MHB Solve a Branching Process: Xn & F(s) - Get Help Now!

  • Thread starter Thread starter Tranquillity
  • Start date Start date
  • Tags Tags
    Process
Tranquillity
Messages
50
Reaction score
0
I attach the following question about branching processes. Xn is the size of the nth generation. F(s) is the pgf of Z, the offspring distribution.

Any hints/help on how to proceed with my proof would be greatly appreciated!

Regards
 

Attachments

Physics news on Phys.org
Your attachment shows your attempt which is great because people don't usually help without seeing effort. Might I suggest though trying to retype your attempt in a new post using Latex? You can take a look at http://www.mathhelpboards.com/showthread.php?27-How-to-use-LaTeX-on-this-site to see how to use Latex on MHB.

Jameson
 
The suggestion is good, though this question involves a proof and a lot of notation is involved, I would prefer to get hints and continue uploading my handwritten attempts, to show every time exactly what I think and where I have a problem.

Are you able to help me with the exact question? Thanks!

Regards
 
Hello,

A bit late, but I can answer. Don't condition by X_1, but by X_{n-1}. Then note that S_n=\sum_{i=1}^{X_{n-1}} Z_i^{(n)} to write the expectation of a product : E\left[\prod_{i=1}^{X_{n-1}} s^{Z_i^{(n)}}\bigg|X_{n-1}\right] and finish it off.
 
Hello, thanks for the reply! I have figured it out! I was asked by the exercise to condition on X1. I have finished my proof which conditions on X1. Thanks again :)
 
Well it'd be nice if you shared it with us, for some people may need to have this kind of proof at hand :) (although I personally don't need it (Evilgrin))
 
Moo said:
Well it'd be nice if you shared it with us, for some people may need to have this kind of proof at hand :) (although I personally don't need it (Evilgrin))

Try this :) It conditions on something similar that was so near to what I was trying to do, If you undestand what I am posting you can derive what I am asked to! Any more help I could provide that! Regards!
 
Back
Top