Entry
Math: Probability:Distribution: Exponential: Can you derive Exponential distribution L . e^(-L . t)?
Feb 16th, 2005 13:47
Knud van Eeden,
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--- Knud van Eeden --- 17 February 2005 - 00:40 am -------------------
Math: Probability:Distribution: Exponential: Can you derive
Exponential distribution L . e^(-L . t)?
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The Exponential distribution is mainly used to model Poisson processes.
E.g. when calculating a Poisson distribution with rate of change L,
and looking for the distribution of waiting times between successive
changes.
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Waiting time distributions are based on the exponential distribution,
which in turn is derived from the Poisson distribution.
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While the Poisson distribution governs the occurrence of random events
in time, space, volume, ..., the intervals between such events is
controlled by the Exponential distribution.
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Consider a problem such as the arrival of telephone calls.
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The total number of occurrences during any particular unit of time is
controlled by the Poisson distribution with mean L per 1 unit time.
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Here is tried to find the distribution of time intervals between
successive arrivals of calls and find the probability that there is a
time interval of length t
between successive arrivals of calls.
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You divide the time interval t in subintervals of length dt such that
there is a small probability p of the occurrence of a call during this
subinterval dt.
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Poisson:
|----|----|----|----|...|----|----|----|----|
0 1 2 3 4 N-1 N
-> number of intervals
---
---
Each of the N intervals has a probability p of success.
p p p p p p p p
|----|----|----|----|...|----|----|----|----|
0 1 2 3 4 N-1 N
-> number of intervals
---
---
Here q is the probability of no occurrence of a call in that
subinterval dt, where p + q = 1 (or thus in other words together they
cover 100%).
Thus
q = 1 - p (1)
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The subinterval dt is assumed small enough to make the probability of
more than one occurrence of a call in this subinterval dt negligible.
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The probability, denoted by dP, that there are N incremental intervals
between successive arrivals of a call is given by:
dP = (probability of a call not arriving) . -+
(probability of a call not arriving) . |
(probability of a call not arriving) . |--> N intervals
... |
(probability of a call not arriving) . -+
(probability of a call arriving) .
= q . q . q . q . q ... q . p
| |
+---- N intervals ----+
---
That is N intervals with no arrivals, and in the (N+1)th interval there
is an arrival
---
N
So dP = q . p (2)
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figure:
The probability of a call coming in after waiting N intervals
<-interval with length t until call arrives ->
q q q q q q q q p
|----|----|----|----|...|----|-----|----|----|----|
0 1 2 3 4 N N+1
^
|
-> number of intervals |
here a call arrives
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Note:
The right side of this structure (2) is called the 'geometric waiting
time distribution', which is in turn a special case of the 'binomial
waiting time distribution'.
---
---
Note:
This process is the same as throwing a dice, and asking yourself what
the
probability is that you throw it N times until say a 5 occurs.
The probability that this occurs is
(probability of not a 5) . --+
(probability of not a 5) . |
(probability of not a 5) . |--> N throws
... |
(probability of not a 5) . --+
(probability of a 5) .
Thus for example, what is the probability that you throw 7 times a dice
and that no 5 occurs.
Now you know that a 5 occurs on average 1 times in 6 times throwing.
Thus its probability is 1 / 6.
Similarly the probability that no 5 occurs is thus 5 times in 6 times
throwing, or thus 5/6, or thus 1 - 1/6.
Thus if you throw a dice 7 times after each other the probability that
no 5 occurs is thus
5/6 . 5/6 . 5/6 . 5/6 . 5/6 . 5/6 . 5/6
or thus
7
(5/6)
The probability that a 5 occurs after throwing 7 + 1, or
8 times is thus
(5/6 . 5/6 . 5/6 . 5/6 . 5/6 . 5/6 . 5/6) . (1/6)
or thus
7
(5/6) . (1/6)
---
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Putting (1) in (2) gives
N
dP = (1 - p) . p (3)
---
What follows will replace p by L, N and dt, from which a differential
equation follows.
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Now if there are on average L arrivals per 1 unit time, correspondingly
the
(mean number of arrivals in subinterval dt)
follows by linearly scaling down from
(old arrivals) (new arrivals)
--------------- = --------------- (4)
(old interval) (new interval)
or thus here:
L x (5)
- = ---
1 dt
from which follows that
x = dt . L
------
1
Thus the (mean number of arrivals in subinterval dt) equals here
dt . L (6)
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Now using that the mean number of arrivals in any interval equals here
(number of trials) times (probability of success at any trial),
thus if you have only only 1 interval dt, then you can write
(mean number of arrivals in subinterval dt) =
( 1 subinterval ) . ( probability of success at any trial)
= 1 . p (7)
As (6) and (7) say the same, you can equate them:
dt . L = 1 . p (8)
As you want to eliminate p from (3), you express (8) in p:
p = L . dt (9)
As you divided here your time interval in N subintervals of size dt,
you can write
t = N . dt (10)
thus
dt = t / N (11)
Putting (11) in (9) gives
p = L . t (12)
-----
N
Putting (9) and (12) in (3) gives:
N
dP = (1 - L . t) . L . dt (13)
-----
N
---
You know maybe that
N
(another) definition of e is:
N
1 1
(1 + ---) = e (14)
N
lim N -> +infinity
An analog expression is:
N
1 -1
(1 - ---) = e (15)
N
lim N -> +infinity
and similarly:
p N +p
(1 + ---) = e (16)
N
lim N -> +infinity
and similarly:
p N -p
(1 - ---) = e (17)
N
lim N -> +infinity
---
Thus here using (17), if N -> infinity, then in (13) you have that
N
(1 - L . t)
-----
N
tends to
-(L . t)
e (18)
Putting (18) in (13) gives
-(L . t)
dP = L . e . dt (19)
Dividing both sides of (19) by dt gives:
-(L . t)
dP = L . e (20)
--
dt
The righthand side is the Exponential distribution, and is the
probability density function for the interval between successive
events, i.e. it is the probability that the time interval lies between
t and t+dt.
