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Jan 11, 2012 at 3:03 vote accept sigma_z_1980
Nov 8, 2011 at 16:22 comment added Max Alekseyev It does not matter how you call it. The formulae still hold.
Nov 8, 2011 at 10:06 comment added sigma_z_1980 No, it's not probability. It's a state, size of population.
Nov 8, 2011 at 8:02 comment added Max Alekseyev So $A(n,x)$ is the probability that in the $n$-th iteration the size of population equals $x$. Then $A(1,x) = \delta_{x1}$ (Kronecker's delta). Correspondingly, $\mathcal{A_1}(x,z) = z^{x-1}$.
Nov 7, 2011 at 20:36 comment added sigma_z_1980 OK, it's not very good notation then. The boundary value would be $A(1)=1$, i.e. in the first iteration the size of population is 1. Each iteration the size either increases by 1 w.p. $p(x), x$ being the size of the population in the previous turn, or stays the same w.p. $q(x)$.
Nov 7, 2011 at 5:52 comment added Max Alekseyev "the boundary constraint is A(1,1)" - and what is its value? And what's about $A(1,x)$ for $x$ not equal 1? To express $A(n,x)$ in terms of $A(1,x)$, take $m=n-1$. $P_m(x,z)$ is a polynomial defined via given $p(x)$ and $q(x)$.
Nov 6, 2011 at 22:45 comment added sigma_z_1980 also, where does the expression for $P_{m}(x,z)$ come from? IS this some determinant?
Nov 6, 2011 at 21:39 comment added sigma_z_1980 the boundary constraint is $A(1,1)$, i.e. in the first iteration the size is 1.
Nov 6, 2011 at 11:38 history edited Max Alekseyev CC BY-SA 3.0
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Nov 6, 2011 at 11:32 history undeleted Max Alekseyev
Nov 6, 2011 at 11:32 history edited Max Alekseyev CC BY-SA 3.0
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Nov 6, 2011 at 11:13 history deleted Max Alekseyev
Nov 6, 2011 at 11:01 history answered Max Alekseyev CC BY-SA 3.0