Chapter 3 Single Random Variables and Probability Distributions (Part 1)

Transcription

1 Chapter 3 Single Random Variables and Probability Distributions (Part 1)

2 Contents What is a Random Variable? Probability Distribution Functions Cumulative Distribution Function Probability Density Function Common Random Variables and their Distribution Functions Transformation of a Single Random Variable Averages of Random Variables Characteristic Function Chebyshev's Inequality Computer Generation of Random Variables

3 What is a Random Variable? Why random variable?: It is easier to describe and manipulate outcomes and events of the chance eperiments using numerical values rather than in words. purpose of a random variable maps each point in S into a point on R 1. This type of transform or mapping is called a function.

4 What is a Random Variable? (Cont.) For eample, = () where is the random variable, is its specific value, and denotes the outcome of a chance eperiment. Figure 3.1: A r.v. is a mapping of the sample space into a real line.

5 What is a Random Variable? (Cont.) NOTE: Events are now described by the random variable, rather than in words, and corresponding probability of the event can be calculated via r.v. Eample: For a chance eperiment of tossing a fair coin, define a r.v. (): (head) = 1 and (tail) = 0 Then, using the equally likely assignment of probability, we have: P( = 1) = ½ and P( = 0) = ½ Value assignment of a r.v. entirely depends on the convenience, i.e. values 0 and 1 would be more convenient to handle than values and e.

6 What is a Random Variable? (Cont.) Eample 3.1 Consider an eperiment of rolling a pair of dice, and find the probability of all possible values of the sum. Solution: Define a r.v. as the sum of the two dice, then referring the Table 3.1 and applying the equally likely assignment of probability.

7 What is a Random Variable? (Cont.) Eample 3.1 (Cont.) Table 3.1: Die 1 Die 2 1 spot 2 spot 3 spot 4 spot 5 spot 6 spot 1 spot spot spot spot spot spot We have: P( = 2) = 1/36, P( = 3) = 2/36=1/18, P( = 4) = 3/36 = 1/12, P( = 5) = 4/36 = 1/9, P(= 6) = 5/36, P( = 7) = 6/36 = 1/6, P( = 8) = 5/36, P(= 9) = 4/36 = 1/9, P( = 10) = 3/36 = 1/12, P( = 11) = 2/36 = 1/18, P( = 12) = 1/36

8 What is a Random Variable? (Cont.) FACT: Functions of r.v. are ALSO r.v.'s. For instance: Y e, W ln, U cos, V 2

9 What is a Random Variable? (Cont.) Categorization of random variables: 1. Discrete random variable assumes only a countable number of values. (e.g.) rolling dice: Eample 3.1 (only 11 values) 2. Continuous random variable can assume a continuum of values. (e.g.) weather vane: angle of the indicator (any value from 0 to 2 radian) 3. Mied random variable is a combination of above two types.

10 Probability Distribution Functions Description of r.v. using probability: In case of discrete r.v.: probability mass distribution tabulate (or plot) probability of its values Two types of general methods of description: (1) Cumulative (probability) distribution function: cdf (2) Probability density function: pdf These apply to and work for all three types of r.v., i.e., continuous, discrete, and mied random variables.

11 Cumulative Distribution Function Definition 3.1 Cumulative Distribution Function: The cdf of a random variable is defined as follows: F ( ) P( )

12 Cumulative Distribution Function (Cont.) Eample 3.3 Consider a chance eperiment of rolling a pair of fair dice, and define a r.v. as the sum of the numbers showing up. Find the cdf of. (refer to Eample 3.1) Solution: Since the cdf is defined as F () = P( ), for instance if = 3, we have: ( ) P 2 3 P 2 P 3 F P 1,1 P1,2 2,1 P1,1 P1,2 P2, Notice that: F (-) = P() = 0, and F () = P(S) = 1.

