Solution to the Galton-Watson Process

We solve the Galton-Watson branching process model by finding the long-term extinction and persistence probability of a family name, and determine the conditions by which a family name is guaranteed a chance to live indefinitely.

[achillesarmstrong]

wow I saw at least 10 videos about branching process, this is the best one. It helps a lot for those who learn by ourselves like me.

[ruanvieira]

I was struggling to understand the Galton-Watson process and you made it very clearer!
Thank you very much!

[porkbrain]

EDIT: I see you might have pointed this out in another comment already.

Thanks for the explanation. I've got a question about the solution given at the end of the video.

e* = f(e*) = 1/6(2e* - 1)(e* - 1)

Why do we look possible values of e* which make the expression 0, that is we solve as if

0 = 1/6(2e* - 1)(e* - 1)

Shouldn't we be looking for the point which make our polynomial intersects line y = x in interval 0 <= x < 1? That would follow from the graphs you presented with the fixture point at 1 and concave up function in that interval. That the above polynomial is equal to 0 at e*=-0.5 and also the line y=x intersects that polynomial at e*=0.5 seems like a coincidence and it doesn't hold for different p.g.f, such as

f(e*) = 1/6 + 3/6e* + 1/6e* ^2 + 1/6e* ^3

[henryhunt]

Isn't 2e*2 + 3e* + 1 factorised to (2e* + 1)(e* + 1)? Not -1, -1? Where have I gone wrong