Every hour, a cell will attempt to clone itself. The cell will succeed with probability $p$ and fail and die with probability $1-p$. On a certain petri dish, there are $n$ cells. I am trying to find the expected number of cells $C_i$ after $i$ hours have elapsed using iterated expectation.

I already know that, say if there is $1$ cell at first, the expected number of cells after $1$ hour is $2(p) + 0(1-p) = 2p$. I also know that if there are $m$ cells at first, the expected number of cells after $1$ hour is $\sum_{i=1}^{m}E[C_1] = 2mp$ since whether each cell duplicates or dies is independent of other cells. Does this necessarily mean that the expected number after $2$ hours would be $E[C_2] = E[C_1]2p = 4mp^2$?

From the formula, the law of iterated expectation states that $E[C_i] = E[E[C_i|Y]]$. I'm a bit lost as to what exactly I should be conditioning $C_i$ on.

  • $\begingroup$ What is $Y$? I don't know much about probability but if $c$ is constant then $E(c)=c$ so $E[C_i] = E[E[C_i|Y]] = E[C_i[Y]]$? $\endgroup$ – Aqua Mar 17 '18 at 8:25
  • $\begingroup$ @ChristianF $Y$ is supposed to be the other random variable that $C_i$ is conditioned on. Hence $C_i | Y$. But I'm a bit confused as to what RV I should define $Y$ as. $\endgroup$ – William Mar 17 '18 at 8:31
  • 2
    $\begingroup$ You know that $E(C_{k+1}\mid C_k)=2pC_k$ hence $E(C_{k+1})=2pE(C_k)$ hence $E(C_k)=(2p)^kE(C_0)$, end of story. $\endgroup$ – Did Mar 17 '18 at 8:32

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