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Chapter 7
Sums of Independent
Random Variables
7.1
Sums of Discrete Random Variables
In this chapter we turn to the important question of determining the distribution of
a sum of independent random variables in terms of the distributions of the individual
constituents. In this section we consider only sums of discrete random variables,
reserving the case of continuous random variables for the next section.
We consider here only random variables whose values are integers. Their distri
bution functions are then deﬁned on these integers. We shall ﬁnd it convenient to
assume here that these distribution functions are deﬁned for
all
integers, by deﬁning
them to be 0 where they are not otherwise deﬁned.
Convolutions
Suppose
X
and
Y
are two independent discrete random variables with distribution
functions
m
1
(
x
) and
m
2
(
x
). Let
Z
=
X
+
Y
. We would like to determine the dis
tribution function
m
3
(
x
)of
Z
. To do this, it is enough to determine the probability
that
Z
takes on the value
z
, where
z
is an arbitrary integer. Suppose that
X
=
k
,
where
k
is some integer. Then
Z
=
z
if and only if
Y
=
z

k
. So the event
Z
=
z
is the union of the pairwise disjoint events
(
X
=
k
) and (
Y
=
z

k
)
,
where
k
runs over the integers. Since these events are pairwise disjoint, we have
P
(
Z
=
z
)=
∞
X
k
=
∞
P
(
X
=
k
)
·
P
(
Y
=
z

k
)
.
Thus, we have found the distribution function of the random variable
Z
. This leads
to the following deﬁnition.
285
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CHAPTER 7. SUMS OF RANDOM VARIABLES
Deﬁnition 7.1
Let
X
and
Y
be two independent integervalued random variables,
with distribution functions
m
1
(
x
) and
m
2
(
x
) respectively. Then the
convolution
of
m
1
(
x
) and
m
2
(
x
) is the distribution function
m
3
=
m
1
*
m
2
given by
m
3
(
j
)=
X
k
m
1
(
k
)
·
m
2
(
j

k
)
,
for
j
=
...,

2
,

1
,
0
,
1
,
2
, ...
. The function
m
3
(
x
) is the distribution function
of the random variable
Z
=
X
+
Y
.
2
It is easy to see that the convolution operation is commutative, and it is straight
forward to show that it is also associative.
Now let
S
n
=
X
1
+
X
2
+
···
+
X
n
be the sum of
n
independent random variables
of an independent trials process with common distribution function
m
deﬁned on
the integers. Then the distribution function of
S
1
is
m
. We can write
S
n
=
S
n

1
+
X
n
.
Thus, since we know the distribution function of
X
n
is
m
, we can ﬁnd the distribu
tion function of
S
n
by induction.
Example 7.1
A die is rolled twice. Let
X
1
and
X
2
be the outcomes, and let
S
2
=
X
1
+
X
2
be the sum of these outcomes. Then
X
1
and
X
2
have the common
distribution function:
m
=
±
123456
1
/
61
/
/
/
/
/
6
¶
.
The distribution function of
S
2
is then the convolution of this distribution with
itself. Thus,
P
(
S
2
=2) =
m
(1)
m
(1)
=
1
6
·
1
6
=
1
36
,
P
(
S
2
=3) =
m
(1)
m
(2) +
m
(2)
m
(1)
=
1
6
·
1
6
+
1
6
·
1
6
=
2
36
,
P
(
S
2
=4) =
m
(1)
m
(3) +
m
(2)
m
(2) +
m
(3)
m
(1)
=
1
6
·
1
6
+
1
6
·
1
6
+
1
6
·
1
6
=
3
36
.
Continuing in this way we would ﬁnd
P
(
S
2
=5
)=4
/
36,
P
(
S
2
=6
)=5
/
36,
P
(
S
2
=7
)=6
/
36,
P
(
S
2
=8
/
36,
P
(
S
2
=9
/
36,
P
(
S
2
=10
)=3
/
36,
P
(
S
2
=11)=2
/
36, and
P
(
S
2
=12)=1
/
36.
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 Spring '10
 Altonji
 Econometrics

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