Talk:Branching process

Branching processes aren't necessarily Markovian

According to N. G. van Kampen's book "Stochastic Processes in Physics and Chemistry", branching processes "need not be Markovian". This contradicts the introductory sentence of this article. Tim 136.186.1.186 (talk) —Preceding undated comment added 05:13, 8 June 2010 (UTC).

Incomprehensible section

Is it me or is the following completely incomprehensible?

> To gain some intuition for this formulation, one can imagine a walk where the goal is to visit every node, but every time a previously unvisited node is visited, additional nodes are revealed that must also be visited. Let Si represent the number of revealed but unvisited nodes in period i, and let Xi represent the number of new nodes that are revealed when node i is visited. Then in each period, the number of revealed but unvisited nodes equals the number of such nodes in the previous period, plus the new nodes that are revealed when visiting a node, minus the node that is visited. The process ends once all revealed nodes have been visited.

I mean, I can understand the idea of a random walk where new nodes keep appearing, but this doesn't bear any obvious relation either to the concept of a branching process or to the equation that precedes this text. Presumably there is some insight here, but it needs a bit more explanation before it can be useful. Nathaniel Virgo (talk) 00:58, 3 May 2017 (UTC)

"consider a process where each individual either has 0 or 100 children with equal probability. In that case, μ = 50, but probability of ultimate extinction is greater than 0.5, since that's the probability that the first individual has 0 children"

I don't think this makes any sense, why probability of extinction is greater than 0.5 if they have equal probability? --Martinerk0 (talk) 09:49, 3 February 2019 (UTC)

But this is correct. On the first generation, the sole individual may have 0 children; prob. 0.5; or 100 children, and then on the second generation, all the 100 may have 0 children each, prob. 0.5 times 0.5¹⁰⁰ added to the former 0.5; and so on. Boris Tsirelson (talk) 10:47, 3 February 2019 (UTC)