13 Cumulative Distribution Function (Cont.) Eample 3.3 (Cont.) Figure 3.2: The cdf of, which is the sum of two dice. F ( ) 0, , , , , , , , , , , , 12

14 Cumulative Distribution Function (Cont.) General properties of cdf: 1. Limiting values: lim F ( ) 0 and lim ( ) 1 2. The cdf is right hand continuous, i.e.: F ( ) lim F ( ) The cdf F () is monotonous non-decreasing function of. 4. The probability of having values b/w 1 and 2 is given by: P F ) F ( ) F 1 2 ( 2 1

15 The CDF (Cont.) Eample 3.4 A coin is tossed three times. Let r.v. be the number of heads in three trials. S = {0, 1, 2, 3} For a small positive number 1 F (1 ) P[ 1 ] P{0 heads} F (1) P[ 1] P{0 or 1heads} 8 8 For a small positive number F (1 ) P[ 1 ] P{0 or 1 heads}

16 The CDF (Cont.) Eample 3.4 (Cont.) CDF Derivative of F () with respect to. 3) ( 8 1 2) ( 8 3 1) ( 8 3 ) ( 8 1 ) ( u u u u F 3) ( 8 1 2) ( 8 3 1) ( 8 3 ) ( 8 1 ) ( f

17 Cumulative Distribution Function (Cont.) Brief verification: 1. Notice that the inverse images of at = and = are respectively: 1 ( ) and 2. This is due to the fact that the cdf is defined as F () P( ) rather than F () P( < ): detailed proof is omitted! To prove this, we need the so called continuity aiom, which is beyond the scope of this class: will be discussed at the graduate level course. 3. Follows from property # 4. 1 ( ) S

18 Cumulative Distribution Function (Cont.) Brief verification (Cont.): 4. Let 1 2, then we have: which is equivalent to: Rearranging the above provides: 0 ) ( ) ( ) ( F F P ) ( ) ( ) ( F P F ) ( P P P P

19 Cumulative Distribution Function (Cont.) Eample 3.4 Find the probability that the sum of two dice is between 3 and 7 inclusive. Solution: From the definition and properties of cdf, and using figure 3.1, we have: P 3 7 F (7) F 1 36 Note that there 20 outcomes favorable to the event, thus applying the equally likely assignment probability, we get 20/36. (3 ) F (7) F (2)

20 Cumulative Distribution Function (Cont.) Eample 3.5 Is F () below is a valid cdf? F ( ) 1 tan 2 Solution: Check if all the properties of cdf are satisfied: self study Figure 3.3: Suitable cdf.

21 Probability Density Function Definition 3.2 Probability Density Function: The pdf of a (continuous) random variable is defined as follows: f df ( ) ( ) d

22 Probability Density Function (Cont.) NOTE: interpretation of pdf!!! Using the definition of derivative, we can rewrite pdf as: For sufficiently small, we can remove the limit, and thus: f P F F ) ( ) ( ) ( F F f ) ( ) ( lim ) ( 0

23 Probability Density Function (Cont.) General properties of pdf: Since the pdf of a r.v. is the derivative of the cdf, cdf F () can be epressed as the integration of the pdf f (), i.e.: F ( ) f ( u) du 1. The pdf is a non-negative function, i.e.: f ( ) 0, 2. The area under pdf is unity, i.e.: f ( ) d 1 3. The probability of having values b/w 1 and 2 is given by: 2 P( 1 2) f ( ) d 1

24 Probability Density Function (Cont.) Brief verification: 1. Since the cdf is non-decreasing, and pdf is the derivative ( or slope) of it, it must be non-negative. 2. Using the relationship b/w the cdf and pdf, we have: f def ( ) d F ( ) 1

25 Probability Density Function (Cont.) Brief verification (Cont.): 3. From the property of the cdf, and using the relationship b/w the cdf and pdf, we have: P( 1 2) F ( 2) F ( 1 ) f f ( u) du ( u) du 1 f ( u) du

26 Probability Density Function (Cont.) Eample 3.6 Obtain the pdf of a r.v. whose cdf is given as: (eample 3.5) F ( ) 1 tan 2 and find the probability of an event in {2 < 5}. Solution: d 1 Since tan d 1 f ( ) , we easily get :

27 Probability Density Function (Cont.) Eample 3.6 (Cont.) Using the property of the cdf, we obtain: P(2 5) tan 5 tan Figure 3.4: The pdf of in Eample 3.6.

