1. ONE DIMENSIONAL RANDOM VARIABLES
UNIT - I
ONE DIMENSIONAL RANDOM VARIABLES
E.MATHIVADHANA, M.Sc.,M.PHIL.
ASSISTANT PROFESSOR
DEPARTMENT OF MATHEMATICS
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2. SYLLABUS:
Random variables – Probability function – Moments – Moment generating functions and their properties – Binomial, Poisson, Geometric, Uniform, Exponential, Gamma and Normal Distributions – Functions of a Random Variable.
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3. PROBABILITY BASIC DEFINITIONS
It is the science of decision-making with calculated risks in the face of uncertainty.
EXPERIMENT:
It is a process which results in some well defined outcomes.
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4. RANDOM EXPERIMENT: Eg:
Tossing a fair coin, Throwing a die ,Taking a card from a pack of cards are Random Experiment.
SAMPLE SPACE:
The set of all possible outcomes of a random experiment is called its Sample Space. It is denoted by S.
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5. Random Experiment Sample Space S Tossing a coin once {H,T}
Eg:
Random Experiment Sample Space S
Tossing a coin once {H,T}
Tossing a coin Twice {HH,TH,HT,TT}
Throwing a die once {1,2,3,4,5,6}
OUTCOME:
The result of a random experiment is called an outcome.
Eg: Head or tail
TRIAL:
Any particular performance of a random experiment is called a trial.
Eg: Tossing a coin
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6. The outcomes of a trial are called events. Eg: Getting head or tail
MATHEMATICAL PROBABILITY:
The Probability of the occurrence of a event A is denoted by P[A] and defined as
P[A] = n[A] / n[S] 0 ≤ P[A] ≤ 1
n[A] = number of favourable cases to A.
n[S] = number of possible cases in S.
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7. CONDITIONAL PROBABILITY:
INDEPENDENT EVENTS:
If the Occurrence of any one of them does not depend on the occurrence of the other.
Eg: Tossing a coin, the event of getting a head in the first toss is independent of getting a head in the second and subsequent throws.
CONDITIONAL PROBABILITY:
The Conditional Probability of an event B ,assuming that the event A has already happened , is denoted by P[B/A] and defined as
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8. EXAMPLE FOR CONDITIONAL PROBABILITY:
When a fair die is tossed ,what is the probability of getting 1 given that an odd number has been obtained.
SOLUTION:
Let S={1,2,3,4,5,6} ; n[S]=6
Let A be the probability of getting a odd number
A={1,3,5} ; n[A]=3
P[A]=n[A] / n[S] =3/6
Let B be the probability of getting 1. B={1}
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9. PROBABILITY DISTRIBUTION FUNCTION OF X :
If X is a random variable ,then the function F[x] defined as
is called the distribution function of X.
PROBABILITY DISTRIBUTION FUNCTION OF Y :
If X is a random variable ,then the function F[y] defined as
is called the distribution function of Y.
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10. Random variables are two types: i) Discrete Random variable
If E is an experiment having sample space S, and X is a function that assigns a real number X(s) to every outcome s є S, then X(s) is called a random variable (r.v.)
Random variables are two types:
i) Discrete Random variable
ii) Continuous Random variable
Eg:
Someone continuously shoot on the same target, until really shot and then stop. s is the number of shots，so
s One shooting Two shooting n shooting......
X(s) n
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11. Random Variable Let S be the sample space.
A random variable X is a function
X: SReal
Suppose we toss a coin twice. Let X be the random variable number of heads
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12. Random Variable (Number of Heads in two coin tosses)X TT TH 1 HT HH 2
We also associate a probability with X attaining that value.
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13. Random Variable (Number of Heads in two coin tosses)
Prob X TT
1/4 TH 1 HT HH 2 X
P(X=x)
1/4 1
1/2 2
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14. Discrete Random variable:
If X is a random variable which can take a finite number or countably infinite number of values, X is called a Discrete Random variable.
Eg:
1. The number shown when a die is thrown.
2. Number of transmitted bits received in error.
Continuous Random variable:
If X is a random variable which can take all values in an interval, then is called a Continuous Random variable.
The length of time during which a vacuum tube installed in a circuit functions is a continuous RV.
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15. Probability Mass Function (or) Probability Function:
If X is a discrete RV with distinct values x1 x2 x3 ,… xn… then P ( X = xi) = pi, then the function p(x) is called the Probability Mass Function.
Provided p(i = 1, 2, 3, …) satisfy the following conditions:
for all i, and 2.
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16. PROBABILITY DENSITY FUNCTION FOR CONTINUOUS CASE:
If X is a continuous r.v., then f(x) is defined the probability density function of X.
Provided f(x) satisfies the following conditions, 1. 2. 3.
