the eperiment is conducted, the conditions are. The Binomial Distribution is characterized by: n = number of Π= probability of success in one independent trial. The values of n and Π are referred to as the parameters of this distribution. They help determine the of success and failure.

59 Page 59 In a binomial distribution, the probabilities of interest are those of receiving a certain number () of in n independent, each trial having the same probability (Π) of success. Note: P(Failure) = 1- P(Success) as there are only 2 possible outcomes. They are complementary outcomes.

60 The Binomial Formula: Page 60 To determine the probability of eactly successes in n repeated Bernoulli trials, each with a probability of success equal to Π, it is necessary to find the probability of any one of outcomes where there are successes. If there are successes in n trials, there must be (n-) failures. The probabilities of interest are those of getting successes in n independent trials, each of which has probability Π of success. P(Event)=P(Outcome #1) + P(Outcome #2)+... = P(Outcomes) * (# of Outcomes) if each outcome has the same probability.

61 We can choose from n in : n! =!(n ) ways. Page 61 So, there are nc outcomes consistent with the event of s and (n-) failures. Each of these (nc) outcomes occurs with probability n [ Π ( 1 Π) ] So:.

62 P(X successes in n trials) n = Π ( 1 Π) ( nc) n! n = n Π ( 1 Π)!( )! = 0, 1, 2, 3,...,n for n = 1, 2, 3,... etc. Page 62 This set of probabilities gives the probability distribution for a Random Variable.

63 Note: If is B(n, Π), it is a random variable: Page 63 P( X) = nc( Π) ( 1 Π) n = 0, 1, 2, 3,...,n n = 1, 2, 3,... is its probability function: So: () i P( X) 0; all. n = 0 ( ii) P( X ) = 1

64 Eample: Π P( 0) = C ( 02. ) ( 08. ) 4 0 = 02. ; n = 4 ; ( 1 Π) = ! 4 = ( 1)( 08. ) = 0. 0!4! Page 64 P( 1) = C( 02. ) ( 08. ) ! 1 3 = ( 02. ) ( 08. ) = 0. 1!3! P( 2) = C ( 02. ) ( 08. ) = 4! (.)(.) = 2!2!

65 P() 3 = C (.)(.) = 4! (.)(.) = 3!1! Page 65 P( 4) = C ( 02. ) ( 08. ) 4 4 = ! 4 0 ( 02. ) ( 08. ) = 0. 4!0! P(X) X

66 Binomial tables assist in the calculations. Page 66 General Characteristics of Binomial Distribution (A) If n is small and Π 0.50: distribution is skewed to the. (B) If n is small and Π > 0.50: distribution is skewed to the. (C) If n is large and/ or Π = 0.50: distribution is.

67 Page 67 Eample: 30% of job applicants are from a minority group. If we take applicants at random, what is the probability of including from the minority group? Let X = minority person n=5; X=2; Π=0.3 n P( X ) = nc( Π) ( 1 Π) Binomial P( 2)! = (. ) (. ) = 0 2! 3!

68 Page 68 Eample: Customers in a bank. One in four customers require service for more than 4 minutes. people are in a queue. What is the probability that eactly 3 customers will take more than 4 minutes each? Π=0.25;n=6; =3 X= customer takes more than 4 minutes. P( X) = nc( Π) ( 1 Π) P() 3 n = (. )(. ). 6! = !!

69 Page 69 What is the probability that at least 3 customers need more than 4 minutes each? I.e. P(3)+ P(4) + P(5) + P(6): P( 4 6! 4 2 ) = (. 0 25)(. 0 75) = 42!! P( 5 6! 5 1 ) = (. 0 25)(. 0 75) = 51!! 6! 6 0 P( 6) = ( 0. 25) ( 0. 75) = 6!0! So the probability is: ( ) = 0. & P(< 3 customers take more than 4 minutes each)= = 0.

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71 Page 71 The Epected Value of a Binomial Random Variable: The number of successes in any given eperiment must equal the number of trials (n) times the probability of a success on each trial ( Π). For eample, the probability that a process produces a defective item is Π = 0. ; Then the mean number of defectives in trials is: (10)(0.25) =2.5.

72 Page 72 Let X be ~ B(n, Π). Then E(X)=n Π : Probability of success = Π number of successes epected from a single trial is: { 1( ) 0( 1 )} Π + Π = Π So, in "n" trials, epected n Π "successes." Binomial mean: E(X Binomial)=

73 Proof: E( X) = P( ) n = 0 n = nc ( Π) ( 1 Π) = 0 n = ( ) = = = = 1 n = 1 n! Π!( n )! n! Π ( 1)!( n )! n ( n 1)! nπ Π ( 1 Π) ( 1)!( n 1 ( 1))! = 1 ( nπ ) y= 0 m! y!( m [ y = 1; m = n 1] So, E( ) = ( nπ)( 1) = nπ m Π y)! n Y= B( m, Π ) ( 1 Π) ( 1 Π) y n n ( 1 Π) 1 ( n 1) ( 1) m y Page 73

74 Similarly, we can show that V(X)= Page 74 So the standard deviation of X: nπ( 1 Π) when X~ B(n, Π).

75 Page 75 Eample: If a manufacturing process is working properly, 10% of the items produced will be defective. We take a random sample of 15 items. Calculate the number of defectives: X = defective ~ B(15, 0.1) So, E() = n Π= (15)(0.1) = 1.5 One or two defectives epected. What is the deviation of the number of defectives? V(X) = n Π (1- Π)=(15)(0.1)(0.9)=1.35 So, the standard deviation (X)= 1.16

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