Chapter 4. Discrete Probability
Distributions
Section 4.2. Expected Values of Random
Variables
Jiaping Wang
Department of Mathematical Science
02/06/2013, Monday
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Outline
Expected Values
Variance and Standard Deviation
Some Properties
Tchebysheff’s Theorem
Homework #4
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Part 1. Expected Values
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Long-run Relative Frequency of
Occurrence
Suppose you and your friend are matching balanced coins. Each
of you flips a coin. If the upper faces match, you win $1; if they
do not match, you lose $1. So on average, how much will you
win per game in a long run?
From these flipping coins, you win +1 and -1
(lose money) in half of the time, so the
average is (-1)(1/2)+(1)(1/2) = 0.
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Definition 4.4
Definition 4.4 The expected value of a discrete random
variable X with probability distribution p(x) is given as
(The sum is over all values of x for which p(x)>0)
We sometimes use the notation
E(X)=μ
for this equivalence.
Note: Not all expected values exist, the sum above must converge absolutely,
∑|x|p(x)<∞.
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Theorem 4.1
Now if we change the wining function, g(x)=3x, which means, if coins are
matched, then you win $3, otherwise, you lose $3. So the expected value
becomes
(3)(1/2)+(-3)(1/2)=0
Theorem 4.1 If X is a discrete random variable with
probability p(x) and if g(x) is any real-valued function of
X, then
E(g(x)=∑g(x)p(x).
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Part 2. Variance and Standard
Deviation
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Definitions 4.5 and 4.6
The variance of a random variable X with expected value
μ is given by
V(X)=E[(X- μ)2]
Sometimes we use the notation
σ2 = E[(X- μ)2]
For this equivalence.
The standard deviation is a measure of variation that maintains the original units of
measure.
The standard deviation of a random variable is the square
root of the variance and is given by
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Cont.
Case I:
μ = (-1.0)(1/2)+(1)(1/2)=0;
σ2=(-1)2(1/2)+(1)2(1/2)=1.
Case II:
μ = (-1.5)(1/2)+(1)(1/4)+
(2)(1/4)=0;
σ2=(-1.5)2(1/2)+(1)2(1/4)+
1)2(1/4)=2.375.
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Cont.
Age
Interval
Age
Midpoint
2000
2100
Under 5
3
6.9%
6.3%
μ=
3(0.069)+8(0.073)+…+(90)(0.033)
=36.6,
5-9
8
7.3%
6.2%
10-19
15
14.4%
12.8%
20-29
25
13.3%
12.3%
and σ2 = (3-36.6)2(0.069)+ (836.6)2(0.073) + … + (90-36.6)2(0.033)
= 510.76
30-39
35
15.5%
12.0%
40-49
45
15.3%
11.6%
50-59
55
10.8%
10.8%
Similarly, we can find for 2100,
μ = 42.5 and σ2 = 691.69
60-69
65
7.3%
9.8%
70-79
75
5.9%
8.3%
80 and
over
90
3.3%
9.9%
Consider the percentages as
probabilities, then for 2000
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Example 4.3
A department supervisor is considering purchasing a
photocopy machine. One consideration is how often
the machine will need repairs. Let X denote the number
of repairs during one year. Base on past performance,
the distribution of X is shown as follows
Number of
Repairs, x
0
1
2
3
p(x)
0.2
0.3
0.4
0.1
1. What is the expected number of repairs during a year?
2. What are the variance and standard deviation of the number of repairs during a
year?
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Solution
1. E(X)=∑xp(x) = (0)(0.2)+(1)(0.3)+(2)(0.4)+(3)(0.1)=1.4
2. V(X)= ∑(x-E(X))2p(x)
= (0-1.4) 2(0.2)+(1-1.4) 2(0.3)+(2-1.4) 2(0.4)+(3-1.4) 2(0.1)
= 0.84.
So the standard deviation is 0.92.
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Part 3. Some Properties
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Theorem 4.2 For any random variable X and constants a
and b.