28 Probability Density Function (Cont.) Remark: The definition of the pdf can generally be applied to both continuous and discrete random variables!!! Motive: Since the cdf of a discrete r.v. can generally be epressed as the sum of the weighted & shifted unit step function u(), i.e.: F N ( ) p u( i1 i i ) Recall that the unit step function is defined as u() 1 for 0 and 0 elsewhere.

29 Probability Density Function (Cont.) Motive (Cont.): The derivative of u() is the unit impulse function () as: du( ) ( ) d the pdf f () of a discrete r.v. can generally be described as the sum of the weighted & shifted unit impulse function (), i.e.: f N ( ) p ( i1 i i )

30 Probability Density Function (Cont.) Eample 3.7 Epress the pdf of the r.v. discussed in Eample 3.3. Solution: The cdf of in terms of u() can be epressed as: ] [ 36 1 ) ( u u u u u u u u u u u F

31 Probability Density Function (Cont.) Eample 3.7 (Cont.) Solution (Cont.): Taking the derivative, we find the pdf to be: ] [ 36 1 ) ( f

32 Probability Density Function (Cont.) Figure 3.5: The pdf of in Eample 3.7

33 Common Random Variables and Their Distribution Functions Commonly occurring and widely used r.v.'s are discussed In terms of their cdf & pdf

34 Common Random Variables and Their Distribution Functions (Cont.) Uniform Random Variable: The uniform random variable is defined in terms of its probability density function as follows: 1 b a, a b, b a f ( ) 0, otherwise By integration, we can derive its cumulative distribution function (cdf) to be: 0, a F ( ) a b a, a b 1, b

35 Common Random Variables and Their Distribution Functions (Cont.) Figure 3.6: Uniform r.v.: (a) pdf, (b) cdf in case of a=0 & b=5

36 Common Random Variables and Their Distribution Functions (Cont.) Eample 3.8 Resistors are known to be uniformly distributed within 10% tolerance range. Find the probability that a nominal 1000 resistor has a value b/w 990 and Solution: Since the tolerance region is bounded in the interval [900; 1100] centered at the nominal value of 1000, the pdf of the resistance R is given by: f R 1 200, ( r) 0, 900 r 1100 otherwise

37 Common Random Variables and Their Distribution Functions (Cont.) Eample 3.8 (Cont.) Solution (Cont.): Therefore, the probability that the resistor has a resistance value in the range of [990; 1010] is then: P( dr R 1010 )

38 Common Random Variables and Their Distribution Functions (Cont.) Gaussian Random Variable: The Gaussian random variable is defined in terms of its probability density function as follows: 2 2 m 2 e f ( ) 2 2 where m and 2 are parameters called the mean and variance respectively. The cdf can be derived by direct integration of the pdf, which cannot be epressed in a closed form, however...

39 Common Random Variables and Their Distribution Functions (Cont.) Gaussian Random Variable (Cont.): Instead, we use either of the following functions defined as: (1) Q function: 1 2 u 2 Q( ) e du 2 whose numerical values are tabulated in Appendi C. (2) Error function: 2 erf ( ) e 0 2 u du

40 Common Random Variables and Their Distribution Functions (Cont.) Gaussian Random Variable (Cont.): In terms of the Q function, the cdf of the Gaussian r.v. is given by: m F ( ) 1 Q derivation: assignment Question: Epress Q and error functions in terms of each other: assignment

41 Common Random Variables and Their Distribution Functions (Cont.) Figure: Illustration of integration for Q and error functions.

42 Common Random Variables and Their Distribution Functions (Cont.) Figure 3.7: Gaussian distribution for m = 5 and = 2: (a) pdf; (b) cdf.

