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At time 0, a cell culture starts with one red cell. At the end of one minute, the red cell dies and is replaced by 2 red cells with probability $\frac{1}{4}$, with 1 red and 1 white cell with probability $\frac{2}{3}$, and with 2 white cells with probability $\frac{1}{12}$. Each red cell lives for one minute and gives birth to offspring in the same way as the parent cell. Each white cell lives for one minute and dies without reproducing.

I need to find a probability generating function for this. I don't have any problems with finding the pgf; however, I am unsure what the offspring distribution is to be used in the pgf summation. What I am thinking is this:

Let $X$ be the number of off-springs. Then $$ \mathbb{P}(X=0) = \frac{1}{12} \qquad \mathbb{P}(X=1) = \frac{2}{3} \qquad \mathbb{P}(X=2) = \frac{1}{4} $$ I am unsure if this is correct because if we have 2 red cells, then won't the total number of offspring be 4? So don't we need an $X=4$ case?

If I am correct with my distribution, then the pgf will be $\frac{1}{12} + \frac{2}{3} s + \frac{1}{4} s^2$.

Thanks for the help.

I am trying to figure out the probability that the entire culture dies out (using the pgf). I am not exactly sure what random variable I need for that...

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I took the liberty of editing your question to use TeX. Please review my edits to make sure the meaning remains as you has intended. – Sasha Jan 27 '12 at 22:16
When you say "I need to find a probability generating function for this", do you mean the pgf for the population size at epoch $n$, or for the red cells count at epoch $n$? – Sasha Jan 27 '12 at 22:18
You need to clarify what is the "this" that you want to find the pgf of. There are a number of random variables involved here. – Robert Israel Jan 27 '12 at 22:19
I am trying to figure out the probability that the entire culture dies out (using the pgf). I am not exactly sure what random variable I need for that... – icobes Jan 27 '12 at 22:23
@Sasha, thanks for the edit. – icobes Jan 27 '12 at 22:23
up vote 4 down vote accepted

This is a two-type branching process hence bivariate generating functions are a well-adapted tool.

Call $R_n$ and $W_n$ the numbers of red cells and of white cells in generation $n$. Let $u$ denote the generating function of the number of descendants of one red cell, hence, for every $s$ and $t$ in $(0,1)$, $$ u(s,t)=\frac14s^2+\frac23st+\frac1{12}t^2. $$ White cells have no descendants hence, conditioning on $(R_n,W_n)$, $$ \mathrm E(s^{R_{n+1}}t^{W_{n+1}}\mid R_n,W_n)=\prod_{k=1}^{R_n}u(s,t)\prod_{\ell=1}^{W_n}1=u(s,t)^{R_n}, $$ Calling $F_n(s,t)=\mathrm E(s^{R_{n}}t^{W_{n}})$ and $f_n(s)=F_n(s,1)=\mathrm E(s^{R_{n}})$, this yields $$ F_{n+1}(s,t)=\mathrm E(u(s,t)^{R_n})=f_n(u(s,t)), $$ and in particular, $$ f_{n+1}(s)=f_n(v(s)),\qquad\text{where}\qquad v(s)=u(s,1)=\frac14s^2+\frac23s+\frac1{12}. $$ Since $(R_0,W_0)=(1,0)$, the initial condition is $F_0(s,t)=s=f_0(s)$, hence $f_n(s)=v^{(n)}(s)$ and $F_n(s,t)=v^{(n-1)}(u(s,t))$. The iteration of $v$ does not yield easy explicit formulas but the extinction probabilities do not require to fully determine $v^{(n)}$. To wit, the probability that the process is extinct at time $n$ is $$ q_n=\mathrm P(R_n=W_n=0)=F_n(0,0)=f_{n-1}(0)=v^{(n-1)}(0). $$ In particular, $q_n\to q$ when $n\to\infty$, where $q=v(q)$ is the smallest fixed point of the function $v$. This yields $q=\frac13$, as you already guessed.

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Look up "branching process". If $\phi(s)$ is the pgf of the number of red offspring of a single red cell, and $\phi'(0)$ (which is the expected number of red offspring) is greater than $1$, then the probability of the culture dying out is the least positive solution of $\phi(p)=p$.

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Correct, that is what I am going for. I was going to do: $\frac{1}{12} + \frac{2}{3} s + \frac{1}{4} s^2$ = s which yields s = 1/3. However, I do not know if my pgf is correct. – icobes Jan 28 '12 at 0:06

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