1. E(aX + b) = aE(X) + b
2. V(aX + b) = a2V(X)
Standardized random variable: If X has mean μ and
standard deviation σ, then
Y=(X – μ)/ σ
has E(Y)=0 and V(Y)=1, thus Y can be called the standardized random
variable of X.
Theorem 4.3 If X is a random variable with mean μ, then
V(X)= E(X2) – μ2
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Example 4.4
The department supervisor in Example 4.3 wants to
consider the cost of maintenance before purchasing
the photocopy machine. The cost of maintenance
consists of the expense of a service agreement and the
cost of repairs. The service agreement can be
purchased for $200. With the agreement, the cost of
each repair is $50. Find the mean, variance and
standard deviation of the annual costs of repair for the
photocopy machine.
Solution: Now the annual cost of the maintenance is 50X + 200. Recall that
E(X)=1.4 and V(X)=0.84 from Example 4.3, so we have
E(50X+200)=50E(X)+200=(50)(1.4)+200=270;
V(50X+200)=2500V(X)= (2500)(0.84)=2100, so the standard deviation is
$45.83.
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Example 4.5
Use the results of Theorem 4.3 to compute the variance of
X as given in Example 4.3.
Number of
Repairs, x
0
1
2
3
p(x)
0.2
0.3
0.4
0.1
Solution: E(X)=1.4, E(X2)=∑x2p(x)=02(0.2)+12(0.3)+22(0.4)+32(0.1)=2.8
Then V(X)= E(X2)-E2(X) =2.8-1.56=0.84.
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Part 4. Tchebysheff’s Inequality
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Theorem 4.4
Tchebysheff’s Theorem. Let X be a random variable with
mean μ and standard deviation σ. Then for any positive k,
P(|X – μ|/ σ < k) ≥ 1/k2
Consider Y=(X- μ)/ σ, then E(Y)=0 and V(Y)=1.
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Example 4.6
The daily production of electric motors at a certain
factory averaged 120 with a standard deviation of
10.
1. What can be said about the fraction of days on
which the production level falls between 100 and
140?
2. Find the shortest interval certain to contain at least
90% of the daily production levels.
Solution: The interval from 100 to 140 means k=2 so that |X-μ|<2σ for μ=120
and σ=10. So based on Chebysheff’s inequality, we have 1-1/4=0.75, which
mean at least 75% of all days will have a total production values that falls in
this interval.
TO find k, we set (1-1/k2)=0.9 , then k=3.16, the interval will be from
120-(3.16)(10) to 120+(3.16)(10), or 88.4 to 151.6.
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Example 4.7
The annual cost of maintenance for a certain photocopy
machine has a mean of $270 and a variance of $2100
(see example 4.4). The manager wants to budget
enough for maintenance that he is likely to go over the
budget amount. He is considering budgeting $400 for
maintenance. How often will the maintenance cost
exceed this amount?
Solution: Let X=400, so k=(400-270)/45.8=2.84, then the interval 270(2.84)(45.8) to 270 + (2.84)(45.8) or 140 to 400 contains at least 11/(2.84*2.84) = 0.88, thus at most 0.12 of the probability mass can exceed
$400.
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Example 4.8
Suppose the random variable X has the probability mass
function given in the table
x
-1
0
1
p(x)
1/8
3/4
1/8
Evaluate Tchebysheff’s inequality for k=2.
Solution: Find mean E(X)=0, find V(X)=1/8 + 1/8=0.25, so σ = 0.5, thus
P(|X-E(X)|<2(0.5))=P(|X-E(X)|<1)=P(X=0)=3/4. By Chebysheff’s inequality, P(|XE(X)|<2 σ) ≥1-1/4=0.75.
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Homework #4
Page 100: 4.2
Page 101: 4.8
Page 102: 4.12
Page 116: 4.17
Page 118: 4.26
Page 119: 4.33
Page 120: 4.42
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