43 Common Random Variables and Their Distribution Functions (Cont.) Eample 3.9 A mechanical component, whose 5mm thickness (T) is known to have a Gaussian distribution w/ m = 5 and = Find the probability that T is less than 4.9mm OR greater than 5.1mm. Solution: We use here the relationship Q() = 1 - Q( ) for < 0. P( T 4.9mm OR T 5.1mm) 1 P(4.9mm T 5.1mm) F Q Q F T T Q 0.05 Q 2 Q2 1 1 Q 2 Q

44 Common Random Variables and Their Distribution Functions (Cont.) Eponential Random Variable 사건들이발생하는시간의간격 장치와시스템의수명을모형화 f e ( ) u( ), 0 F e ( ) (1 ) u( )

45 Common Random Variables and Their Distribution Functions (Cont.) Gamma Random Variable: The Gamma random variable has its probability density function as follows: f ( ) b c ( b) b1 e c u( ), b and c 0 where (b) is the gamma function given by the integral: 0 b1 y ( b) y e dy It can be shown by evaluation that (1) = 1 and (½ )= ½, and by replacing b by b+1, we can also show that (b+1) = b(b), which in turn provides that (n+1) = n! in case of b an integer n.

46 Common Random Variables and Their Distribution Functions (Cont.) Special cases: 1. Chi-square pdf: (statistics) where b and c are replaced by b = n/2 and c = 2 2. Erlang pdf: (queuing theory) where b = n is an integer. ) ( 2) ( 2 1 ) ( u e n f n n ) ( 1! ) ( 1 u e n c f c n n

47 Common Random Variables and Their Distribution Functions (Cont.) Figure 3.8: Plots of (a) chi-square pdf and (b) Erlang pdf.

48 Common Random Variables and Their Distribution Functions (Cont.) Cauchy Random Variable: The Cauchy random variable has its probability density function as follows: f ( ) 2 2 Corresponding cdf is given by: F ( ) tan 2 Note: Eample 3.5 with figure 3.3 is the case of Cauchy r.v. for = 1.

49 Common Random Variables and Their Distribution Functions (Cont.) Binomial Random Variable: The r.v. is binomially distributed if it takes on the non-negative integer values 0, 1, 2,, n with probabilities: where p + q = 1. Corresponding pdf and the cdf, using the unit impulse function, are given by: n k k n k k q p k n f 0 ) ( n k q p k n k P k n k, 2, 0,1,, ) ( k k n k q p k n F ) (

50 Common Random Variables and Their Distribution Functions (Cont.) Remark: This is the distribution describing the number (k) of heads occurring in n tosses of a fair coin, in which case p = q = 0.5. Figure 3.9: The pdf of binomial r.v. for n = 10: (a) p = q = 1/2 ; (b) p =1/5, q = 4/5.

51 Common Random Variables and Their Distribution Functions (Cont.) Geometric Random Variable: Suppose we flip a biased coin w/ probability of a head is p whereas the probability of tail is 1- p. Then, the probability of getting the head (success) for the first time at the k-th toss is given by: P(first succes at trial k) P( k) where p is the probability of success. called the geometric distribution k1 1 p p, k 1, 2, corresponds to the geometric mean of k - 1 and k + 1 values.

52 Common Random Variables and Their Distribution Functions (Cont.) Poisson Random Variable: The r.v. is Poisson with parameter a if it takes on the non-negative integer values 0, 1, 2, with probabilities: k a a P( k) e, k 0,1, k! 2, Corresponding cdf of a Poisson r.v. is given by: F ( ) k a e k! k a

53 Common Random Variables and Their Distribution Functions (Cont.) Figure 3.10: Poisson pdf (a) a = 0.9; (b) a = 0.2.

54 Common Random Variables and Their Distribution Functions (Cont.) Limiting Forms of Binomial and Poisson Distributions: Theorem 3.1 De Moivre-Laplace theorem: For n sufficiently large, the binomial distribution can be approimated by the samples of a Gaussian curve properly scaled and shifted as: p k q nk where m = np and 2 = npq respectively. Proof: omit e km 2 2 2, n 1

55 Common Random Variables and Their Distribution Functions (Cont.) Corresponding cdf of a binomial r.v., based on the De Moivre-Laplace theorem, can be approimated as: np F ( ) 1 Q npq where Q() is the Q-function.

56 Common Random Variables and Their Distribution Functions (Cont.) Figure 3.11: Binomial cdf and De Moivre-Laplace approimation for n = 10: (a) p = 0.2; (b) a = 0.5.

57 Common Random Variables and Their Distribution Functions (Cont.) Limiting Forms of Binomial and Poisson Distributions (Cont.): Theorem 3.2 : The Poisson distribution approaches to a Gaussian distribution for a >> 1 as: k a e k! a Proof: e ka 2 2a 2a for n 1, This is due to the fact that a binomial distribution approaches the Poisson distribution with a = np if n >> 1 and p << 1, and applying the De Moivre-Laplace theorem proves it! p 1, np a

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59 Common Random Variables and Their Distribution Functions (Cont.) Eample 3.11 In a digital communication system, the probability of an error is What is the probability of 3 errors in transmission of 5000 bits? Solution: This has a binomial distribution, and the Poisson approimation with n = 5000, p = 10-3, and k = 3 shows: P(3 errors) e ! whereas the eact value of the probability using the binomial distribution is:

60 Common Random Variables and Their Distribution Functions (Cont.) Poisson Points and Eponential Probability Density Function: Consider a finite interval T (fig. 3.12), with the probability of k occurrence of an event ( ni : arrival of electrons at certain point) in this interval obeys a Poisson distribution, i.e.: k T T P( k) e, k 0,1, k! 2, where is the number of events per unit time.

61 Common Random Variables and Their Distribution Functions (Cont.) Poisson Points and Eponential Probability Density Function (Cont.): Then, the interval from an arbitrarily selected point and the net event is a r.v. W following an eponential distribution with its pdf as: f W w ( w) e u( w) Figure 3.12: Poisson points on a finite interval T.

62 Common Random Variables and Their Distribution Functions (Cont.) Poisson Points and Eponential Probability Density Function (Cont.): Proof: First, find the cdf of W as: w FW ( w) P( W w) 1 P( 0) 1 e, w 0 {W w} corresponds to the event that there is at least one Poisson event in the interval (0, w). Differentiating it, we obtain: f W d w ( w) FW ( w) e u( w) dw

63 Common Random Variables and Their Distribution Functions (Cont.) Eample 3.12 Raindrops impinge on a tin roof at a rate of 100/sec. What is the probability that the interval between adjacent raindrops is greater than 1(msec), and 10(msec)? Solution: This can be approimated by a Poisson point process with = 100, and therefore: P( W P( W ) 1 P( W 1 ) 1 P( W e e ) ) e e

64 Common Random Variables and Their Distribution Functions (Cont.) Pascal Random Variable: The r.v. has a Pascal distribution if it takes on the positive integer values 1, 2, 3, with probabilities: n 1 k nk P( k) p q for k 1, 2,, k 1 and q 1 p

65 Common Random Variables and Their Distribution Functions (Cont.) Note: 1. For eample, in a sequence of coin tossing, the probability of getting the k-th head on the n-th toss obeys a Pascal distribution, where p = prob. of a head. 2. Reasoning: We get the first k-1 heads in any order in the first n-1 tosses, which is binomial, and then must get a head on the n-th toss. 3. Geometric distribution is a special case of the Pascal distribution. Eample 3.13 Self study

66 Common Random Variables and Their Distribution Functions (Cont.) Hypergeometric Random Variable: Consider a bo containing N items, K of which are defective. Then, the probability of obtaining k defective items in a selection of n items without replacement follows the hypergeometric distribution: K N K k n k P ( k), k 0,1, 2,, n N n Eample 3.14 Self study