EE 322: Probabilistic Methods for Electrical Engineers. Zhengdao Wang Department of Electrical and Computer Engineering Iowa State University

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1 EE 322: Probabilistic Methods for Electrical Engineers Zhengdao Wang Department of Electrical and Computer Engineering Iowa State University Discrete Random Variables 1

2 Introduction to Random Variables Intuitively: A random variable takes a value every time an experiment is performed. Examples: the number of cars that passes by within a minute Example: the height of a randomly picked student The voltage of a received signal from antenna at a certain time Two kinds of random variables: discrete and continuous 2

3 PMF: Probability Mass Function Example: coin flipping. Getting Head and Tail with 1/2 chance each. Map H to 0 and T to 1. Denote the result X. P (X = 0) = P (getting a Head) = 1/2. P (X = 1) = P (getting a Tail) = 1/2. PMF of X

4 PMF: 6-sided die 6-sided die, get 1 to 6 with equal probability. Let X denote the result. Then P (X = i) = 1/6 if i = 1,..., 6. 1/

5 PMF: number of heads in two flips HH, HT, TH, TT: each with 1/4 probability. Let N h be the number of heads. P (N h = 0) = P ({T T }) = 1/4. P (N h = 1) = P ({HT, T H}) = 1/2. P (N h = 2) = P ({HH}) = 1/4. So, the PMF is... 5

6 Histogram A graphical display of frequencies numbers, 6 zeros and 4 ones. Frequency of zero is 6/10. Frequency of one is 4/10. The histogram is 6/10 4/

7 In Matlab Use hist(). Example: x=randint(1, 1e6, 100); hist(x, 100); Caution: matlab does not divide by the total number of samples. 7

8 PDF: probability density function PDF tells us how the total probability 100% is distributed across the values Waiting time T at a bus station: uniformly distributed between 0 and 10. f T (t) = 1/10 for t [0, 10]. Voltage X across a resistor: thermal noise Gaussian distributed. f X (x) = 1 ) ( 2πσ 2 exp x2 2σ 2. where σ 2 depends on the bandwidth of the voltage meter, the resistance, and temperature. 8

9 Property of PDF P (a < X < b) = b a f X(x)dx f X(x) = 1 f X (x) 0, x 9

10 Random Variables: Definition An assignment of a real number to each point in sample space. A function mapping sample space into the real line. Random variable X Sample Space Ω x Real number line Idea: Randomness is in the experiment. A random variable just assigns a number to each sample point. Each run of the experiment yields a specific sample point ω, which produces a sample value, say x = X(ω), of the random variable. 10

11 Random variable vs. probability law Probability Law Sample Space Ω 0 1 A Probability Law maps an event (subset of Ω) to [0, 1] 11-1

12 Random variable vs. probability law Probability Law Sample Space Ω Random variable X 0 1 A Probability Law maps an event (subset of Ω) to [0, 1] A Random Variable maps an outcome (element of Ω) to (, ) 11-2

13 Example of Random variable Throw two 4-sided dice Random variable X: the maximum of the two throws 4 3 2nd roll Real Number Line 1st roll Random Variable X = max. roll 12-1

14 Example of Random variable More Examples: Let ω = {ω 1, ω 2 }, ω i = value of the ith roll X = product of two faces, X(ω) = ω 1 ω 2 Y = (difference) 2, Y (ω) = (ω 1 ω 2 ) 2 R = outcome of first roll, R(ω) = ω 1. S = outcome of second roll, S(ω) = ω

15 Probability Mass Functions Abbreviation: PMF; Suitable for discrete RVs For continuous RVs, will use Probability Density Function (PDF) Definition and Notation:p X (x) = P ({X = x}) Probability Law Sample Space Ω Random variable X 0 1 Notation: upper case letters for RVs and lower ones for their values 13

16 Calculation of PMF of a Random Variable For each of the value x the RV X can take, do the following Collect all the possible outcomes that give rise to {X = x} Obtain p X (x) = P r({x = x}) Example: Let X be the number of heads obtained from two independent tosses of a fair coin p X (x) p X (x) = 1/4, if x = 0 or x = 2 1/2, if x = 1 0, otherwise 1/4 0 1/ /4 x 14

17 Remarks about PMF PMF Properties 0 p X (x) 1 for all x x p X(x) = 1 Random variables are often referred to according to their PMF s. E.g., we say that a random variable is a Bernoulli RV if its PMF is { p, if x = 0 p X (x) = 1 p, if x = 1 15