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17. CUMULATIVE DISTRIBUTION FUNCTION (OR) DISTRIBUTION FUNCTION OF D.R.V.:
The cumulative distribution function F(x) of a discrete r.v. X with probability distribution p(x) is given by
CUMULATIVE DISTRIBUTION FUNCTION OF C.R.V.:
The cumulative distribution function of a continuous r.v. X is
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18. EXPECTED VALUE OF X FOR D.R.V:
The mean or expected value of the discrete random variable of X, denoted as μ or E(X) is
The variance of X denoted as σ2 or V(X) is
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19. PROBLEMS UNDER DISCRETE RANDOM VARIABLES:
TYPES
GIVEN
TO FIND
TYPE I
P[x]
F[x]
TYPE II
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20. A r. v. X has the following probability functions
PROBLEM 1: [TYPE I]
A r. v. X has the following probability functions X 1 2 3 4 5 6 7
p(x) k 2k 3k k2
2k2
7k2 + k
Find (i) value of k
(ii) P(1.5 < X < 4.5 / X > 2)
(iii) if P(X ≤ a) > ½, find the minimum value of a.
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21. Solution:
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22. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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23. X 1 2 3 4 5 6 7
p(x)
1/10
2/10
3/10
1/100
2/100
17/100
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24. A random Variable X has the following probability distribution. x -2
PROBLEM 2
A random Variable X has the following probability distribution. x -2 -1 1 2 3
p(x)
0.1 k
0.2 2k
0.3 3k
Find
the value of k
Evaluate P[X<2] and P[-2<X<2]
(iii)Find the cumulative Distribution of X.
(iv)Evaluate the Mean and Variance of X.
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25. Solution:
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26. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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27. (iii) The Cumulative Distribution of X:
F(x) = P(X ≤ x) -2
F(-2)=P(X≤-2)=P(X=-2)=0.1=1/10 -1
F(-1)=P(X≤-1)=F(-2)+P(-1)=1/10+k=1/10+1/15=0.17
F(0)=P(X≤0)=F(-1)+P(0)=1/6+2/10=1/6+1/5=0.37 1
F(1)=P(X≤1)=F(0)+P(1)=11/30+2/15=15/30=1/2 2
F(2)=P(X≤2)=F(1)+P(2)=1/2+3/10=(5+3)/10=8/10=4/5 3
F(3)=P(X≤3)=F(2)+P(3)=4/5+3/15=(12+3)/15=1
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28. (iv) Mean:
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29. Variance:
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30. PROBLEM 3 : [TYPE II]
Obtain the probability function or probability distribution function from the following distribution x 1 2 3
F[x]
0.1
0.4
0.9
1.0
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31. Check that the above is a p.d.f. Compute P(X ≤ ½ | ⅓ ≤ X ≤ ⅔)
Problem 3:
The diameter, say X of an electric cable, is assumed to be a continuous r.v. with p.d.f. :
f(x) = 6x(1-x), 0 ≤ x ≤ 1
Check that the above is a p.d.f.
Compute P(X ≤ ½ | ⅓ ≤ X ≤ ⅔)
Determine the number k such that
P(X< k) = P(X > k)
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32. Solution:
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33. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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34. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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35. rth MOMENT ABOUT ORIGIN
The rth moment about origin of a r.v. X is defined as the Expected value of the rth power of the X
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36. MOMENT GENERATING FUNCTION
The moment generating function of a r.v. X (about origin) whose probability function f(x) is given by
t is a real parameter and the integration or summation being extended to the entire range of x.
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37. Moment Generating Function
Problem: Let the random variable X has the p.d.f
Find the Moment Generating Function,
(i) mean,
(ii) variance of X and also
(iii) find P(x>1/2)
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38. Solution:
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39. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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40. (ii) VARIANCE:
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41. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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42. STANDARD DISTRIBUTIONS
DISCRETE DISTRIBUTION
CONTINUOUS DISTRIBUTION
1.Binomial Distribution
1. Uniform Distribution
2. Poisson Distribution
2. Exponential Distribution
3. Geometric Distribution
3. Gamma Distribution
4. Negative Binomial
Distribution
4. Normal Distribution
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43. DISCRETE DISTRIBUTION
BINOMIAL DISTRIBUTION
A random variable X is said to follow Binomial Distribution if it assumes only non- negative values and its probability mass function is given by
P(X=x) =P(x) = ncxpxqn-x, x=0, 1, 2, 3…..n, q=1-p
, otherwise
Where n and p are called parameter of the Binomial distribution.
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44. Mean of Binomial Distribution. Mean = E(X) = np
Variance of Binomial Distribution.
