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kai_sikorski
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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?
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?
Last edited: