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24) is the average time spent in the area (0, x) and the second term in the same formula is the time spent in the other part (x, 1). 19) correspond to the two different parts in which a path stays, (see Fig. 4). 51 -4Nx log x x -4N(1-x) log (I-x) Fig. 4 Diagram illustrating a sample path in the regions above x and below x. 9). 9 and fey) where now fey) = 0 otherwise. 9) \N(l - x) dy + / /. 9 4Nx dy . 9 T (x 1 O. 1, O. 1, O. 9) = 2 log 9. It is interesting to note that if Xo > 0 and xl < 1, and if the population is large, the time spent in an interval (x O' xl) is independent of the population size N.

10), we must have T(x) because f(x) = 1 for all x. 19). 24) is the average time spent in the area (0, x) and the second term in the same formula is the time spent in the other part (x, 1). 19) correspond to the two different parts in which a path stays, (see Fig. 4). 51 -4Nx log x x -4N(1-x) log (I-x) Fig. 4 Diagram illustrating a sample path in the regions above x and below x. 9). 9 and fey) where now fey) = 0 otherwise. 9) \N(l - x) dy + / /. 9 4Nx dy . 9 T (x 1 O. 1, O. 1, O. 9) = 2 log 9. It is interesting to note that if Xo > 0 and xl < 1, and if the population is large, the time spent in an interval (x O' xl) is independent of the population size N.

41) H(x, 1) = 4Nx(2 - x) . Therefore the heterozygosity associated with this return process is not significantly affected when the boundary approaches 1. 41). is x = The limit- In particular if the initial frequency 1/2N 2 (2 - 2~) z 4 whereas the corresponding value for the case with two absorbing boundaries at x = 0 and x = 1 is 2. It is interesting that the heterozy- gosities in the two situations differ by a factor of 2. 38) it is clear that the quantity which vanishes as y approaches 1 will not be affected by the location of the return point.