18 Binomial PMF Binomial: Toss a coin n times, and let X denote the number of heads ( ) n p X (k) = P (X = k) = p k (1 p) n k, k = 0, 1,..., n k Binormial PMF n=9, p=0.5 Binormial PMF n=90, p=0.1 16

19 Geometric and Poisson Geometric PMF: Number of tosses for a head to show up p X (k) = (1 p) k 1 p, k = 1, 2,... p Geometric PMF, p=0.2 Poisson PMF λ λk p X (k) = e, k = 0, 1, 2 k!... Good for modeling the total effect of independent small probability events Used extensively in queuing theory 17

20 Poisson PMF (cont d) A good approximation of Binomial PMF for large n and small p λ λk e k! n! k!(n k)! pk (1 p) n k, k = 0, 1,..., n where λ = np, n is large and p is small Poisson, λ=0.5 Poisson, λ=

21 Functions of Random Variables Suppose we have a RV X defined on an experiment We can define functions of the random variable Y = g(x), such as X 2, X, cos(x), e 3X. For each function, Y will be a random number that we can get every time we run the experiment. These are functions of a random variable, and are randomvariables themselves. PMF of Y, p Y (y), can be found from the original probabilitymodel or from the PMF of X 19

22 FUNCTIONS OF RANDOM VARIABLES Sample Space set of possible values of X set of possible values of Y B(y) A(y) y B(y)= set of outcomes corresponding to y B(y) = {x g(x) = y} p Y (y) = P (B(y)) = P (X A(y)) p Y (y) = P ({ω : Y (ω) = y}) = P (ω) = P (ω) = ω:g(x(ω))=y x:g(x)=y ω:x(ω)=x x:g(x)=y p X (x) 20

23 EXAMPLE X = number of heads in three tosses of a fair coin PMF of X is P X (x) = 1/8 x = 0, 3 3/8 x = 1, 2 0 otherwise Y = g(x) = 3 X (number of tails) Find the PMF of Y P Y (y) = P r(y = y) = P r(3 X = y) = P r(x = 3 y) = P X (3 y) So P Y (0) = P X (3) = 1/8, P Y (1) = P X (2) = 3/8, etc 21

24 Exercise X = number of heads in three tosses of a fair coin Y = X 2 P Y (y) = P r(y = y) = P r( X 2 = y) Find the PMF of Y 22

25 Expectation, Mean, Variance OUTLINE Review of Concepts of Random Variables and PMF Functions of Random Variables Expectation, Mean, Variance Reading: Bertsekas & Tsitsiklis, 2.3,

26 Expectation The expectation or mean of a random variable X, with PMF p X (x), is defined as E[X] = x xp X (x) center of gravity c = mean = E[X] x E[X]=Sum of product of values and their probabilities It is the way you compute your GPA. 24

27 Variance, Moments, Standard deviation 2nd Moment: E[X 2 ], nth Moment: E[X n ] Variance var(x) = σ 2 X = E [ (X E[X]) 2] Always non-negative Standard deviation σ X = var(x) 25

28 Expected Value Properties Linearity of the expectation E[g 1 (X) + g 2 (X)] = E[g 1 (X)] + E[g 2 (X)] Variance in terms of moments: Var(X) = E[X 2 ] (E[X]) 2 Mean and variance of a linear function of a RV. Let Y = ax + b E[Y ] = ae[x] + b, Var(Y ) = a 2 Var(X) 26

29 Mean and Variance of Bernoulli RV Bernoulli PMF: p X (x) = { 1 p, if x = 0 p, if x = 1 Mean, second moment, and variance E[X] = 1 p + 0 (1 p) = p E[X 2 ] = 1 2 p (1 p) = p Var[X 2 ] = E[X 2 ] (E[X]) 2 = p p 2 = p(1 p) Note: The mean may not be an outcome. 27

30 Six-Sided Die Let X denote the number we get from a single roll of a 6-sided die. { 1/6, if x = 1, 2, 3, 4, 5, 6 p X (x) = 0, otherwise E[X] = 3.5 E[X 2 ] = 1 6 ( ) = 91 6 Var[X] = = σ X = Var[X] = 35/12 28

31 A generalization: n-sided die Let X denote the RV that takes values between 1 and n, inclusive, equally likely { 1/n, if x = 1, 2, 3,..., n p X (x) = 0, otherwise E[X] = 1 + n 2 E[X 2 ] = 1 n k 2 = n Var[X] = k=1 (n + 1)(2n + 1) 6 (n + 1)(2n + 1) 6 ( ) n = n

