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Chapter 3 Discrete Random Variables. 主講人 : 虞台文. Content. Random Variables The Probability Mass Functions Distribution Functions Bernoulli Trials Bernoulli Distributions Binomial Distributions Geometric Distributions Negative Binomial Distributions Poisson Distributions - PowerPoint PPT Presentation
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Chapter 3Discrete Random Variables:
ContentRandom VariablesThe Probability Mass FunctionsDistribution FunctionsBernoulli TrialsBernoulli DistributionsBinomial DistributionsGeometric DistributionsNegative Binomial DistributionsPoisson DistributionsHypergeometric DistributionsDiscrete Uniform Distributions
Chapter 3Discrete Random VariablesRandom Variables
Definition Random VariablesA random variable X of a probability space (, A, P) is a real-valued function defined on , i.e.,
Definition Random VariablesA random variable X of a probability space (, A, P) is a real-valued function defined on , i.e.,
Example 1
Example 11
Example 12
Example 1
Example 2
Example 2
Notations
Example 1
Example 1
Example 1
Definition Discrete Random VariablesA discrete random variable X is a random variable with range being a finite or countable infinite subset {x1, x2, . . .} of real numbers R.
Definition Discrete Random VariablesA discrete random variable X is a random variable with range being a finite or countably infinite subset {x1, x2, . . .} of real numbers R.What is countablity?The set of all integersThe set of all real numbers
Example 1X, Y and Z are discrete random variables. All are finite
Example 2X, Y and Z are not discrete random variables. All are uncountable
Example 2X, Y and Z are not discrete random variables. All are uncountableIn fact, they are continuous random variables.
Chapter 3Discrete Random VariablesThe Probability Mass Functions
Definition The Probability Mass Function (pmf)The probability mass function (pmf) of r.v. X, denoted by pX(x), is defined as
Example 4
Example 4
Example 4
Example 4
Properties of pmfsx
Chapter 3Discrete Random VariablesDistribution Functions
Cumulative Distribution Function (cdf)cdf
Cumulative Distribution Function (cdf)cdfpmf
Cumulative Distribution Function (cdf)cdfpmf
Cumulative Distribution Function (cdf)cdfpmfxpX(x)x1x2x3x4
Cumulative Distribution Function (cdf)cdfpmfxpX(x)x1x2x3x4
Cumulative Distribution Function (cdf)cdfpmfxpX(x)x1x2x3x41pX(x1)pX(x2)pX(x3)pX(x4)
Example 5
Example 5
Example 5
Example 5
Example 5
Example 5
Properties of cdfs xMonotonically nondecreasing.
Properties of cdfs xMonotonically nondecreasing.
F(b)
F(a)
Chapter 3Discrete Random VariablesBernoulli Trials
Bernoulli TrialsSuppose an experiment consists of n trials, n > 0.The trials are called Bernoulli trials if three conditions are satisfied:Each trial has a sample space {S =1, F =0} (two outcomes), S to be called success and F to be called failure.For each trial P(S) = p and P(F) = q, where 0 p 1 and q = 1 p.The trials are independent.
Example 6Tossing a die ten times, the actual face number in each toss is unnoted. Instead, the outcome of 1 or 2 will be considered a success, and the outcome of 3, 4, 5, or 6 will be considered a failure. What is the sample space of the experiment?Is this experiment to performing Bernoulli Trials?Why?
DiscussionWhat probabilities may interest us on performing Bernoulli Trials?
Chapter 3Discrete Random VariablesBernoulli Distributions
Bernoulli DistributionsLet r.v. X denote the outcome of a Bernoulli trial, and let the probability of success equal to p.
Then, we havecdfpmf
Bernoulli DistributionsLet r.v. X denote the outcome of a Bernoulli trial, and let the probability of success equal to p.
Then, we havecdfpmf
Bernoulli DistributionsLet r.v. X denote the outcome of a Bernoulli trial, and let the probability of success equal to p.
Then, we havecdfpmf
Bernoulli DistributionsLet r.v. X denote the outcome of a Bernoulli trial, and let the probability of success equal to p.
Then, we havecdfpmf
Bernoulli DistributionsLet r.v. X denote the outcome of a Bernoulli trial, and let the probability of success equal to p.
Then, we havecdfpmf
Bernoulli Distributionscdfpmf
Chapter 3Discrete Random VariablesBinomial Distributions
pnXI(X)=?P(X=x)=?
