2.
(
Independence
) For disjoint sets
A
1
, A
2
, . . . , A
n
the random vari-
ables
N
(
A
1
)
, N
(
A
2
)
, . . . , N
(
A
n
) are independent.
Quick exercise
12.3 Suppose that the locations of defects in a certain type of
material follow the two-dimensional Poisson process model. For this material
it is known that it contains on average five defects per square meter. What is
the probability that a strip of length 2 meters and width 5 cm will be without
defects?
In Figure 7.4 the locations of the buildings the architect wanted to distribute
over a 100-by-300-m terrain have been generated by a two-dimensional Poisson
process. This has been done in the following way. One can again show that
given the total number of points in a set, these points are uniformly distributed
over the set. This leads to the following procedure: first one generates a value
n
from a Poisson distribution with the appropriate parameter (
λ
times the
area), then one generates
n
times a point uniformly distributed over the 100-
by-300 rectangle.
Actually one can generate a higher-dimensional Poisson process in a way that
is very similar to the natural way this can be done for the one-dimensional
process. Directly from the definition of the one-dimensional process we see
that it can be obtained by consecutively generating points with exponentially
distributed gaps. We will explain a similar procedure for dimension two. For
s >
0, let
M
s
=
N
(
C
s
)
,
where
C
s
is the circular region of radius
s
, centered at the origin. Since
C
s
has area
πs
2
,
M
s
has a Poisson distribution with parameter
λπs
2
. Let
R
i
denote the distance of the
i
th closest point to the origin. This is illustrated
in Figure 12.2.
Note that
R
i
is the analogue of the
i
th arrival time for the one-dimensional
Poisson process: we have in fact that
R
i
≤
s
if and only if
M
s
≥
i.
In particular, with
i
= 1 and
s
=
√
t
,
P
(
R
2
1
≤
t
)
= P
R
1
≤
√
t
= P
(
M
√
t
>
0
)
= 1
−
e
−
λπt
.
In other words:
R
2
1
is
Exp
(
λπ
) distributed. For general
i
, we can similarly
write
P
(
R
2
i
≤
t
)
= P
R
i
≤
√
t
= P
(
M
√
t
≥
i
)
.

12.4 Higher-dimensional Poisson processes
175
.
.
R
1
.
.
R
2
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
.
.
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................
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........................
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Fig. 12.2.
The Poisson process in the plane, with the two circles of the two points
closest to the origin.
So
P
(
R
2
i
≤
t
)
= 1
−
e
−
λπt
i
−
1
j
=0
(
λπt
)
j
j
!
,
which means that
R
2
i
has a
Gam
(
i, λπ
) distribution—as we saw on page 157.
Since gamma distributions arise as sums of independent exponential distribu-
tions, we can also write
R
2
i
=
R
2
i
−
1
+
T
i
,
where the
T
i
are independent
Exp
(
λπ
) random variables (and where
R
0
= 0).
Note that this is quite similar to the one-dimensional case. To simulate the
two-dimensional Poisson process from a sequence
U
1
, U
2
, . . .
of independent
U
(0
,
1) random variables, one can therefore proceed as follows (recall from
Section 6.2 that
−
(1
/λ
) ln(
U
i
) has an
Exp
(
λ
) distribution): for
i
= 1
,
2
, . . .

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