This is the first of two videos on the topic of birth processes, which corresponds to Chapter
3 in the lecture notes.
So before we turn to our full generalisation of Poisson processes, which will be continuous
time Markov chains, in the next chapter.
We're going to first generalise one aspect of the Poisson process.
So suppose we want to model a growing population in continuous time.
So if new individuals are born or arrive at a constant rate, then we could just use a
Poisson process.
If, on the other hand, the birth rate depends on the number of individuals who are present
already in the population,
then the Poisson process is not going to be a good model. And that motivates the following
definition.
So let's fix a sequence of parameters lambda_n for n in the naturals.
So this is going to be a sequence such that every element is finite and is greater than
or equal to zero for each n greater than or equal to zero.
We got to fix a starting state somewhere in the naturals,
and we're going to let Z_1, Z_2 and so on... be independent random variables which are
exponentially distributed,
with Z_n being exponential with the shifted parameter,
so lambda_k+n-1 for all n greater than or equal to 1.
So then we define the process X_t by the following:
We're going to start in state k and then we're going to add on to that the number of values
n
such that this sum of exponential random variables with different parameters is less than or
equal to t
But note that these exponential random variables are independent of one another.
So a process which is constructed in this way is called a simple birth process started
from the state k.
So let's think a little about what's going on here.
So X is a counting process, which when it first reaches the state n, waits an length
of time which is distributed is exponential lambda_n, and then jumps to n+1.
So here the parameter lambda_n is referring to the fact that we're in state n, rather
than this is the nth jump that we've made,
because we started in some general state k.
And the word simple here is referring to the fact that no two births are occurring at the
same time.
So it's going to be useful for the rest of our discussion of both processes to have in
our hands the following proposition.
So this is about competing exponentials.
So let's let E_2, E_2, E_n and so on be IID this time with the same parameter,
so they're IID exponential lambda where we have the same parameter lambda,
thought of as the times until n alarm clocks ring.
So you've got n alarm clocks. Each of them has an exponential lambda time until it rings.
So then M, which is defined to be the minimum of those, so that's the time until the first
of the alarm clocks rings,
has exponential n lambda distribution.
And moreover, if we take K to be the index of the first alarm clock that rings then K
is uniformly distributed on 1 up to n, it's equally likely to be each of the alarm clocks,
and conditionally on it having been alarm clock number k, then the random variables
M and then E_j minus M for the other indices are independent.
So let's interpret this, so E_j minus M is the remaining time until the j-th alarm clock
rings.
OK. And these random variables are independent and each exponential lambda distributed.
So you can see straightforwardly here that this is going to be a consequence of the memoryless
property of the exponential.
So a careful proof of this is a special case of a question on Problem Sheet 1.
So I'm going to leave that to you. But we're going to use this proposition in a moment.
So Section 3.1 is a worked example.
This is the prototypical example of a birth process, which is called the Yule process.
So let's consider a population in which each individual gives birth after an exponential
lambda time, independently and repeatedly.
So individuals don't die. They just keep going forever.
And every exponential lambda length of time, they give birth.
So in particular, if there are n individuals present at a specific time,
then each of them waits an exponential lambda time until it next gives birth.
And so, in particular, if I'm starting with n people, the first birth is going to occur
after an exponential n lambda time,
because this is the minimum of n independent exponentials. So in particular then we have
n plus 1 individuals.
And by the proposition that we just saw, the process just begins over from the beginning
again.
We've got the n minus 1 individuals who didn't reproduce.
Those must be each waiting a further exponential lambda time, by the proposition.
Then there's the individual which did give birth and that gets a new exponential lambda
time until it next gives birth,
and then there's the individual who was just born, who of course, gets a fresh exponential
lambda until it first gives birth.
So in particular, the size of the population in this case is performing a simple birth
process with rates lambda_n equals n lambda.
So when we have n individuals present, we have an exponential n lambda time until the
next birth.
And so, as I said, this is often known as a Yule process.
So let's suppose that we start from a single individual and let's let Y_t be the number
of individuals that are alive at time t greater than zero.
So what I'd like to do now is think about how the mean of this process behaves through
time.
So let's define m of t to be the expectation of Y_t.
So the proposition, which we're going to prove, says this m of t, so the mean population size
at time t,
behaves like E to the lambda t, for t greater than or equal to 0.
So how do we prove this. So this is a very important technique in this course.
So let's firstly define T, which is going to be the first birth time.
So let's let this be the infinum of times greater than or equal to zero, such that,
we're starting starting from Y_0 equals one, so this is the first time that Y is equal
to two.
Okay. And the thing you should notice is that after the first birth has occurred,
we now have two individuals, each of which is behaving like the original individual did.