Binomial DistributionsConsider n Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #successes in the experiment.Then,cdfpmfn trials
x successes
nx fails
ppp(1p)(1p)(1p)......
Binomial DistributionsConsider n Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #successes in the experiment.Then,cdfpmfn trials
x successes
nx fails
ppp(1p)(1p)(1p)......
Binomial DistributionsConsider n Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #successes in the experiment.Then,cdfpmfn trials
x successes
nx fails
ppp(1p)(1p)(1p)......
Binomial DistributionsConsider n Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #successes in the experiment.Then,cdfpmf
Binomial DistributionsConsider n Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #successes in the experiment.Then,cdfpmfwithout closed-form
Binomial DistributionsConsider n Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #successes in the experiment.Then,cdfpmf
Binomial Distributionscdfpmf
Binomial Distributions
Example 7Verify that b(x; n, p) is a valid pmf.
Example 810% of IC chips from an IC manufacturer are known to be defective. Taking a sample of ten IC chips from the manufacture. Find the probabilities of1. no IC chip is defective;2. at least 2 IC chips are defective.
Example 810% of IC chips from an IC manufacturer are known to be defective. Taking a sample of ten IC chips from the manufacture. Find the probabilities of1. no IC chip is defective;2. at least 2 IC chips are defective.Let X denote #defectives
Chapter 3Discrete Random VariablesGeometric Distributions
pI(X)=?P(X=x)=?!!!X
Geometric DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote the number of trials up to and including the first success.Then,cdfpmfx trials
x1 fails
(1p)(1p)(1p). . .First successp
Geometric DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote the number of trials up to and including the first success.Then,cdfpmf
Geometric DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote the number of trials up to and including the first success.Then,cdfpmfThe parameter of the experiment.
Geometric Distributionscdfpmf
Chart4
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0.06561
0.059049
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x
pX(x)
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200.50.250.75
00.00000095370.00317121190
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200.000000953700.0031712119
111
x
b(x;20,0.5)
b(x;20,0.25)
x
x
b(x;20,0.75)
Example 9Tossing a fair die, find:the probability of the first appearing of 1 is in the 5th toss;the probability of the first appearing of 1 is in the first five tosses.
Example 9Tossing a fair die, find:the probability of the first appearing of 1 is in the 5th toss;the probability of the first appearing of 1 is in the first five tosses.Let X denote #tossing to reach the 1st 1
Memoryless or Markov Property12m?
Memoryless or Markov Property12mm+1m+2m+n12n
Memoryless or Markov PropertyX
Memoryless or Markov Property
Memoryless or Markov PropertyA r.v. X is said to have memoryless or Markov property if it satisfies
Theorem 1Let r.v. X have image 1, 2,. Then, XG(p) P(X > m + n|X > m) = P(X > n)where m, n be any positive integers.
Theorem 1PF)
Theorem 1PF)Define
Theorem 1Let r.v. X have image 1, 2,. Then, XG(p) P(X > m + n|X > m) = P(X > n)where m, n be any positive integers.
Geometric DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote the number of trials up to and including the first success.Then,cdfpmfModifiedYY0,1,2,Yyy = 0,1,y+1Yyy < 00 yyfailures
Modified Geometric DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. Y denote the number of failures up to the first success.Then,cdfpmf
Modified Geometric Distributionscdfpmf
Chapter 3Discrete Random VariablesNegative Binomial Distributions
pI(X)=?P(X=x)=?!!!rX:r
Negative Binomial DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #trials up to and including the rth success.Then,cdfpmfrX = x
x1r1
Negative Binomial DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #trials up to and including the rth success.Then,cdfpmf
Negative Binomial DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #trials up to and including the rth success.Then,cdfpmfwithout closed-form
Negative Binomial DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #trials up to and including the rth success.Then,cdfpmfwithout closed-form
Negative Binomial DistributionscdfpmfFact:
Negative Binomial Distributionscdfpmf
Chart1
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pX(x)
X~NB(5,0.2)
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b(x;20,0.5)
b(x;20,0.25)
x
x
b(x;20,0.75)
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x
pX(x)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
50.00032
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x
pX(x)
X~NB(5,0.2)
Negative Binomial Distributionspmf
Negative Binomial DistributionspmfVerify that nb(x;r,p) is a valid pmf.Exercise
pI(X)=?P(X=x)=?!!!rX:r
pI(Y)=?P(Y=y)=?!!!rY:rFact: Y = X r
Negative Binomial DistributionspmfModifiedFact: Y = X rYYyyy =0,1,2, yyy
Modified Negative Binomial DistributionspmfFact:
Chapter 3Discrete Random VariablesPoisson Distributions
A Toll Station
Arriving/Failure Rate:
Poisson DistributionsConsider a highway toll station.Assume that, on average, vehicles pass the station per unit time interval (e.g., an hour). Let X denote #vehicles passing in a time interval of duration t, i.e., (0, t].: Arriving rateX: #vehicles passing in (0, t]I(X) = ?P(X=x) = ?{0, 1, 2, }
Poisson Distributions: Arriving rateX: #vehicles passing in (0, t]P(X=x) = ? ?
