Probability generating functions & branching processes

Question

Looking at a cancer model.

Let $Z(t)$ be the number of cells. The birth and death rate of the cells is $a$ and $b$ respectively.

Then in an interval $[0, \Delta t]$ with $\Delta t$ small and $Z(0) = 1$, then to linear order in $\Delta t$, $Z(t + \Delta t)$ is either:

1. $0$ with probability $b \Delta t$, so a single death event

2. $1$ with probability $(1 - (a+b)\Delta t)$, so no event

3. $2$ with probability $a \Delta t$, so a single birth event.

The probability generating function for $Z(\Delta t) = 0, 1, 2$ is then $1$, $F(x, t)$ and $F(x, t)^2$, assuming the cells are independent.

Question: why are the probability generating functions as he states them to be?

I don't know much about probability generating functions, as I'm doing a course on mathematical biology and trying to find my feet. He defines it as $F(x, t) = \mathbb{E}(x^{Z(t)}) $ but I don't really know how I can use that...

Answer 1

Let $Z(t)$ be the number of cells. The birth and death rate of the cells is $a$ and $b$ respectively. Let $Z(0)=1$ and $\Delta t$ small.

Since $\Delta t$ is small, we assume only one event has time to take place. So, only 1 of 3 things can happen:

1. $Z(\Delta t) = 0$, meaning 1 death occurred. Deaths at a rate of $b$ per unit time, so over a period of $\Delta t$, the probability of death is $b\Delta t$.

2. $Z(\Delta t) = 2$, meaning 1 birth occurred. (Same reasoning).

3. $Z(\Delta t) = 1$, meaning nothing happens. Since probabilities sum to $1$, this must occur with probability $1 - a\Delta t - b\Delta t$.

Another way to see this is to look at the governing differential equation $$ \partial_t P(Z(t)=k)=aP(Z(t)=k-1) + bP(Z(t)=k+1) - (a+b)P(Z(t)=k) $$ So, over a small step $\Delta t$, \begin{align} \frac{P(Z(t+\Delta t)=k) - P(Z(t)=k)}{\Delta t} &\approx aP(Z(t)=k-1) + bP(Z(t)=k+1) - (a+b)P(Z(t)=k) \\[5mm] \therefore P(Z(t+\Delta t)=k) &= a\Delta tP(Z(t)=k-1) + b\Delta tP(Z(t)=k+1) \\ &\;\;\;\;\;- \Delta t(a+b)P(Z(t)=k) + P(Z(t)=k) \\ &= a\Delta t \;p_{k-1}(t) + b\Delta t\; p_{k+1}(t) - (1-a\Delta t - b\Delta t)p_k(t) \end{align} But at $t=0$, we have $Z(0) = 1$, so $p_k(0)$ is $1$ for $k=1$ and zero otherwise, $p_{k+1}(0)$ is $1$ for $k=0$ and $0$ otherwise, and $p_{k-1}(0)$ is $1$ for $k=2$ and $0$ otherwise.

Using indicator functions $\mathcal{I}$: $$ P(Z(\Delta t)=k) = a\Delta t \;\mathcal{I}[k=2] + b\Delta t\; \mathcal{I}[k=0] - (1-a\Delta t - b\Delta t)\;\mathcal{I}[k=1] $$ Notice only $k=0,1,2$ have non-zero probability mass. So we can ignore all other cases.

Anyway, all of this is to say that $$ Z(\Delta t) = \begin{cases} 0 & \text{with prob }\;\; p_0(\Delta t) = b\Delta t \\ 1 & \text{with prob }\;\; p_1(\Delta t) = 1 - a\Delta t - b\Delta t \\ 2 & \text{with prob }\;\; p_2(\Delta t) = a\Delta t \\ \end{cases} $$

Let's figure out the probability generating function (PGF) now. By definition, given a discrete random variable $Z$, the PGF of $Z$ is given by: $$ \mathfrak{G}_Z(\xi) = \mathbb{E}\left[\xi^Z\right] = \sum_{z=0}^\infty P(Z=z) \xi^z $$

What if you have a discrete random variable that depends on time, like our $Z$? Simply compute the PGF at each time point: $$ \mathfrak{G}_{Z(t)}(\xi) = \mathbb{E}\left[\xi^{Z(t)}\right] = \sum_{z_t=0}^\infty P(Z(t)=z_t) \xi^{z_t} $$ Using the notation in your question, then $$ F(x,t) = \mathbb{E}\left[x^{Z(t)}\right] = \mathfrak{G}_{Z(t)}(x) $$ Great, so everything matches up. Let's actually compute the PGF now, but for $Z(\Delta t)$. (I noticed you are mixing up $Z(t),Z(\Delta t),Z(t+\Delta t)$; try to be careful with that).

Now, as derived above, $Z(\Delta t)$ can only take on 3 values: $0,1,2$. So \begin{align} \mathfrak{G}_{Z(\Delta t)}(x) &= \sum_{z_t=0}^2 P(Z(\Delta t)=z_{t}) x^{z_{t}} \\ &= \underbrace{P(Z(\Delta t)=0)}_{b\Delta t} \underbrace{x^{0}}_1 + \underbrace{P(Z(\Delta t)=1)}_{1-a\Delta t-b\Delta t} x^{1} + \underbrace{P(Z(\Delta t)=2)}_{a\Delta t} x^{2} \\ &= b\Delta t + (1-a\Delta t-b\Delta t)x + a\Delta t x^2 \\ &= F(x,\Delta t) \end{align} according to your notation.

Honestly, the sentence

The probability generating function for $Z(Δt)=0,1,2$ is then $1$, $F(x,t)$ and $F(x,t)^2$, assuming the cells are independent.

is somewhat confusing because the PGF is a function of $x$ and $t$, for a given random variable indexed by $t$, NOT for specific values of that random variable. Furthermore, it's not clear how the PGF of $Z(\Delta t)$ relates to $F(x,t)$ because the former quantity is not related to $t$, while the latter depends on it. Perhaps you've written something down wrong.

Hopefully that helps. A nice resource to look at might be this paper.