x[t+1] = (1-) x[t]

y[t+1] = x[t] + y[t].

These are linear equations that can be described by a transition matrix: {{(1-), 0},{,1}}.

The eigenvalues of this matrix are (1-) and 1 and their associated eigenvectors are {1,-1} and {0,1}.

We can perform the matrix operation: AD^tA.{x[0],y[0]} to find the values of x[t] and y[t]:

x[t] = x[0] (1- )^t

y[t] = x[0] (1-(1- )^t)+y[0]

Since x[0]=1 and y[0]=0, these equal

x[t] = (1- )^t

y[t] = (1-(1- )^t)

NOTE: You could also figure this out on the basis of probability reasoning. For a cell to be non-mutant at cell generation t, it had to not undergo any mutations in any of the t cell divisions, which will happen with probability (1- )^t. 1 minus this quantity is therefore the probability that the cell will be mutant.

When = 10^-7, y[100] is very nearly 10^-5, i.e. there will be very few mosaic cells at a particular gene.

Across all of the genes in the genome with = 10^-2 mutations per cell division, y[100] is about 0.63, i.e. there will be very many mosaic cells that contain changes at some gene in the genome.