Lecture 4-Discrete Random Variables .pdf

The lecture "Discrete Random Variables in Probability, Statistics & Random Processes" at HCMIU introduces the concept of discrete random variables and their applications. It covers probability mass functions, expected values, and variances for distributions like binomial, Poisson, and geometric. Students learn to model real-world scenarios, such as customer arrivals or equipment failures, using these distributions. The lecture emphasizes calculations and problem-solving, with practical examples and tools like Python or R for analyzing discrete random processes.

1. Discreterandomvariables
October19,2023
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2. Objectives
1Understandrandomvariables2FordiscreterandomvariablesaDetermine probabilities from probability mass
functions and the reverse
bDetermine probabilities from cumulative
distribution functions and cumulative
distribution functions from probability mass
functions, and the reverse
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3. Intro
•Samplespaceandeventsarebasic
componentsofprobability
•Similartonumbersincalculus
•Studytherelationsbetweennumberswe
usefunctions
•Inprobabilityweuserandomvariables
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4. Definition
ArandomvariableXdefinedonasamplespace
Ωisaquantitythatiscalculatedbytheoutcomes
-afunctionofoutcomes
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5. Example
•Tossacointhreetimes
•Xbethenumberoftimesthattailsappear
•SamplespaceΩ =
{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
•X(HHH) =0,X(HHT) =1 ...
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6. wHHHHHTHTHHTTTHHTHTTTHTTT
X(w)0 1 1 2 1 2 2 3
All possible values ofXare 0, 1, 2, 3. We sayRange(X) =
{0,1,2,3}
P(X=0) =P(there is no tail) =P(HHH) =
1
8
P(X=1) =P(there is 1 tail) =P({HHT,HTH,THH}) =
3
8
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7. Example
•TossingafaircoinuntilaHeadappear.
Thesamplespaceis
Ω ={H,TH,TTH,TTTH,...}
•X: thenumberoftossing
•Range(X) ={1,2,3,......}
•X=1ifandonlyifthefirstcointurnshead
soP(X=1) =P(firstcointurnsH) =0.5
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8. Question
Evaluate
P(X=2)
and
P(1<X≤4)
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9. TypesofRV
BasedonrangeoftherandomvariableX
•IfthesetofpossiblevaluesofXisfiniteor
countablelike{0,1,2,3},{1,2,...}then
XiscalleddiscreteRV
•IfthesetofpossiblevaluesofXis
uncountable(liketheinterval[a,b],
[a,∞))thenXiscalledcontinuousRV
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10. Probabilityassociatedwitha
randomvariableX
P(X=a),P(X≥a),P(X>a),
P(X≤b),P(X<b),
P(a≤X≤b),P(a<X≤b),
P(a≤X<b),P(a<X<b)
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11. Cumulativedistributionfunction
(cdf)
cdfofF(.)oftherandomvariableXisafunction
definedby
F(b) =P(X≤b)
istheprobabilitythatXtakesonvalueslessthan
orequaltob
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12. Use cdf to answer questions about a random variable
P(X≤b) =F(b)
P(X<b) = lim
h→0
+
P(X≤b−h) = lim
x→b
−
F(x) =F(b
−
)
P(X>a) =1−P(X≤a) =1−F(a)
P(a<X≤b) =P(X≤b)−P(X≤a) =F(b)−F(a)
P(a≤X≤b) =F(b)−F(a
−
)
P(a<X<b) =F(b
−
)−F(a)
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13. DiscreteRandomVariables
Randomvariablecantakeonatmostcount-
ablenumberofpossiblevalues
Range(X)={x
1,x
2,...}
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14. Probabilitymassfunction
Theprobabilitymassfunction(p.m.f)ofthedis-
creterandomvariableXisdefinedas
p(xi) =P(X=xi)forallxi∈Range(X)
theprobabilitythatthevalueofXisequaltoxi
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15. Example
Rolltwofairdice
Ω ={(1,1,),...,(6,6)}={(i,j) :1≤i,j≤6}
LetXbe the largest of numbers on two dice, i.e
iftherollingresultis(i,j)then
X(i,j) = max(i,j)
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16. TablevaluesofX
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17. AllpossiblevaluesofXis
Range(X) ={1,2,3,4,5,6}
soXisadiscreterandomvariable.
In order to determine the pmf ofX, we need to
findalltheprobabilities
P(X=1),P(X=2),...,P(X=6)
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18. X=1ifandonlyiftheoutcomeis(1,1). So
P(X=1) =P((1,1)) =1/36
X=2 if and only if the outcomes is one of
(1,2),(2,2),(2,2). So
P(X=2) =P({(1,2),(2,1),(2,2)})
=P((1,2))+P((2,1))+P((2,2) =3/36
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19. Similar,wehaveP(X=3) =5/36,P(X=4) =
7/36,P(X=5) =9/36,P(X=6) =11/36.
Wecansummaryp.m.fofXinthe
x 123456
P(X=x)
1
36
3
36
5
36
7
36
9
36
11
36
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20. Similar,wehaveP(X=3) =5/36,P(X=4) =
7/36,P(X=5) =9/36,P(X=6) =11/36.
Wecansummaryp.m.fofXinthe
x 123456
P(X=x)
1
36
3
36
5
36
7
36
9
36
11
36
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21. Illustrationtocalculatepmf
For each possible valuex, we collect all the outcomes that
give rise toX=xand add their probabilities to obtain
pX(x) =P(X=x). 20 / 38