32 Geometric RV Geometric PMF The number of tosses needed for a head p X (k) = (1 p) k 1 p, k = 1,..., n In order to compute the mean, we will need the following fact ka k 1 = k=1 1 (1 a) 2 which can be obtained by taking the derivative w.r.t. a the following p Geometric PMF, p= a k = k=1 a 1 a E[X] = 1 p, Var[X] = 1 p p 2 30

33 Expected Value Rule Let Y = g(x) be a function of the RV We can verify that E[Y ] = y yp Y (y) = x g(x)p X (x) Discrete case E[g(X)] = x g(x)p X (x) Continuous case E[g(X)] = g(x)f X (x)dx 31

34 Example of ER To understand the ER, we may use a small example. Let X be a RV that is equal to 1, 2, 3, and 4 with equal probability 1/4 (so there is a 4-face die). Let Y = (X 2) 2 which is a function of the RV X. x p X (x) (x 2) 2 1 1/ / / /4 4 We want to find E[Y ]. By definition of expectation, we know E[Y ] = y yp Y (y), where p Y (y) = 1/4 y = 0 1/4 y = 4 1/4 + 1/4 y =

35 Example of ER Therefore, E[Y ] = But tracing back to where 4, 0, 1 and 1 4 and 1 2 come from, we know that E[Y ] can also be written as E[Y ] = (2 2) (4 2) (1 2) (3 2)2 1 4 and this is the ER. The ER is actually very intuitive from the table: there is 1/4 probability of X = 2, and Y = (2 2) 2, there is 1/4 probability of X = 4, and Y = (4 2) 2, there is 1/4 probability of X = 1, and Y = (1 2) 2, there is 1/4 probability of X = 3, and Y = (3 2) 2. Summing up all the four possibilities of Y (two of them have the same Y value), weighted by the probability 1/4, we have the expectation (average) of Y. 32-2

36 ER for Conditional Expectation In addition to ER for unconditional expectation, we also have conditional expectation, where one would replace the unconditional PDF or PMF with the conditional ones: The ER for conditional expectation as follows: E[g(X) Y = y] = x g(x)p X Y (x y) discrete case (1) E[g(X) Y = y] = g(x)f X Y (x y)dx continuous case (2) 33

37 Multiple Random Variables Reading: Bertsekas & Tsitsiklis, We have seen that we can define many RVs on a given experiment. So far, we considered PMF for only one RV at a time. Ideas extend to two (or more) RVs. Given an experiment with two RVs X and Y defined on it, the joint PMF is defined by p X,Y (x, y) = P (X = x and Y = y) = P ({ω : X(ω) = x and Y (ω) = y}). Each performance of the experiment produces a sample (x, y), a random pair of numbers. From the additivity and normalization axioms of probability p X,Y (x, y) = 1. x,y 34

38 Intuitive Example (from C. Bishop) A fiendish murder has been committed Two suspects: the Butler or the cook There are three possible murder weapons: a Knife a Pistol a fireplace Poker Source: slides.pdf 35

39 Prior Distribution Butler has served family well for many years Cook hired recently, rumours of dodgy history 36

40 Conditional Distribution Butler is ex-army, keeps a gun in a locked drawer Cook has access to lots of knives Butler is older and getting frail 37

41 Joint Distribution What is the probability that the Cook committed the murder using the Pistol? Likewise for the other five combinations of Culprit and Weapon 38-1

42 Joint Distribution x y Sum = 100% p(x, y) = p(x)p(y x) Product Rule 38-2

43 Marginal Distribution of Culprit x y P (x) = y P (x, y) Sum Rule 39

44 Marginal Distribution of Weapon x y P (y) = x P (x, y) Sum rule 40

45 Posterior Distribution We discover a Pistol at the scene of the crime 20% 80% 41

46 Bayes Theorem P (x, y) = P (x)p (y x) = P (y)p (x y) likelihood Prior P (x y) = P (y x)p (x) P (y) posterior Prior belief before making a particular observation Posterior belief after making the observation Posterior is the prior for the next observation Intrinsically incremental 42

47 The Rules of Probability Sum Rule P (x) = y P (x, y) Product Rule P (x, y) = P (x)p (y x) Bayes Rule Denominator P (x y) = P (y x)p (x) P (y) P (y) = x P (x)p (y x) 43

48 Multiple Random Variables - Example Our usual 2-dice problem: Let X be the outcome of the first roll, Y of the second; then p X,Y (x, y) = 1 16, x = 1, 2, 3, 4; y = 1, 2, 3, 4. Same experiment: let S =sum of two faces, T =product. Here a table might help: the rows correspond to toss 1, the columns to toss 2. Each entry shows the sample values of (S, T ) resulting from the underlying sample point: (2,1) (3,2) (4,3) (5,4) 2 (3,2) (4,4) (5,6) (6,8) 3 (4,3) (5,6) (6,9) (7,12) 4 (5,4) (6,8) (7,12) (8,16) 44-1

