Journal of Applied Probability
31
, 262267, 1994
A Generalization of “Expectation Equals Reciprocal of
Intensity” to Nonstationary Exponential Distributions
Eugene A. Feinberg
SUNY at Stony Brook
June 1992
Revised: November 1992 and January 1993
Abstract
An observer watches one of a set of Poisson streams. He may switch from one stream to
another instantaneously. If an arrival occurs in a stream while the observer is watching
another stream, he does not see the arrival. The experiment terminates when the observer
sees an arrival. We derive a formula which states essentially that the expected total time
that the observer watches a stream is equal to the probability that he sees the arrival in
this stream divided by the intensity of the stream. This formula is valid independently of
the observation policy. We also discuss applications of this formula.
Keywords:
Nonstationary exponential distribution, Poisson arrivals, intensity.
AMS (MOS) subject classification:
60E05, 90B25, 90B30, 90C40.
1
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1.
Introduction.
If
X
is a random variable having an exponential distribution with
parameter
λ,
then
EX
=
λ

1
.
In this note we describe a generalization of this formula to
nonstationary exponential random variables.
Let us consider the following example. There are two Poisson streams with positive
arrival intensities
λ
i
, i
= 1
,
2
.
At any epoch
t
∈
R
+
= [0
,
∞
)
,
an observer watches one of
the streams and may switch from one to another instantaneously. If an arrival occurs in
stream
i
while he is watching stream
j, j
6
=
i,
then the observer does not see the arrival.
Such models arise in multiarmed bandit problems in continuous time; e.g., Presman and
Sonin (1983, 1990).
Let
X
be the time before the observer sees the first arrival. Let
X
i
, i
= 1
,
2
,
be the
amount of time during [0
, X
] when the observer watches stream
i,
so that
X
1
+
X
2
=
X.
He may see this arrival either while watching stream 1 or while watching stream 2. Let
p
i
, i
= 1
,
2
,
be the probability that the observer sees the first arrival while he is watching
stream
i, p
1
+
p
2
= 1
.
If
X
2
= 0 a.s., the observer watches stream 1 all the time.
Then
p
2
= 0
, p
1
= 1
,
and
X
1
is an exponential random variable.
So, in this case,
EX
1
=
λ

1
1
.
Theorem 1
below implies that
EX
i
=
p
i
/λ
i
, i
= 1
,
2
,
independently of the rule according to which
the observer watches the streams. If there is a finite or countable number of streams, for
mula
EX
i
=
p
i
/λ
i
remains valid. Theorem 1 below gives such a result for the case where
the set of streams may be an arbitrary measurable space.
In addition to multiarmed
bandit problems, possible applications of Theorem 1 include continuoustime Markov de
cision processes, the analysis and optimization of production systems, and discrete event
simulation.
2.
Results.
Let (
A,
A
) be a measurable space,
λ
be a measurable realvalued function
defined on
A
such that 0
≤
λ
(
a
)
<
∞
,
and
φ
: [0
,
∞
)
→
A
be a measurable mapping.
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 Fall '11
 EugeneA.Feinberg
 Probability, Probability theory, Exponential distribution, Lebesgue integration, Pπ

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