22. One can use p.m.f of the discrete random vari-
ableXtoansweranyquestionofXsuchas
P(1<X<4) =P(X=2orX=3)
=P(X=2)+P(X=3) =3/36+5/36
usingadditiverulefordisjoinset
P(A∪B) =P(A)+P(B)ifA∩B=∅
forA={X=2},B={X=3}
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23. Practice
Probabilitymassfunctionofdiscreterandomvari-
ableXis
x -2-1012
p(x) =P(X=x)1/82/82/82/81/8
Determine
1P(X≤ −1orX=2)2P(−1≤X≤1)
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24. Practice
A shipment of 20 similar laptop computers to a
retail outlet contains 3 that are defective. If a
school makes a random purchase of 2 of these
computers,find the probability mass function
(p.m.f)for.
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25. Practice
An urn contains 11 balls, 3 white, 3 red, and 5
blue balls. Take out 3 balls at random, without
replacement. You win $1 for each red ball you
select and lose a $1 for each white ball you se-
lect. Determinethep.m.fofyourloss/profitX.
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26. Propertiesofpmf
•Xisdiscrete
→Range(X) ={x1,...,xn...}
•p(xi) =P(X=xi)≥0
•P(X∈A) =
X
xi∈A
p(xi)
•Normalization
P(−∞<X<∞) =1⇒
∞
X
i=1
p(xi) =1
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27. Propertiesofpmf
•Xisdiscrete
→Range(X) ={x1,...,xn...}
•p(xi) =P(X=xi)≥0
•P(X∈A) =
X
xi∈A
p(xi)
•Normalization
P(−∞<X<∞) =1⇒
∞
X
i=1
p(xi) =1
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28. Propertiesofpmf
•Xisdiscrete
→Range(X) ={x1,...,xn...}
•p(xi) =P(X=xi)≥0
•P(X∈A) =
X
xi∈A
p(xi)
•Normalization
P(−∞<X<∞) =1⇒
∞
X
i=1
p(xi) =1
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29. Example
SupposeXhas3values1,2,3and
p(1) =
1
2
,p(2) =
1
3
thenwhatisp(3)?
p(3) =1−p(1)−p(2) =1/6.
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30. Example
SupposeXhas3values1,2,3and
p(1) =
1
2
,p(2) =
1
3
thenwhatisp(3)?
p(3) =1−p(1)−p(2) =1/6.
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31. Graphofp(x)
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32. Practice
Suppose that the pmf of random variableXis
givenby
p(x) =c(x+5),x=0,1,2,3,4
FindcandP(0<X<2.5).
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33. Cumulativedistributionfunction
(cdf)
ProbabilitythatXdoesnotexceedagivenvalue
F(b) =P(X≤b) =
X
xi≤b
P(X=xi)
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34. Example
SupposethatpmfofXisgivenbyp(1) =
1
2
,p(2) =
1
3
,p(3) =
1
6
then
F(0.5) =P(X≤0.5) =0
F(2.4) =P(X≤2.4) =P(X=1)+P(X=2) =
5
6
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35. TheformulaofthecdfofXis
F(x) =











0,x<1
1
2
,1≤x<2
5
6
,2≤x<3
1,x≥3
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36. GraphofF(x)
Remark
Jumpsizeat1isP(X=1),...
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37. Example
Determinethep.m.fofXfromthec.d.f
F(x) =











0 ifx<−2
0.2 if−2≤x<0
0.7 if0≤x<2
1 ifx≥2
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38. Solution
Graph ofF(x)is
P(X=a) =F(a)−F(a
−
)isnonzeroatthepoints−2,0,2.
The p.m.f at each point is the change (jump size) of c.d.f
at the point 34 / 38

39. p.m.fofXisgivenby
p(−2) =P(X=−2) =F(−2)−F(−2
−
)
=0.2−0=0.2
p(0) =P(X=0) =F(0)−F(0
−
)
=0.7−0.2=0.5
p(2) =P(X=2) =F(2)−F(2
−
)
=1−0.7=0.3
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40. PropertiesofcdfofadiscreteRV
•lim
x→−∞
F(x) =0andlim
x→∞
F(x) =1
•FXhasapiecewiseconstantand
staircase-likeform.
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41. Practice
c.d.fofdiscreterandomvariableXisgivenby
F(x) =





0 ifx<1
0.7 if1≤x<3
1 ifx≥3
Compute
1P(X≤2)andP(X>2)2P(1≤X≤2)
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42. Keywords
•pmf of a discrete RV with range{x1,...,xn,...}
p(xi) =P(X=xi)
•0≤p(xi)≤1
•
P
p(xi) =1
•P(X∈A) =
P
xi∈A
p(xi)
•cdf
F(x) =P(X≤x) =
X
xi≤x
p(xi)
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