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A given cell can replicate at rate $\lambda$ and die at rate $\mu$. Upon replication, the cell divides into 2.

The question asks: how many cells will be produced by this cell before it dies?

My interpretation:

Case 1) I only count the cells which are born directly by this cell, e.g. if it replicates, I don't count the replications of the 2nd cell.

Case 2) I consider the total number of cells generated.

Case 1: With probability $\lambda / \mu$ the cell will replicate before it dies, with probability $(\lambda / \mu)^2$ it will replicate twice and so on using the memoryless property of the exponential. Each time we get an extra cell, hence we have $E(X)=\sum_1^\infty (\lambda /\mu)^i \times i$

Case 2: I don't know how to include all possible divisions of daughter cells. Maybe it's not required?

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    case 2 can get out of hand very very quickly if μ < λ , because at that point it should keep replicating at an exponential rate, which could be modelled, but case 1 should be relatively simple – 2017-02-23

1 Answers 1

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For your case 1, your calculation is not quite correct. The probability of no offspring is $\dfrac{\mu}{\lambda+\mu}$, of exactly one offspring $\dfrac{\lambda\mu}{(\lambda+\mu)^2}$, of two $\dfrac{\lambda^2\mu}{(\lambda+\mu)^3}$ etc. giving the expectation $$\sum_{i=0}^\infty i\dfrac{\lambda^i\mu}{(\lambda+\mu)^{i+1}} = \sum_{i=1}^\infty \dfrac{\lambda^i}{(\lambda+\mu)^{i}} = \dfrac{\lambda}{\mu}$$

An alternative way would be to assume the expectation was $x$ and then using memorylessness to solve $x=\dfrac{\lambda}{\lambda+\mu}(1+x)$ to give $x=\dfrac{\lambda}{\mu}$ again

The latter approach works for case 2 too. Suppose its expectation is $y$. You would then want to use memorylessness to solve $y=\dfrac{\lambda}{\lambda+\mu}(1+2y)$ to give $y=\dfrac{\lambda}{\mu-\lambda}$, at least when $\mu \gt \lambda$