We now follow a program that parallels our development
for the case of the Bernoulli process.
We will study the time until the first arrival,
a random variable that we denote by T1.
We're interested in finding the probability
distribution of this random variable.
And later on, we will continue and try
to study the time until the kth arrival.
Now T1 is a continuous random variable,
because the Poisson process runs in continuous time.
And therefore, it has a PDF.
But instead of finding the PDF directly,
we will first find the CDF of this random variable.
So we fix a certain time, T. And we're
asking for the probability that the first arrival happens
during this interval.
Now this is 1 minus the probability
that the first arrival happens outside this interval.
So we can write this probability as 1 minus the probability
that T1 is bigger than t.
But what is this event?
The first arrival occurring after time, little t,
is the same as saying that there were no arrivals in the time
interval from 0 to little t.
And this probability of 0 arrivals
in a time interval of length t is
something for which we already have a formula.
Take this formula and replace k by 0, tau by t.
When k is equal to 0, this term is
something to the 0-th power equal to 1.
Using our convention, that 0 factorial is equal to 1,
we're left just with e to the minus lambda t.
And this is the answer for the CDF of the time
until the first arrival.
We then take the derivative.
And we find that the PDF of the time until the first arrival
has this form, which is the PDF of an exponential random
variable.
Of course, this calculation here is
only valid for t's that are non-negative.
For negative t's, the PDF of T1 is, of course, 0.
For the exponential random variable,
we have seen that it has certain memorylessness properties.
Namely, if I condition on an event
that nothing has occurred until a certain time,
t, and I am interested in the time from now
until the first arrival occurs, this remaining
until the first arrival is again an exponential distribution.
That is, looking ahead from this time,
I will still wait an exponentially distributed
amount of time until I see the first arrival.
Whatever happened in the past and how long
I have been waiting doesn't matter.
Starting from this time, I will still
wait an exponentially distributed amount of time.
This is essentially an expression
of a fresh start property of the Poisson process, which
is analogous to the fresh start properties for the Bernoulli
process.
And we will be discussing this fresh start
property a lot more.
Having figured out the distribution
of the time of the first arrival,
let us now study the time of the k-th arrival, a random variable
that we denote by Y sub k, similar to the case
of the Bernoulli process.
This random variable is a continuous one,
because arrivals happen in continuous time,
so it takes continuous values.
And therefore, it will be described by a PDF.
And this is what we want to find.
In order to find it, we will make use of the Poisson PMF
that we have already derived for the number of arrivals
during an interval of a fixed length.
One approach to finding the PDF of Yk
is the usual program, similar again
to what we did for the case of the first arrival time.
We can first find CDF, and then differentiate to find the PDF.
So what is the CDF?
We want to calculate the probability
that Yk is less than or equal to some number, little y.
Now what is this event?
The k-th arrival occurs by time y.
This means that by time y, we've had at least k arrivals.
We've had k arrivals, or maybe k plus 1, or maybe k plus 2.
We've had some number of arrivals, n,
in an interval of length, y.
And this is an event that happens with this probability.
But we need to take into account all of the possible values of n
that are at least as large as k.
Now we have a formula for this probability, the probability
of n arrivals in an interval of given length.
This is the Poisson PMF with appropriate changes of symbols.
So we can take this expression, substitute it here, and then
differentiate to do some algebra and find the answer.
Instead of carrying out this algebra,
however, we will proceed in a more intuitive way that
will get us there perhaps faster.
And the derivation that we would follow actually
parallels the one that we went through
in the case of the Bernoulli process.
The intuitive argument that we will use
will rest on the interpretation of a PDF
in terms of probabilities of small intervals.
So the PDF evaluated at some particular point,
y, times delta, is approximately the probability
that our random variable falls within a delta interval
from this number, little y, that we're considering.
So here's time 0, here's time y, and here's time y plus delta.
We want to find or to say something
about the probability of falling inside this small interval.
Now what does it mean for the k-th arrival
to fall inside this interval?
This is an event that can happen as follows.
The k-th arrival falls in this interval,
and we've had k minus 1 arrivals during the previous interval.
What is the probability of this event?
A basic assumption about the Poisson process
is the independence assumption.
Therefore, having k minus 1 arrivals in this interval
and having one arrival in this interval
are independent events.
Therefore, the probability of this scenario
is the product of the probabilities
that we've had k minus 1 arrivals in an interval
of length, y, times the probability
that we've had one arrival in an interval of length delta.
And that latter probability is approximately
equal to lambda times delta.
So I should write here an approximate equality instead
of an exact equality, to indicate
that there are other terms, order of delta squared,
for example, but which are much smaller compared to the delta.
However, this is not the only way
that we can get the k-th arrival in this interval.
There's an alternative scenario.
We might have had k minus 2 arrivals during this interval,
and then two arrivals during that little interval.
In this case, the k-th arrival again
occurs within that little interval.
So we need to also calculate the probability of this scenario.
The probability of that scenario is
the probability of k minus 2 arrivals in an interval
of length, y, times the probability of two arrivals.
But the probability of two arrivals
is something that's order of delta squared.
And order of delta squared is much smaller
than this term, which is linear in delta.
And so this term can be ignored as long as we're just
keeping track of the dominant terms,
those are linear in delta.
And then, they would be similar expressions.
For example, the scenario that we have three arrivals up
to time y, and then three more arrivals
during that little interval, which is again
an event of probability, order of delta
squared, that we get three arrivals there.
And all of those terms are insignificant,
and we can ignore them.
And we end up with an approximate equality
between this term and this expression here.
Delta shows up on both sides, so we can cancel delta.
And therefore, we have ended up with a formula for the PDF.
In particular, the PDF is equal to this probability times
lambda.
What is this probability?
We have a formula for it.
But we just need to substitute.
Put k minus 1 in the place of k, and put y in the place of tau.
This gives us lambda y to the power k minus 1, e
to the minus lambda y, divided by k minus 1 factorial.
And then we have the extra factor
of lambda, which can be put together with this lambda
to the k minus 1 here.
And we end up with this final formula for the PDF of Yk.
The distribution that we have here
is called an Erlang distribution.
But actually, it's not just one distribution.
We have different distributions depending
on what k we're considering.
The distribution of the time of the third arrival
is different from the distribution
of the 10th arrival.
So if we fix a particular k, then
we say that we have an Erlang distribution of order k.
For the case where k is equal to 1,
this term here disappears, k minus 1 is equal to 0.
And the denominator term disappears,
and we end up with lambda times e to the minus lambda y.
But this is the exponential distribution
that we had already derived with a different method earlier.
As you increase k, of course, you
get different distributions.
And these tend to shift towards the right.
This makes sense.
The time of the second arrival is
likely to take certain values.
The time of the third arrival is likely to take
values that are higher.
And the more you increase k, the more the distribution
will be shifting towards the right.