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I managed to reduce a problem in my research to the following: Suppose that we have a population of cells which starts with a single cell that has zero mutations, and at each time step each cell divides in two. After each division, one of the daughter cells can gain a mutation with probability $\mu$, and lose a mutation with probability $\beta$, where $\beta < \mu$ (a cell with zero mutations cannot lose any). After a long time, what is the fraction $f_i$ of cells that have $i$ mutations.

I thought that we would have $f_i = (1-\mu-\beta)f_i + \mu f_{i-1} + \beta f_{i+1}$, since the fraction of cells with $i-1$ mutations could mutate with probability $\mu$ to have $i$ mutations, the fraction of cells with $i+1$ mutations could mutate with probability $\beta$ to have $i$ mutations, and the fraction of cells with $i$ mutations would neither lose nor gain any mutations with probability $1-\mu-beta$. This seems to be wrong though, because $f_i = c$ for any constant $c$ satisfies this equation. What is the correct way to solve this problem? Thanks.

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Imagining a population size with two alleles ($a$, $b$) and diploid, the difference equation for the proportion of $a$ alleles, conventionally $p$, with forward ($a \rightarrow b$), $\mu$, and reverse ($b \rightarrow a$), $\nu$, is:

$$p_{t+1} = p_t - \mu p_t + \nu (1 - p_t).$$

To find the long-term behavior, we can imagine a proportion of a population withe allele $a$, $p$, reaching a point where it no longer changes. In other words, when $p_{t+1} = p_t$; we'll call this $p^*$. We can do this by setting $p_t$s and $p_{t+1}$ to $p^*$:

$$p^* = p^* - \mu p^* + \nu (1 - p^*).$$

Algebraically solving for $p^*$, we find that the long-term behavior is that

$$p^* = \frac{\nu}{\mu + \nu}.$$

This is true of small populations, large populations, populations that are shrinking, or populations that are growing. With small populations or changing populations there are sampling effects that trivially deviate from this solution. But in general, this would be the mean expectation.

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