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Hi.
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In this problem, we're going
to look at random incidence
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under Erlang arrivals.
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First, let's parse
what that means.
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In a Poisson process, remember,
the time between
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arrivals, or the inter-arrival
time, is distributed as an
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exponential random variable.
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And random incidence for a
Poisson process refers to the
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somewhat surprising result
that when you consider a
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specific time, say, T-star,
then the length of the
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inter-arrival interval that
contains that time T-star is
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not distributed as an
exponential random variable.
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It's actually distributed as an
Erlang random variable of
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order 2 or it's distributed as
a sum of two exponential
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random variables.
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And the reason for
that is that it
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comprises of two parts.
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One is the time since the last
arrival until T-star, which is
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exponentially distributed, and
the time from T-star until the
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next arrival, which is also
exponentially distributed.
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So that brings us to a review
of what Erlang random
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variables are.
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An Erlang random variable of
order k is just the sum of k
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independent and identically
distributed
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exponential random variables.
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So to be more specific, if Ti
is an exponential random
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variable with parameter lambda,
then if you take kiid
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copies of Ti and add them up,
and call that Yk, then Yk is
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an Erlang random variable
of order k.
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And one other fact is that the
mean of Yk, the mean of an
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Erlang random variable of order
k, is just k, the order,
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over lambda, which is the rate
of the underlying exponential
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random variables.
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So as a consequence, if you have
an Erlang random variable
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of order two and that random
variable also has a mean of
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two over lambda, we can
interpret that random variable
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as just being the sum of two
exponential random variables.
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2 iid exponential random
variables, T1 and T2, where
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each one takes exponential
with the rate in lambda.
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OK, so in this problem now,
we're dealing with the random
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incidence not under Poisson
processes, but under something
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else, which we call here an
Erlang process with Erlang
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arrival times.
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So to be more specific, what
we're saying is that, instead
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of inter-arrival time being
exponentially distributed, in
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this process, and inter-arrival
time is actually
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distributed as an Erlang random
variable of order 2
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with mean 2 over lambda.
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So to be explicit, this is no
longer a Poisson process.
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It's some other process because
the inter-arrival
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times are not exponential.
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So let's make use of this fact
that we talked about earlier
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because now we know that the
inter-arrival times of this
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Erlang process are
Erlang order 2
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with mean 2 over lambda.
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But we know that that can just
be re-interpreted as a sum of
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two simple exponentials, each
with parameter lambda.
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So let's just draw another
picture and imagine that for
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each of these arrivals, so say
we have three sample arrivals
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in this Erlang process, we can
fill in, kind of, the gaps
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between these with additional
arrivals.
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And then think of each one of
these times as all being
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exponential with parameter
lambda.
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So this is a valid
interpretation because when we
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connect these, these
inter-arrival times correspond
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to the combination of two
inter-arrival times, which we
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know we can split that into
just two exponentials.
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So each one of these is an
exponential random variable.
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And when you combine them, you
get an Erlang order of 2.
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But the nice thing about this
is that if we look at this
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diagram, it actually is just
exactly a Poisson process with
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a rate lambda because
now, what we're
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dealing with are exactly--
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the inter-arrival times
are now exactly
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exponential random variables.
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And so this is in fact, now,
just a simple Poisson process.
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And we can also just think of
it as we take the Poisson
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process, and take every other
arrival, say, all the
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even-numbered arrivals, and make
those corresponds to be
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arrivals in the Erlang
process.
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OK, so now let's think about
some specific time T-star.
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We want to know what is the
distribution of the length of
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this to be specific
inter-arrival interval that
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T-star is in.
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Well, what we can do is take it
down to the level of this
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Poisson process and look
at it from there.
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Well, we do that because, for
a Poisson process, we know
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about random incidence for
Poisson processes.
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And we know how to deal with
Poisson processes.
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So let's think about this now.
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Well, T-star is here.
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And what we know from random
incidence for a Poisson
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processes is that the length of
this inter-arrival interval
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for the Poisson process,
we know that this is an
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exponential plus
an exponential.
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So combined, this is
Erlang order 2.
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But that only covers
from here to here.
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And what we want is actually
from here to there.
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Well now, we tack on an extra
exponential because we know
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that the inter-arrival times--
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the time between this arrival
and that arrival in the
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Poisson process is just
another exponential.
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And now all of these are in
[INAUDIBLE] time intervals.
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And they're all independent.
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And so the time of this
inter-arrival interval in the
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Erlang process is just going
to be the sum of three
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independent exponentials
within the
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underlying Poisson process.
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And so to answer here is
actually, it's going to be an
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Erlang of order 3.
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Now this is one possible
scenario for
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how this might occur.
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Another scenario is actually
that T-star is somewhere else.
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So let's draw this again.
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And suppose now, in this case,
T-star landed between an even
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numbered arrival in the Poisson
process and an odd
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numbered arrival.
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Now it could also arrive between
an odd numbered and an
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even numbered arrival.
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So it could be that T-star
is actually here.
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Well, but in this case, it's
actually more or less the same
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thing because now what we want
is the length of this entire
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inter-arrival interval, which,
in the Poisson world, we can
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break it down into random
incidence within this
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interval, this inter-arrival
interval, which is two
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exponentials, or an Erlang of
2, plus this interval, which
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is just a standard inter-arrival
time, which is
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another exponential.
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So in this case as well, we
have the sum of three
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independent exponential
random variables.
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And so, in either case, we have
that the inter-arrival
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time in the Erlang process
is an Erlang of order 3.
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And so the final answer is, in
fact, that the inter-arrival
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for random incidence under these
Erlang-type arrivals is
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an Erlang of order 3.
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OK, so in this problem we looked
at the random incidence
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under a different type of an
arrival process, not Poisson,
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but with Erlang random
variables.
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But we used the insight that
Erlang really can be
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re-interpreted as the sum of
independent and identically
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distributed exponential
random variables.
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And exponential random variables
can be viewed as one
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way of interpreting and viewing
a Poisson process.
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And so by going through those
steps, we were able to use
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what we know about random
incidence under Poisson
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processes to help us solve
this problem of random
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incidence its Erlang arrivals.
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So I hope that was helpful.
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And I'll see you next time.
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