Using a series to model an infinite set of consecutive dependent probabilities

• February 18th 2013, 04:48 AM
blaisem
Using a series to model an infinite set of consecutive dependent probabilities
Hello everyone.

I am currently trying to model a simple system in programming using a geometric series. The problem at hand is:

For every one unit of "x" I receive, I have a 30% chance of obtaining a second "x". This second "x" has in turn a 30% chance to spawn a third x, which can further propagate to 4, 5, and 6 to infinity, each with a 30% chance of initiating the subsequent term. If at any point the probability condition is not satisfied, the propagation stops with that term.

In order to determine how many "x" I end up with, I wanted to model the system as the geometric series:

$\sum\limits_{n = 0}^{\infty} {(0.3)^{n}x }$

This converges to 1.428

My questions are:

1. I'm having trouble interpreting the meaning of the converging value. Is it accurate to say the "true" probability of spawning an additional x is 42.8%, rather than 30%, because the series converges to that? If not, what is the significance of this converging value? My qualm here is that the n = 1 term is still dependent on a 30% chance to occur, and not a 42.8% chance, which leads me to my next question.

2. Since each additional term in the series is not arbitrarily present but dependent on a previous term occurring, I don't know if it is correct to model it as a geometric series. Factoring in this probability has given me trouble. In light of this, I have also considered the following:

Assuming I have ten units, I can expand the series into the following terms:

1st term: 10 units
2nd term: 0.3 * 10 units
3rd term: 0.3 * (0.3 *10 units) = ${(0.3)^{2}}*10$

Since the 3rd term is dependent on the probability of the 2nd term occurring, I could rewrite the 3rd term as 3rd'' term:

3rd'' term = 3rd term * probability of 2nd term occurring = $[(0.3)^{2}*10]*0.3$

The corresponding series would be:

x + $\sum\limits_{n = 1}^{\infty} {(0.3)^{n}x\,(0.3)^{n-1} }$ = x + $\sum\limits_{n = 1}^{\infty} {(0.3)^{2n-1}x }$ = $\sum\limits_{n = 0}^{\infty} {(0.3)^{2n}x }$

Using the same derivation for convergance of a geometric series, I got that the above series converges to $\frac{a}{1 - r^{2}} = \frac{1}{1 - 0.3^{2}} = 1.099$

I have no idea what the significance of this convergance would be.

Is this going in the right direction? Thank you for any advice/help!