t
n
n
Poisson Distributions: Arriving rateX: #vehicles passing in (0, t]P(X=x) = ? ?
t
n
n()?010101001000110001010
Poisson Distributions: Arriving rateX: #vehicles passing in (0, t]P(X=x) = ? ?
t
n
nn
p = ?
Poisson Distributions: Arriving rateX: #vehicles passing in (0, t]P(X=x) = ? ?
t
n
nn
p = ?np
Poisson Distributions: Arriving rateX: #vehicles passing in (0, t]P(X=x) = ? ?
t
n
nn
p = ?
Poisson Distributions: Arriving rateX: #vehicles passing in (0, t]
=1, n
=1, n
Poisson Distributions: Arriving rateX: #vehicles passing in (0, t]Let
Poisson Distributions: Arriving rateX: #vehicles passing in (0, t]
Poisson DistributionsConsider a highway toll station.Assume that, on average, vehicles pass the station per unit time interval (e.g., an hour). Let X denote #vehicles passing in a time interval of duration t, i.e., (0, t].: Arriving rateX: #vehicles passing in (0, t]cdfpmfwithout closed-form
Poisson DistributionsConsider a highway toll station.Assume that, on average, vehicles pass the station per unit time interval (e.g., an hour). Let X denote #vehicles passing in a time interval of duration t, i.e., (0, t].cdfpmfwithout closed-formin an interval the same interval
Poisson DistributionscdfpmfVerify that p(x;) is a valid pmf.Exercise
Poisson DistributionscdfpmfVerify that p(x;) is a valid pmf.Exercise
Chart1
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Sheet3
0
0
0
0
0
0
0
0
0
0
0
0
0
Sheet4
00
00
00
00
00
00
00
00
Sheet5
200.50.250.75
00.00000095370.00317121190
10.00001907350.02114141290.0000000001
20.00018119810.06694780760.0000000016
30.00108718870.13389561520.000000028
40.00462055210.18968545490.0000003569
50.01478576660.20233115190.0000034265
60.03696441650.16860929320.0000256987
70.0739288330.11240619550.0001541923
80.12013435360.06088668920.0007516875
90.16017913820.02706075080.0030067501
100.1761970520.00992227530.0099222753
110.16017913820.00300675010.0270607508
120.12013435360.00075168750.0608866892
130.0739288330.00015419230.1124061955
140.03696441650.00002569870.1686092932
150.01478576660.00000342650.2023311519
160.00462055210.00000035690.1896854549
170.00108718870.0000000280.1338956152
180.00018119810.00000000160.0669478076
190.00001907350.00000000010.0211414129
200.000000953700.0031712119
111
Sheet5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
x
b(x;20,0.5)
Sheet6
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
b(x;20,0.25)
x
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
x
b(x;20,0.75)
10.1
20.09
30.081
40.0729
50.06561
60.059049
70.0531441
80.04782969
90.043046721
100.0387420489
110.034867844
120.0313810596
130.0282429536
140.0254186583
150.0228767925
160.0205891132
170.0185302019
180.0166771817
190.0150094635
200.0135085172
210.0121576655
220.0109418989
230.009847709
240.0088629381
250.0079766443
260.0071789799
270.0064610819
280.0058149737
290.0052334763
300.0047101287
310.0042391158
320.0038152042
330.0034336838
340.0030903154
350.0027812839
360.0025031555
370.00225284
380.002027556
390.0018248004
400.0016423203
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
x
pX(x)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
50.00032
60.00128
70.003072
80.0057344
90.00917504
100.0132120576
110.0176160768
120.0221459251
130.0265751101
140.0307090162
150.0343940981
160.0375208343
170.0400222233
180.0418694028
190.0430656714
200.0436398804
210.0436398804
220.04312647
230.042168104
240.0408364797
250.0392030205
260.03733621
270.0352996895
280.0331510127
290.0309409452
300.0287131971
310.0265044897
320.0243448646
330.0222581619
340.0202626026
350.0183714263
360.0165935464
370.0149341917
380.0133955174
390.0119771685
400.0106767902
410.0094904802
420.0084131824
430.0074390245
