Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider a branching process $X=\{X_n, n=0,1,\dotsc\}$ where $X_n=\sum\nolimits_{i = 1}^{{X}_{n-1}}{Z_i }$ , $X_0=1$, and let $Z_i$ be such that $P[{Z_i=0]}=1/2$, $P[Z_i=1]=1/4$, $P[Z_i=2]=1/4$.

How to find the probability of extinction $\pi_0=P[\bigcup_n(X_n=0)\mid X_0=1]$?

Thanking you in anticipation

share|improve this question
    
I fixed the formatting of the upper bound of the sum; please check whether this is what you intended. –  joriki Feb 5 '13 at 12:19
    
Thankyou , yeah i intended the same. –  QAK Feb 5 '13 at 12:21
    
I also fixed some other formatting stuff; you may want to take a look at the source for future reference. The braces you'd sprinkled the code with had no effect; they grouped things there was no reason to group. Also, it's generally not a good idea to $\TeX$ only those symbols where you need subscripts and the like and to use normal text otherwise; that creates a jarring mixture of fonts. Usually all mathematical symbols should occur in a $\TeX$ environment. Also note that $\TeX$ treats the vertical bar as a norm bar; to get the right spacing in a context like this you need to use \mid. –  joriki Feb 5 '13 at 12:23
    
What's $U_n$? If this is to be the probability of extinction, you'd need something like an infinite product there? –  joriki Feb 5 '13 at 12:27
    
I meant "if $\pi_0$ is to be the probability of extinction", not $U_n$. Anyway, "probability of extinction" is clear enough, so you might want to remove the unclear formula. –  joriki Feb 5 '13 at 12:35

1 Answer 1

The probability of extinction $\pi_0$ satisfies the recurrence

$$\pi_0=\frac12\cdot1+\frac14\cdot\pi_0+\frac14\cdot\pi_0^2\;.$$

The solutions of $\pi_0^2-3\pi_0+2=0$ are $\pi_0=1$ and $\pi_0=2$. Since $\pi_0=2$ isn't a valid probability, it follows that extinction occurs with probability $1$.

share|improve this answer
    
Thanks :), I got it, $\pi_{o}=\sum_{j=0}^{\infty}\pi_{o}^{j}P_{j}$ –  QAK Feb 5 '13 at 17:15
    
As with every subcritical (mean $\lt1$) or critical (mean $=1$) branching process. Here the mean is $E(Z)=0\cdot\frac12+1\cdot\frac14+2\cdot\frac14=\frac34\lt1$. –  Did Feb 7 '13 at 12:53
    
@Did: I'd consider the process with constant population $1$, which has mean $1$ and extinction probability $0$, as a special case of a branching process. –  joriki Feb 7 '13 at 13:41
    
Sure. And yet you know what I mean. –  Did Feb 7 '13 at 13:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.