PROBABILITY AND DECISION MAKING
45730
QUIZ 2 (OFFLINE)
Problem 1
(
Expectation
)
.
Consider the following random variables:
The random variable
X
takes the values
{
0
,
1
,
2
}
with equal probabilities.
The random variable
Y
takes the value 9 with probability
1
3
and 12 with probability
2
3
.
The random variable
Z
takes the values
{
2
,
4
,
6
,
8
}
with equal probabilities.
(1.1) What is the expected value of
X
?
Solution.
E
[
X
] =
1
3
·
0 +
1
3
·
1 +
1
3
·
2 = 1.
/
(1.2) What is the expected value of
Y
?
Solution.
E
[
Y
] =
1
3
·
9 +
2
3
·
12 = 11.
/
(1.3) What is the expected value of
Z
?
Solution.
E
[
Z
] =
1
4
·
2 +
1
4
·
4 +
1
4
·
6 +
1
4
·
8 = 5.
/
(1.4) What is the expected value of the random variable
W
= 2
X

Y
+ 4
Z
?
Solution.
By the linearity of expectation we have that
E
[
W
] =
E
[2
X

Y
+ 4
Z
]
Therefore
E
[
W
] = 2
·
E
[
X
]

E
[
Y
] + 4
·
E
[
Z
] = 2

11 + 20 = 11.
/
Problem 2
(
Variance
)
.
Consider the following four random variables. Which of these has the highest
variance?
A: Binomial random variable with parameters
p
=
1
2
, n
= 4.
B: Poisson random variable with mean = 1.
C: Discrete random variable which taking one of the values 0
,
1
,
2
,
3
,
4 with equal probabilities.
D: Discrete random variable which takes value 0 with probability
1
2
and value 4 with probability
1
2
.
(2.1) What is the variance of
A
?
Solution.
V ar
(
A
) = 4
·
1
2
·
1
2
.
/
(2.2) What is the variance of
B
?
Solution.
V ar
(
B
) = 1.
/
(2.3) What is the variance of
C
?
Solution.
V ar
(
C
) =
1
5
·
[2
2
+ 1
2
+ 0
2
+ 1
2
+ 2
2
] = 2.
/
(2.4) What is the variance of
D
?
Solution.
V ar
(
D
) =
1
2
·
[2
2
+ 2
2
] = 4.
/
Problem 3
(
Expected value & Variance
)
.
A vending machine sells snacks at $0
.
90 each. What is
the expected value and variance of daily revenue
Y
from the machine, if
X
, the number of snacks sold
per day, has
E
(
X
) = 230 and
V ar
(
X
) = 40 ?
(3.1) Can we compute the expected value of
Y
? If so, what is its value?
Solution.
We have that the random variable
Y
is given by
Y
= 0
.
90
·
X
. So
E
(
Y
) = 0
.
90
·
E
(
X
) =
0
.
90
·
230 = 207.
/
(3.2) Can we compute the variance of
Y
? If so, what is its value?
1
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QUIZ 2 (OFFLINE)
Solution.
V ar
(
Y
) = 0
.
90
·
0
.
90
·
V ar
(
X
) = 0
.
90
·
0
.
90
·
40 = 32
.
40.
/
(3.3) Let
T
be a random variable with variance
V ar
(
T
) = 20 and the covariance of
X
and
T
is
Cov
(
X, T
) = 30. Can we compute the variance of the random variable
Z
= 3
X

T
? If so,
what is its value?
Solution.
Note that
var
(
a
·
X
+
b
·
T
) =
a
2
·
var
(
X
) +
b
2
·
var
(
T
) + 2
·
a
·
bCov
(
X, T
). In our
case we have
var
(
Z
) =
var
(3
·
X

T
) = 360 + 20

180 = 200.
/
Problem 4
(
Expected Value
)
.
An insurance company issues a policy on a used car under the con
ditions:
In case of total loss it will pay $10000. This event occurs with probability 0.01.
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 Spring '08
 ravi
 Decision Making, Probability theory, insurance company, $100, $0.90, $770

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