3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Alex Robinson Feb 23 '17 at 15:09
1
$\begingroup$

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 $\mu \gt \lambda$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.