Variance = Var(X) = npq.
Moment Generating Function (M.G.F) of a Binomial Distribution.
M G F = Mx ( t ) = E ( etx ) = ( p et + q )n
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45. (iv) children of both sexes.
PROBLEM
Out of 800 families with 4 children each, how many families would be expected to have
(i) two boys and 2 girls,
(ii) atleast 1 boy
(iii) atmost 2 girls and
(iv) children of both sexes.
Assume equal probabilities for boys and girls.
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46. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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47. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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48. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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49. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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50. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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51. POISSON DISTRIBUTION Where λ is parameter of the Poisson distribution.
A random variable X is said to follow Poison distribution if it assumes only non -negative values and its probability mass function is given by
Where λ is parameter of the Poisson distribution.
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52. Mean of Poisson Distribution. Mean=E(X) = λ
Variance of Poisson Distribution.
Variance=Var(X) = λ
Moment Generating Function (M.G.F) of a Poisson Distribution.
M.G.F = Mx ( t ) = E ( etx ) =
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53. With only one breakdown and With atleast one breakdown.
Problem:
The number of monthly breakdowns of a computer is random variable having a Poisson distribution with mean equal to 1.8. Find the probability that this computer will function for a month
Without a breakdown
With only one breakdown and
With atleast one breakdown.
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54. Let X denotes the number of break downs of the computer in a month.
Solution:
Let X denotes the number of break downs of the computer in a month.
X follows a Poisson distribution with mean λ = 1.8
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55. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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56. GEOMETRIC DISTRIBUTION
Suppose that independent trails, each having a probability p, 0 < p < 1, of being a success, are performed until a success occurs. If we let X equal the number of trails required, then
P(X = x) = (1-p)x-1p = qx-1 p, x=1,2,3…
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57. Mean of Geometric Distribution. Mean =E(X) = 1/p
Variance of Geometric Distribution.
Variance = Var(X) = q/p2
Moment Generating Function (M.G.F) of a Geometric Distribution.
M.G.F = Mx ( t ) = p/e-t-q
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58. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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59. PROBLEM :
If the probability that a target is destroyed on any one shot is What is the probability that it would be destroyed on 6th attempt?
SOLUTION:
Let `X’ be the R.V denoting the number of attempts required for the first success.
Given p = 0.5 ∴ q= 1 – 0.5 = 0.5.
We know that P(X = x) = qx-1p
P(X = 6) = q6-1p = q5p = (0.5)5(0.5) [∵p=0.5, q=0.5, n=6]
= (0.5)6 =
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60. Problem: A die is tossed until 6 appears
Problem: A die is tossed until 6 appears. What is the probability that it must be tossed more than 4 times. Solution: Let X be a Random variable denoting the number of times 6 appears. Here p = 1/6, q = 1- p = 1-1/6 = 5/6 w.k.t. P(X = x) = qx-1p x=1,2,3…..  P(X = x) = (5/6)x-1(1/6)
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61. = 1{P(x = 1) + P( x= 2) + P(x = 3) + P(x = 4)}
Here P(x> 4) = 1-P(x ≤ 4)
= 1{P(x = 1) + P( x= 2) + P(x = 3) + P(x = 4)}
= 1{(1/6) + (5/6)(1/6) + (5/6)2(1/6) (5/6)3(1/6)}
= 1 - 1/6 (3.1065) =
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62. NEGATIVE BINOMIAL DISTRIBUTION
A random variable X is said to follow Negative Binomial Distribution if its probability mass function is given by
(or)
Where p is the probability of success in a trail.
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63. Mean of Negative Binomial Distribution . Mean = E(X) = rP
Variance of Negative Binomial Distribution .
Variance = Var(X) = rPQ
Moment Generating Function (M.G.F) of a Negative Binomial Distribution.
M.G.F = Mx ( t ) = ( Q-P et ) -r
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64. CONTINUOUS DISTRIBUTION
Uniform Distribution
A random variable X is said to have a continuous Uniform Distribution over a interval (a,b) if its probability density function is given by
(i.e.)
Where `a’ and `b’ are the two parameters of the uniform distribution.
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65. Mean of Uniform Distribution
Variance of Uniform Distribution
Moment Generating Function (M.G.F) of a Uniform Distribution
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