So we can actually think of the population now as being composed of people descending
from
the first of those individuals and people descended from the second of those individuals.
And you can see that each of those is just a little Yule process in its own right.
So let's split the expectation according to whether T has occurred or not.
OK, so you've got m of t, this is the expectation of Y_t,
so that this I can split up as the expectation of Y_t on the events that the time capital
T occurs before little t,
plus the expectation of Y_t on the event that the first birth has not yet occurred, so capital
T is bigger than t.
Okay, so we can go a little bit further with this.
So let's condition on precisely what the value of T is.
So, we've got an integral between 0 and t,
and let's have the expectation of Y_t given that capital T equals u,
and then we need to integrate over the density of T, which is just an exponential lambda
random variable.
So it has density e to the minus lambda u... du.
And then in the second term. Well if the first birth has not yet occurred then we still have
the single individual that we started with.
And so in particular, this expectation here is just equal to the probability that T is
greater than t,
since the Y_t is always equal to one in the situation.
So here, this is equal to one,
and so in particular, that whole term is equal to the probability that leaves T is bigger
than t.
So that's just the tail probability for an exponential lambda distribution.
So we have that once we've had our first birth, so in this situation here where we know that
the first birth is occurring at time T equals u,
so then the remaining evolution, so we've got time T minus u left to go, and we've just
got to independent Yule processes, as we said,
because we've got now two individuals, each of which are giving rise to their own little
Yule processes.
So in particular, if T is equal to u, then we have that from time u onwards,
this process Y_u+s, for s greater than equal to 0, evolves as the sum of two independent
copies of the original process.
So that tells me that my expectation of Y_t, given that T equals u, and u which is something
that is earlier than little t,
so this is equal to 2 times the expectation of the original yule process but now run for
time t minus u.
But of course, that is just the definition of m of t minus u.
So 2 ... m ... t minus u.
So if we insert that back into this equation that we've already got, so let's call it star,
star is equal to the integral between 0 and t of 2 lots of m times t minus u.
Of course, times the density of our exponential variable.
And then we've got the tail of that exponential number which has just e to the minus lambda
t.
So that's I can get rid of this star and just say that this is equal to m of t.
So this gives us an integral equation for the quantity m of t at different times,
and it's easiest to solve this by converting it into a differential equation.
And that will be rendered easier if we do, first, a change of variables in the integral.
So let's let s be t minus u.
So then changing variable, we get that m of t is equal to...
let's bring some things out the front...
the integral between 0 and t... so I've got now m of s,
and then here I get a lambda e to the minus t, and then an e to the lambda s.
and then I still got my e to the lambda t term.
And things will be facilitated by multiplying this equation through by e to the lambda t.
So I get e to the lambda t... times m of t, is equal to 2 lambda,
the integral between 0 and t of m of s, e to the lambda s, d s... plus 1.
Okay, so let's now differentiate both sides of this equation.
So that's going to give us... I get a lambda e to the lambda t, times m of t,
plus an e to the lambda t times m primed,
and then differentiating the other side I get, obviously, my two lambda...
I get m of t, e to the lambda t, and differentiating the constant 1 gives 0.
And so if I rearrange this, I say that I've got lambda... e to the lambda t... m of t
term which cancels off the right hand side.
I can also cancel the e to the minus lambda t's because they're non-negative
And so this gives that m primed of t is equal to, simply, lambda times m of t.
Okay. So we also have an initial condition here.
We have put the expectation of Y_0 as simply 1, because we will start with a single individual.
So m of 0 is equal to 1.
And together with that initial condition, you recognise the differential equation, which
is simply solved by the exponential function.
And that's the unique solution.
So this implies that m of t is simply e to the lambda t.
That's the end of the proof.
So we have a population which, at least in terms of its mean, is growing exponentially,
and on Problem Sheet 1 there's a question which will show you that this is true not
just in the terms of the mean,
but also in an almost sure sense.
So let's now turn to the Markov property for a birth process.
So in the same way as we saw for the Poisson process there was a Markov property,
this is also true for birth purposes in general and for much the same reason.
So proposition.
So let's let X be a simple birth process with, as before, this sequence of rates lambda_n,
for n greater than or equal to 0.
Started again from some state k, X_0 equals k.
So let's fix a time and a second state i which is greater than or equal to k.
So then given that at time t we're in state i,
the process up until time t and the process from time to onwards are conditionally independent
and the conditional distribution of the process from time t onwards is that of a simple birth
process with the same rates,
but simply now started from i rather than from k.
And the proof of this proposition procedes in exactly the same way as for Poisson processes,
we just replace lambda by lambda_n for the appropriate n.
So I'm not going to go through that again and I'll leave you to fill in the details.