440.0065616011
450.0057742089
460.0050700371
470.0044423182
480.0038844457
490.0033900617
500.0029531204
510.0025679308
520.0022291825
530.0019319582
540.0016717352
550.0014443792
560.0012461311
570.0010735899
580.0009236924
590.0007936913
600.0006811314
610.0005838269
620.0004998378
630.0004274475
640.0003651416
650.0003115875
660.0002656156
670.0002262016
680.0001924509
690.0001635833
700.00013892
710.0001178715
720.0000999269
730.0000846439
740.0000716407
750.0000605875
760.0000512007
770.0000432362
x
pX(x)
X~NB(5,0.2)
00
10.0000000003
20.0000000043
30.0000000362
40.000000226
50.0000011302
60.0000047092
70.0000168185
80.0000525578
90.000145994
100.000364985
110.0008295113
120.0017281486
130.0033233628
140.0059345764
150.0098909606
160.0154546259
170.0227273911
180.0315658209
190.0415339749
200.0519174686
210.0618065102
220.0702346707
230.0763420334
240.0795229515
250.0795229515
260.0764643764
270.0708003485
280.0632145969
290.0544953422
300.0454127851
310.0366232138
320.0286118858
330.0216756711
340.0159379934
350.011384281
360.0079057507
370.0053417234
380.0035142917
390.0022527511
400.0014079694
410.000858518
420.0005110226
430.0002971062
440.0001688103
450.0000937835
460.0000509693
470.0000271113
480.0000141205
490.0000072043
500.0000036022
510.0000017658
520.0000008489
530.0000004004
540.0000001854
550.0000000843
560.0000000376
570.0000000165
580.0000000071
590.000000003
600.0000000013
610.0000000005
620.0000000002
630.0000000001
640
650
660
670
680
690
700
710
720
730
740
750
760
770
780
790
800
810
820
830
840
850
860
870
880
890
900
910
920
930
940
950
960
970
980
990
1000
x
pX(x)
Example 12On average, a job arrives for CPU service every 6 seconds. Find the probability that there will be less than or equal to 4 arrivals in a given minute? = 1/6 job/sec.Let X denote #arrivals in the minute.t = 60 secs. = t = 10 jobs.
Poisson ApproximationThe binomial distribution is important, but the probability values associated with it are hardly evaluated.
Poisson Approximationppnp
Poisson Approximationpp
Poisson Approximationpppp
Poisson Approximation
Poisson ApproximationnpE.g., n 20 and p 0.05.
Example 13A manufacture produces IC chips, 1% of which are defective. A box contains 100 chips. Find the probability that1. the box contains no defective.2. the box contains less than or equal to 2 defectives.Let X denote the number of defectives.
Chapter 3Discrete Random VariablesHypergeometric Distributions
Hypergeometric Distributions+= N
d
NdI(X)=?P(X=x)=?
Hypergeometric DistributionsI(X)=?P(X=x)=?cdfpmfwithout closed-form
Hypergeometric Distributionspmf
Example 14Compute the probability of obtaining three defectives in a sample of size ten taken without replacement from a box of twenty components containing four defectives.Let X denote the number of defectives.
Hypergeometric Distributionspmf
Example 15A box contains 200 red balls and 800 black balls. Now 10 balls are taken without replacement. Find the probability of obtaining none red ball.Let X denote #red balls taken.
Chapter 3Discrete Random VariablesDiscrete Uniform Distributions
Discrete Uniform DistributionsA r.v. X is said to possess a discrete uniform distribution if it has a finite image {x1, x2, , xN} and has the pmf
Example 16One ball is drawn from a box containing 10 balls numbered 1,2,. . . ,10. Find the probability that the ball number is less than 4.Let r.v. X denote the ball number.Is uniform distribution really that simple?
ReviewBernoulli DistributionsBinomial DistributionsGeometric DistributionsNegative Binomial DistributionsPoisson DistributionsHypergeometric DistributionsDiscrete Uniform Distributions
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