49 Multiple Random Variables - Example Since each of the original sample points has probability 1/16, the new joint PMF can be evaluated by suitable summing, e.g. 1 16, for (s, t) = (2, 1), (4, 4), (6, 9), (8, 16), p S,T (s, t) = 1 8, for (s, t) = (3, 2), (4, 3), (5, 4), (5, 6), (6, 8), (7, 12). It can be checked that this PMF sums to

50 Marginal PMFs Suppose X and Y are two random variables defined on a common experiment and that they have a joint PMF p X,Y (x, y). As before, we can also determine their individual or marginal PMFs p X (x) and p Y (y). How are these related? Using total probability: p X (x) = P (X = x) = y P (X = x and Y = y) = y p X,Y (x, y). The Sum Rule: p X (x) = y p X,Y (x, y) p Y (y) = x p X,Y (x, y) 45

51 Joint and Marginal PMFs Joint PMF p X,Y (x, y) in tabular form y x 1 x 2 x 3 x 4 Row sums: marginal PMF p Y (y) y 1 0 1/20 1/20 1/20 3/20 y 2 1/20 2/20 3/20 1/20 7/20 y 3 1/20 2/20 3/20 1/20 7/20 y 4 Column sums: marginal PMF p X (x) 1/20 1/20 1/20 0 3/20 6/20 8/20 3/20 3/20 x 46

52 Function of Multiple random variables Z = g(x, Y ) p Z (z) = {(x,y) g(x,y)=z} p X,Y (x, y) Expected value rules: E[g(X, Y )] = x,y g(x, y)p X,Y (x, y) E[g 1 (X, Y ) + g 2 (X, Y )] = E[g 1 (X, Y )] + E[g 2 (X, Y )] Extension to more than two RVs: [ n E i=1 a i X i ] = n a i E[X i ] i=1 47

53 Example: mean of the binomial Binomial RV X: the number of heads in n tosses Mean: E[X] = n ( ) n k p k (1 p) n k (not easy) k k=0 Define the indicator RVs { 1 if the ith toss is a head (prob. p), X i = 0 if the ith toss is a tail (prob. 1 p), Then X = X X n. Now E[X] = n E[X i ] = i=1 n p = np i=1 48

54 Summary of Joint PMF Joint PMF of X and Y p X,Y (x, y) = P ({X = x} {Y = y}). Marginal PMF from joint PMF p X (x) = y p X,Y (x, y). Expected value rule: E[g(X, Y )] = x,y g(x, y)p X,Y (x, y) Linearity: [ n E i=1 a i X i ] = n a i E[X i ] i=1 49

55 Conditioning Conditional PMF of X given Y = y (hence, A = {Y = y}) p X Y (x y) = p X,Y (x, y) p Y (y) Conditional PMF of X given that event A occurs p X A (x A) = P (X = x A) Properties immediately from results for conditional probabilities: p X,Y (x, y) = p Y (y)p X Y (x y) Product Rule p X (x) = y p Y (y)p X Y (x y) Total Probability for RVs. (Sum Rule) 50

56 Slice View of Conditional PMF P X Y (x 3) y Jont PMF P X Y (x 2) x y=3 y=2 y=1 x P X Y (x 1) x x 51

57 Conditional PMF: Example Joint PMF p X,Y (x, y) in tabular form y x 1 x 2 x 3 x 4 Row sums: marginal PMF p Y (y) y 1 0 1/20 1/20 1/20 3/20 y 2 1/20 2/20 3/20 1/20 7/20 y 3 1/20 2/20 3/20 1/20 7/20 y 4 Column sums: marginal PMF p X (x) What is P X Y (x y)? 1/20 1/20 1/20 0 3/20 6/20 8/20 3/20 3/20 x 52

58 Conditional PMF: Example P X Y (x y) = P r(x = x Y = y) = P X,Y (x,y) P Y (y). P X Y (x y 1 ) = 0 3, 1 3, 1 3, 1 3, for x = x 1, x 2, x 3, x 4. P X Y (x y 2 ) = 1 7, 2 7, 3 7, 1 7, for x = x 1, x 2, x 3, x 4. P X Y (x y 3 ) = 1 7, 2 7, 3 7, 1 7, for x = x 1, x 2, x 3, x 4. P X Y (x y 4 ) = 1 3, 1 3, 1 3, 0 3, for x = x 1, x 2, x 3, x 4. 53

