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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

joriki
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QAK
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  • I fixed the formatting of the upper bound of the sum; please check whether this is what you intended. – joriki Feb 05 '13 at 12:19
  • Thankyou , yeah i intended the same. – QAK Feb 05 '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 05 '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 05 '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 05 '13 at 12:35
  • I've edited the expression for the probability of extinction to clarify that this is the union of the events $X_n=0$. – joriki Feb 05 '13 at 17:58
  • Note that you could have guessed the answer to be 1 since the average branching is less than one. You can actually prove that $\pi_0=1$ if this is the case and some number in $(0,1)$ if it (the mean) is greater or equal to $1$ – Jean-Sébastien Feb 06 '13 at 06:19
  • @Jean-Sébastien No. Mean 1 implies almost sure extinction (except if P(Z=1)=1). – Did Feb 07 '13 at 12:54
  • @Did yes my bad, still what I said holds taken what you said into account. – Jean-Sébastien Feb 07 '13 at 15:40

1 Answers1

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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 $1$.

joriki
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  • Thanks :), I got it, $\pi_{o}=\sum_{j=0}^{\infty}\pi_{o}^{j}P_{j}$ – QAK Feb 05 '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 07 '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 07 '13 at 13:41
  • Sure. And yet you know what I mean. – Did Feb 07 '13 at 13:52