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28 Jan. 2011
Juan Fung
1
Review: Functions of discrete random variables
Let’s go over some basic properties of the expectations operator.
Suppose
X
takes values in
{
X
1
,...,X
N
}
, with respective probabilities
{
p
1
,...,p
N
}
,
∑
i
p
i
=
1. Then the expected value of
X
is
E
[
X
] =
N
X
i
=1
p
i
X
i
,
and the variance of
X
is
V ar
[
X
] =
E
(
X

E
[
X
])
2
=
N
X
i
=1
p
i
(
X
i

E
[
X
])
2
.
Let
Y
be another discrete random variable. Then the covariance between
X
and
Y
is
Cov
[
X,Y
] =
E
(
X

E
[
X
])(
Y

E
[
Y
])
.
1. Let
a
k
∈
R
be constants. Use your thorough knowledge from Econ 506 to ﬁnd the
following:
(a)
E
[(
a
1
X
)
2

a
2
1
Y
+
a
3
] =
a
2
1
E
[
X
2
]

a
2
E
[
1
Y
] +
a
3
.
(b)
V ar
[(
a
1
X
)
2

a
2
1
Y
+
a
3
] =
a
4
1
V ar
[
X
2
] +
a
2
2
V ar
[
1
Y
]

2
a
2
1
a
2
Cov
[
X
2
,
1
Y
].
Lesson
: be careful with
nonlinear
transformations!
2. Consider the joint probability distribution of
X
and
Y
given below:
X
5
10
15
20
25
Y
10
8
10
15
12
Prob
1/5
1/5
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 Spring '08
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