66. (a) less than 5 min for a bus and (b) at least 12 min for a bus.
PROBLEM :
Buses arrive at a specified stop at 15 min. interval starting at 7 AM, That is, they arrive at 7, 7.15, 7.30, 7.45 and so on. If a passenger arrives at the stop at random time that is uniformly distributed between 7 and 7.30 A.M. find the probability that he waits.
(a) less than 5 min for a bus and
(b) at least 12 min for a bus.
Solution:
Let X denotes the time that a passenger arrives between 7 and 7.30 a.m. Then X∼U(0,30)
Then
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

67. (a) Passenger waits less than 5 minutes,
(i.e.) he arrives between or
P (waiting time less than 5minutes)
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

68. (b) Passenger waits at least 12 minutes
(i.e.) he arrives between or
P (waiting time at least 12 minutes)
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

69. THE EXPONENTIAL DISTRIBUTION
A continuous random variable X is said to follow an Exponential Distribution with parameter λ > 0 if its probability density function is given by
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

70. Mean of Exponential Distribution
Variance of Exponential Distribution
Moment Generating Function (M.G.F) of Exponential Distribution
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

71. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

72. PROBLEM
The time (in Hours) required to repair a machine is exponentially distributed with parameter λ = ½. What is the probability that the repair time exceeds 2 h? What is the conditional probability that a repair takes at least 10 h given that its duration exceeds 9h?
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

73. Solution:
The conditional probability that a repair takes at least 10 h given that its duration exceeds 9h is given by,
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

74. THE GAMMA DISTRIBUTION
A continuous random variable X is said to follow an Gamma Distribution with parameter λ > 0 if its probability function is given by
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

75. Mean of Gamma Distribution
Variance of Gamma Distribution
Moment Generating Function (M.G.F) of Gamma Distribution
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

76. Problem:
Find the p.d.f of Gamma distribution Find the M.G.F, Mean and variance. Solution: A continuous random variable X taking non-negative values is said to follow gamma distribution, if its probability density function is given by Where α is the parameter of the distribution.
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

77. Moment generating function
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

78. Mean and variance of Gamma Distribution
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

79. Normal Distribution
A random variable X is said to have a normal distribution with parameters μ and σ2, if its p.d.f is given by the probability law:
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

80. Mean of Normal Distribution
Variance of Normal Distribution
Moment Generating Function (M.G.F) of Gamma Distribution
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

81. NORMAL DISTRIBUTION CURVE
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

82. X is a normal variate with mean 30 and S.D. 5. Find P(26 ≤ X ≤ 40)
Problem:
X is a normal variate with mean 30 and S.D. 5. Find
P(26 ≤ X ≤ 40)
P(X ≥ 45)
P(|X -30| > 5)
Solution:
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

83. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
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84. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2

85. IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M. E
IFETCE/H&S- II/MATHS/MATHIVADHANA/IYEAR/ M.E.(CSE)/I-SEM/MA7155/APPLIED PROBABILITY AND STATISTICS /UNIT–I/PPT/VER1.2