1.3.
EXAMPLES OF DISTRIBUTIONS
29
Given
A
∈ B
(
R
) we write
μ
λ
(
A
) =
A
p
λ
(
x
)
dx.
The exponential distribution can be considered as a
continuous time
version
of the geometric distribution. In particular, we see that the distribution does
not have a memory in the sense that for
a, b
≥
0 we have
μ
λ
([
a
+
b,
∞
)

[
a,
∞
)) =
μ
λ
([
b,
∞
))
with the conditional probability on the lefthand side. In words: the proba
bility of a realization larger or equal to
a
+
b
under the condition that one has
already a value larger or equal
a
is the same as having a realization larger or
equal
b
. Indeed, it holds
μ
λ
([
a
+
b,
∞
)

[
a,
∞
))
=
μ
λ
([
a
+
b,
∞
)
∩
[
a,
∞
))
μ
λ
([
a,
∞
))
=
λ
∞
a
+
b
e

λx
dx
λ
∞
a
e

λx
dx
=
e

λ
(
a
+
b
)
e

λa
=
μ
λ
([
b,
∞
))
.
Example 1.3.4
Suppose that the amount of time one spends in a post office
is exponential distributed with
λ
=
1
10
.
(a) What is the probability, that a customer will spend more than 15 min
utes?
(b) What is the probability, that a customer will spend more than 15 min
utes from the beginning in the post office, given that the customer
already spent at least 10 minutes?
The answer for (a) is
μ
λ
([15
,
∞
))
=
e

15
1
10
≈
0
.
220.
For (b) we get
μ
λ
([15
,
∞
)

[10
,
∞
)) =
μ
λ
([5
,
∞
)) =
e

5
1
10
≈
0
.
604.
1.3.7
Poisson’s Theorem
For large
n
and small
p
the Poisson distribution provides a good approxima
tion for the binomial distribution.
Proposition 1.3.5
[Poisson’s Theorem]
Let
λ >
0
,
p
n
∈
(0
,
1)
,
n
=
1
,
2
, ...
, and assume that
np
n
→
λ
as
n
→ ∞
. Then, for all
k
= 0
,
1
, . . .
,
μ
n,p
n
(
{
k
}
)
→
π
λ
(
{
k
}
)
,
n
→ ∞
.