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In this segment, we go through
another random incidence

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example, one that does not
involve the Poisson process,

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but a much simpler
arrival process.

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We do this for two reasons.

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One is because of the
simplicity of the example,

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perhaps the intuition will
be a little more transparent.

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And the second reason
is to illustrate

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that we're dealing with
a phenomenon that's

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not specific to the
Poisson process,

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but is much more general.

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The example is as follows.

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We have an arrival
process, in which

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arrivals happen at random.

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And the consecutive
interarrival times

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are independent, identically
distributed, random variables.

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However, unlike the
Poisson process,

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these interarrival times are not
exponential random variables.

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But instead, we assume
that they are either

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5 or 10 minutes with
equal probability.

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So we have an arrival.

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The next arrival may
happen five minutes later.

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The next arrival may again
happen five minutes later.

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Then maybe we get an
interarrival interval of length

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10, then maybe another
interarrival interval of length

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10, followed by one
of five, and so on.

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What is the expected value
of the kth interarrival time?

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Well, an interarrival time is
with probability 1/2 of length

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five and with probability
1/2 of length 10.

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So the average
interarrival time is 7.5.

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Now, you show up
at a random time.

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And by random we
mean a time that's

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completely uncoordinated
with the arrival process.

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You show up at
some point in time,

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and you look at the interarrival
interval in which you fall.

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And you're interested
in the length

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of that particular
interarrival interval.

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What is the probability that
you fall inside a five minute

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interarrival interval?

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Since intervals of
length five are as likely

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as intervals of length
10, in the long run,

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there's going to be roughly
as many five minute intervals

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as there will be 10
minute intervals.

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On the other hand, the
10 minute intervals

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occupy twice as much
space on the real line.

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And for this reason, it
is 2 times more likely

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that you will fall in a
10 minute interval rather

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than a five minute interval.

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In other words, the probability
of falling in a five minute

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interval is going to be 1/3.

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Whereas the probability of
following in a 10 minute

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interval is going to be 2/3.

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For this reason,
the expected length

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of the interarrival
interval that you

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get to see when you arrive is
equal to, with probability 1/3,

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you see a five.

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And with probability
2/3, you see a 10.

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And this number evaluates
approximately to 8.3,

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which is indeed larger than 7.5.

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The conclusion from this
example is similar to the one

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that we had for the
Poisson process.

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Although the average
interarrival time

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is 7.5, when you show
up at a random time

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you are more likely to
fall in a larger interval.

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And for that reason,
on the average,

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you will be seeing longer
interarrival intervals.

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The calculations that we carried
through in that simple example

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can be generalized for
more general arrival

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processes, called
renewal processes.

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In a renewal process, once more
the consecutive interarrival

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times are independent,
identically distributed,

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random variables.

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But they have a
general distribution.

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For this case, there's
a formula again

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that tells us the
length or the expected

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length of the interarrival
interval that you get to see.

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But the main message
from this example

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is that it does make a
difference how you sample,

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how you choose what to
watch and what to measure.

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It makes a difference
whether you

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decide to measure the kth
interarrival time and it's

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average value or to
decide to measure

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an interarrival
time that's chosen

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by showing up at a random time.

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The two methods of sampling
give you different results.

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And we will see
next a few examples

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that have this
particular flavor.