Probability transitions for branching process


by kai_sikorski
Tags: branching, probability, process, transitions
kai_sikorski
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#1
Feb27-12, 01:30 PM
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Working through a paper that uses this result about branching processes. Can't seem to figure out a way to connect the dots. Anyone have any suggestions?

Let Zt be a branching process where each individual gives birth to a single offspring at rate β > 1 and dies at rate 1. The transition probabilities for Zt can be computed exactly and are given by

[tex]P_{10}(t) = \frac{1 - e^{-(\beta-1)t}}{\beta - e^{-(\beta-1)t}}[/tex]
[tex]P_{1k}(t) = (1 - P_{10}(t))(1 - \eta(t))\eta(t)^{k-1}[/tex]
[tex]\eta(t) = \frac{1 - e^{-(\beta-1)t}}{1 - \frac{1}{\beta}e^{-(\beta-1)t}}[/tex]

I know that P1i satisfy the following system of equations

[tex] P_i'(t) = Q P(t) [/tex]

Where Q is the transition rate matrix (i think that's what it's called) and is given by

[tex] Q = \left(
\begin{array}{ccccc}
0 & 1 & 0 & \ldots & \\
0 & -(1+\beta ) & 2 & \ddots & \vdots \\
0 & \beta & -2(1+\beta ) & 3 & \\
\vdots & \ddots & 2\beta & -3(1+\beta)& \ddots \\
& \ldots & & \ddots & \ddots \\
\end{array}
\right) [/tex]

I have verified the above solution works, but I'm not sure how one would have gotten it. Any ideas?
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kai_sikorski
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#2
Feb27-12, 01:41 PM
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Oh sorry, the IC is pi(0) =δi1
kai_sikorski
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Feb28-12, 03:07 AM
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Oh I figured it out, mostly

Let G be the generating function for Zt

[itex] G[s,t] = p_{10}(t) + p_{11}(t)s + p_{12}(t)s^2 + ... [/itex]

By using the differential equations for the ps, and manipulating some indices you can get a PDE for G

[itex] \partial_t G[s,t]-(1-s(1+\beta))+\beta s^2) \partial_s G[s,t] = 0[/itex]
[itex] G[s,0] = s [/itex]

I can plug this into DSolve and it gives the right answer. However I'm having some trouble doing the method of characteristics on this. For the family of characteristics I get

[itex] s[t] = \frac{e^{t \beta +c}+e^{t+\beta c}}{e^{t \beta +c}+e^{t+\beta c} \beta }[/itex]

But I don't know how to pick c, so that I can set an initial condition on s[t]. Any ideas?


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