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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.

EDIT:
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.2012-01-27
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    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 $n$?2012-01-27
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    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.2012-01-27
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    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...2012-01-27
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    @Sasha, thanks for the edit.2012-01-27
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    Since a red cell always gives birth to two cells and a white cell gives birth to none, the total number of cells at a given time is always twice the number of red cells at the previous time, which means you only need to keep track of the number of red cells. However, the pgf you have is just the pgf for the number of red cells at time $1$. If you want to keep track of all the relevant information in the system, you will need a pgf in two variables, one for the number of red cells and one for time.2012-01-27
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    Which will I need to figure out the probability that the culture dies out?2012-01-27

2 Answers 2

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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.2012-01-28