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Kind: captions
Language: en
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Now I am going for the stationary increment.
The distribution of N(t-s) depends only on
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the length of the interval t minus s and does
not depends on the value of s. That means,
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during the interval delta t, the one arrival
is going to lambda times delta t, order of
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delta t that will tends to zero as delta t
tends to zero.
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That means, the stationary increment means
if you find out the rate that means you find
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out the average per unit of time, then that
is going to be constant. So this is the assumption
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we have taken it in the car insurance problem,
the average rate per unit day, that is going
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to be constant and that is an assumption we
have taken at going to be a constant throughout
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the year and also the different times of a
day.
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So here also we will get whenever we have
a Poisson process, then the average rate is
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going to be a constant because of the stationary
increment.
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The next property, suppose you have a Poisson
process of a one arrival and you have a Poisson
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process of the other arrival, that means one
type of arrival is a Poisson process with
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the parameter lambda 1 and another type of
arrival who that is also Poisson process with
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the parameter lambda 2. As long as both are
independent, the arrivals are independent,
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then the together superposition.
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That is going to be again Poisson with the
parameter lambda 1 plus lambda 2. You can
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add the parameter. That means for fixed t
that is going to be a Poisson distributed
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random variable with the parameter lambda
1 plus lambda 2 times t. Whenever you have
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a two independent or more than one independent
Poisson process arrival, then the merging
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or the superposition will be again Poisson
process as long as they are mutually independent
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with the parameter is nothing but the sum
of those parameters.
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That is, you
can combine many Poisson process as streams
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into one stream and that is going to be a
Poisson stream with the parameter sum of parameters
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lambda 1 to lambda n. So this is possible,
this is used in many telecommunication application,
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that means, suppose, you have a Poisson arrival
of a packets from the different streams and
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all the streams are mutually dependent, the
arrival are independent.
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Then the total number of packets arriving
into the particular switch or router, whatever
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it is. Then the multiplexed one, that is going
to be always Poisson process, that arrival
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follows a Poisson process with the parameters
are sum of, parameter is nothing but the sum
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of these parameters as long as they are Poisson
as well as independent.
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The next property, decomposition, suppose
if you have a one Poisson stream, you can
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decompose into many Poisson streams with the
sum proportion. So that proportions are the
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p1, p2 and pn’s. So one Poisson stream can
be split into n Poisson streams with the parameter
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lambda times p1, lambda times p2 where each
pi’s are greater than zero. The summation
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of pi’s has to be one.
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That means these are all the probabilities.
With these probabilities you can split one
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Poisson stream into many Poisson streams.
So here I have made a n Poisson streams, that
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means the same arrival is with some probability
p1, it lands up here with some probability
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p1, this put up here, with some probability
pn is put up here. So the split of one Poisson
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stream into n Poisson streams is allowed.
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That means the same example, if you have a
one router and from the router if the arrival
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is splitted into many streams with a probability
p1, it goes to the first stream with the probability
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p2 it goes to the second stream and with the
probability pn it goes to the last stream.
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Then each one is going to be a Poisson process
with the parameter lambda times p1 and lambda
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times p2 and so on, lambda times pn.
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So the split is possible as well as the super
imposition is also possible from the Poisson
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process. So this also has a many more applications
in the telecommunication networks. One type
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of packet arrival can be splitted into n proportions
p1, p2, pn’s and each one is going to be
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a Poisson process.
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Now I am going to give the first example to
illustrate the Poisson process. Consider the
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situation of a waiting for a bus in a bus
stand. Assume that the bus arrivals in minutes
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follow a Poisson process with the parameter
5. With the rate, the parameter here that
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is nothing but the intensity or rate. Suppose
you come to the bus stand at some time, what
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is the average waiting time to get the bus.
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When you land up bus stand, there is a possibility
the bus would have come before some time,
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the time in which the next bus is about to
come, you are going to take that bus and till
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that time, you are going to wait in the bus
stand, that is the waiting time. So the waiting
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time is a random variable. So that is a continuous
random variable. The question is what is the
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average waiting time.
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One can find out the distribution of the waiting
time also. Here the question is what is the
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average waiting time. So then, what I can
do, I can use the Poisson concept here.
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The arrival follows the arrival of a bus follows
a Poisson process. Suppose at some time you
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come to the bus stand and suppose the bus
is going to come at this time, your waiting
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time is this much. So suppose, you make W
is going to be your waiting time, W is going
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to be your waiting time, the question is what
is an average waiting time. Just now I have
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explained the Poisson process as the property,
the inter-arrival times are exponential distribution.
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The inter-arrival times are exponential distribution
and all the times are, all the inter-arrival
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times are independent also. Therefore, this
X1 and this is X2 and this is X3, so X1, X2,
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X3 like that so many, all the inter-arrival
times, that is going to follow exponential
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distribution with the parameter lambda. Since
the waiting time is going to be the remaining
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time of arrival of the third bus.
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So the W, the waiting time is same as a the
remaining or residual time of the third bus
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to come into the bus stand. So X3 is exponential
distribution with the parameter lambda. The
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residual lifetime of X3, suppose I make it
as a notation X3 bar, the residual lifetime,
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residual time of arrival, not life time, residual
arrival time of the third bus coming to the
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bus stand, that is also going to be exponential
distribution.
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This is because of the memoryless property,
the residual time is also, whenever, some
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time is exponentially distributed, some random
variable time is exponentially distributed,
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then the residual time is also going to be
exponentially distributed using the memoryless
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property. Therefore, residual arrival time
of bus to come to the bus stand, that is also
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exponential distribution with the parameter
same lambda.
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So this is same as the W. The waiting time
W is same as a residual time. Therefore, W
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is always going to be exponentially distributed
with the parameter lambda. That means the
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waiting time for the bus to come to the bus
stand to catch. So the W is exponentially
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distributed therefore the question is what
is a average waiting time. So average waiting
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time is nothing but one divided by the parameter.
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So here it says the Poisson process with the
intensity 5, that rate is lambda, that is
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the mean inter-arrival times between the buses
is 5 minutes, that means the mean inter-arrival
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times between the buses is 5 minutes is nothing
but it is exponentially distributed with the
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parameter that is average 5 minutes therefore
that is the same thing. Therefore, that is
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equal to 5 minutes.
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Because the way I have given the clue, the
mean interval between the buses is 5 minutes,
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that means the average of X i’s that is
equal to 5 minutes. So that is as same as
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your waiting time because it is exponential
distribution therefore the residual is also
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exponential distribution, therefore you can
use the same value, therefore the average
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is going to be 5 minutes. So using Poisson
process one can find out the different results
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related to the number of arrivals.