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