The expectation is, similar to an archimedes lever, where you multiply
the arm by the mass at that distance, given by in general by:
t->infinity
-
|
|
|(independent variable).(dependent variable).d(independent variable)
|
|
-
t=0
thus here the expectation is given by:
t->infinity
-
|
| -(L . t)
| t . L . e . dt
|
|
-
t=0
Integrating by parts gives
--+ t->infinity
-(L . t) -(L . t) |
-t . L . e - e |
------------------------------ |
L |
--+ t=0
Filling in the begin and end points gives:
-(0 . t) -(0 . t)
0 - ( -0 . L . e - e
------------------------------ )
L
=
0 - ( 0 - 1 )
-------------
L
or thus
expectation = 1 (21)
---
L
This result is what you should expect. If L is the expected number of
events in unit time (from the original Poisson distribution), then you
could also expect that the mean interval between events is 1/L.
In other words if L, the average total amount of occurrences or thus
calls in the given interval gets larger and larger, the average waiting
time for an occurrence of a call will get smaller and smaller. Thus
here they are inversely proportional.
---
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To further find the total waiting time between 2 events, you
integrate both sides of (19), giving
- -
| |
| | -(L . t)
| dP = | L . e . dt (22)
| |
| |
- -
or thus the probability P for a waiting time equals:
-
|
| -(L . t)
P = | L . e . dt (22)
|
|
-
to be more precise:
t->infinity
-
|
| -(L . t)
P = | L . e . dt (22)
|
|
-
t=t
end
---
---
For example:
In a telephone call center, incoming telephone calls can be modeled as
a Poisson process with a mean of 25 incoming telephone calls per hour.
What is the probability that there are no telephone calls in an
interval
of 6 minutes?
First convert units.
Now 25 calls per hour
is thus
25 calls per 60 minutes.
Thus L = 25
--
60
The probability that there are no telephone calls in the first
6 minutes can be calculated as the probability of getting
telephone calls between 6 minutes and infinity.
t->infinity
-
|
| -(25 . t)
| --
| 60
|
P = | (25) . e . dt (23)
| --
| 60
-
t=6
working out, by integrating the right side of (23) gives:
--+ t->infinity
-(25 . t) |
-- |
60 |
|
P = e | (24)
--+ t=6
-(25 . 6)
--
60
P = 0 - ( - e ) (25)
-(25 . 6)
--
60
P = e (26)
P = 0.082 (27)
or thus in about 8 times in 100 times you will have to wait
6 minutes until the next call arrives.
---
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For example:
At a bus stop, arriving buses can be modeled as
a Poisson process with a mean of 2 incoming buses per hour.
You arrive at the busstop, just seeing the current bus leave.
What is the probability that there is no bus coming in an interval
of 15 minutes?
First convert units.
Now 2 buses per hour
is thus
2 busses per 60 minutes.
Thus L = 2
--
60
The probability that there are no buses arriving in the first
15 minutes can be calculated as the probability of buses
arriving between 15 minutes and infinity.
t->infinity
-
|
| -(2 . t)
| --
| 60
|
P = | (2 ) . e . dt (28)
| --
| 60
-
t=15
working out, by integrating the right side of (23) gives:
--+ t->infinity
-(2 . t) |
-- |
60 |
|
P = e | (29)
--+ t=15
-(2 . 15)
--
60
P = 0 - ( - e ) (30)
-(2 . 15)
--
60
P = e (31)
P = 0.607... (32)
or thus in about 6 times in 10 times you will have to wait
15 minutes until the next bus arrives.
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Book: see also:
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[book: Montgomery, Douglas, C. / Runger, George C. / Hubele, Norma,
F. - engineering statistics - publisher: John Wiley - ISBN 0-471-17026-
7 - p. 100 'Exponential distribution', 'in a large corporate network,
user log-ons to the system can be modeled as a Poisson process with a
mean of 25 log-ons per hour. What is the probability that there are no
log-ons in an interval of 6 minutes?]
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[book: Hastings, Kevin J. - Probability and statistics - ISBN 0-201-
59278-9 - p. 135 'Exponential density' (suppose that the time interval
between arrivals of buses to a particular stop is a random variable T
(in hours) with the exp(2) distribution. You arrive, panting, to the
stop just after the bus leaves. What is the probability that you will
have to wait at least 15 minutes for the next bus?']
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Internet: see also:
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[Internet: source: Waiting Time Distributions:
http://mathforum.org/library/drmath/view/52162.html]
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Math:Probability:Distribution:Exponential: Can you describe
Exponential distribution L . e^(-L . t)?
http://www.faqts.com/knowledge_base/view.phtml/aid/33099/fid/815
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Math: Probability: Distribution: Link: Overview: Can you give an
overview of links?
http://www.faqts.com/knowledge_base/view.phtml/aid/32917/fid/815
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