59 Example: passing the exam n times A student takes a test repeatedly until he passes, or fail the test after n trials. Probability of pass in each test is p. Geometric: p X (k) = p(1 p) k 1 What is the PMF for the number of attempts given that he passes? Let A = {X n} P (A) = n (1 p) m 1 p m=1 The conditional PMF p X A (k) = { (1 p) k 1 p n m=1 (1 p)m 1 p, if k = 1,..., n 0, otherwise 54

60 Example: passing the exam n times p X (k) p p X A (k) p/p(a) n k 0 1 n k 55

61 Conditional Expectation Conditional expectation of X given event A E[X A] = x xp X A (x A) Conditional expectation of X given Y = y E[X Y = y] = x xp X Y (x y) 56

62 Conditional Expectation - Properties Total expectation theorem for RVs E[X] = y p Y (y)e[x Y = y] The result follows from the definitions and total probability. p Y (y)e[x Y = y] = p Y (y) xp X Y (x y) (3) y y x = x x y p Y (y)p X Y (x y) (4) = x xp X (x) = E[X]. (5) 57

63 Conditional Expectation The conditional expectations of course depends on what value Y takes. For a different y, we have a different conditional expectation. That is E[X y] is a function of y. Replacing the argument of any function with a RV will give us a function of a RV. So, E[X Y ] is a function of Y. Using the ER, with g(y ) replaced with E[X Y ], we can evaluate E[ E[X Y ] ]. 58

64 Independence Two random variables X and Y are called independent if p X,Y (x, y) = p X (x) p Y (y) for all x, y i.e. if the events {X = x} and {Y = y} are independent in the original sense for all possible x and y. It follows that p X (x) = p X Y (x y) If X and Y are independent and g(x) and h(y) are functions,then E[g(X)h(Y )] = x,y g(x)h(y)p X,Y (x, y) = x,y g(x)h(y)p X (x)p Y (y) = ( g(x)p X (x) )( h(y)p Y (y) ) x y = E[g(X)]E[h(Y )] 59

65 Independence - Example Y Z 4 1/20 2/20 2/ /30 2/30 1/30 2/30 3 2/20 4/20 1/20 2/20 3 2/30 4/30 2/30 4/ /20 3/20 1/20 0 1/ X Are X and Y independent? 2 1 1/30 2/30 1/30 2/30 1/30 2/30 1/30 2/ W Are Z and W independent? 60

66 Sums of Independent RVs If X 1, X 2,..., X n are independent E[X X n ] = E[X 1 ] + + E[X n ] var[x X n ] = var[x 1 ] + + var[x n ]. Previously, we also had E[aX + b] = ae[x] + b, and Var[aX + b] = a 2 Var[X] The mean result is true regardless ofindependence The variance result makes use of the independence to get rid of the cross terms. Let s prove the variance result for n = 2 case 61

67 Sums of Independent RVs Example: Mean and variance of the Binomial where X = X X n { 1 if the ith toss is a head (prob. p), X i = 0 if the ith toss is a tail (prob. 1 p), n n E[X] = E[X i ] = p = np var[x] = i=1 i=1 n var[x i ] = np(1 p). i=1 62

68 The Sample Mean An important special case of sums of random variables occurs when a i = 1/n for all i and the random variables X i have a common mean µ and variance σx 2. The sample mean is defined by S n = 1 n X i. n This is also called the sample average. From the previous formulas E[S n ] = n i=1 var[s n ] = 1 n 2 (recall that var[ax] = a 2 var[x].) i=1 1 n E[X i] = µ, n i=1 var[x i ] = σ2 X n. 63

69 The Sample Mean This is actually a very important result since it tells us that the variance of the S n shrinks to zero as the number of samples goes to infinity. This will be a key fact in the proof of the law of large numbers. 64

70 The Sample Mean - Example We want to reduce the error of a measurement of some quantity m by doing the measurement independently multiple times, and then average. We know that for each measurement the error standard deviation is δ. How many measurement do we need to reduce the error (standard deviation) to 0.1 δ? Model: Let X i denote the result in the ith measurement. We modelx i as a random variable due to measurement error. We model the mean of X i as m. The variance is δ 2 The question is: For what n is the standard deviation of n S n = 1 n i=1 X i equal to 0.1δ. 65

71 Review of the topics in Chapter 2 Concept of random variables Probability mass function: definition, calculation, properties Useful PMFs: Binomial, Geometric, Poisson Expectation mean and variance expected value rule linearity rule 66-1

72 Review of the topics in Chapter 2 Multiple RV joint and marginal PMFs conditional PMF conditional expectation independence expectation of product of independent RVs expectation and variance of sum of independent RVs 66-2