Probability and Markov Chains

Question

Suppose a family of worms initially contains two worms, a mommy worm and a dady worm.. In any given month, the worms can either have no children, have n new children, or lose m children. Suppose they have no new children 10% of the time, and the probability of them having each new child is the same as them losing a child ( so for instance the probability of them having 2 baby worms is the same a the probability of them losing 2 baby worms). Finally, suppose the probability of having n baby worms is one half the probability of having n-1 baby worms. For simplicity assume either worms are killed or nothing happens or new worms are born. Compute the transition probability p_2,10 (the probability of going from 2 to 10 worms in the family in one month).

Answer 1

We will compute the 1-stage transition probability of the worm family increasing in size from 2 to 10 worms in one month.

The Markov chain under the discussion is one with a countably infinite state space, and this state space is the set of all possible values of the size of the worm family, namely, the set of non-negative integers, {0, 1, 2, 3, 4, ...}. This makes sense because the size of the worm family is initially equal to 2, but it's statistically possible for it to be equal to any non-negative integer the following month.

To compute the desired 1-stage transition probability P_2,10 we need to consider all the 1-stage transition probabilities P_{2, i} where i is any non-negative integer. If there are 2 worms in the family at any particular month, then the probability that any of the parents die by the following month is P_2,0 + P_2,1. The probability that both parents survive but they have no children by the following month is P_2,2, so the probability that no child worms are born by the following month is P_2,0 + P_2,1 + P_2,2 = 0.10.

Next, recall that it was assumed that the transition probability of n ≥ 1 child worms being added to the family at the end of the month is 0.5 the probability of only n-1 child worms being added. Thus, P_2,3 = (0.5)P_2,4 = (0.5)²P_2,5 = (0.5)³P_2,6 = ..., etc.

Therefore, 1 = ∑_i=0^∞P_2,i = 0.1 + ∑_i=3^∞P_2,i = 0.1 + P_2,3∑_i=0^∞(0.5)ⁱ = 0.1 + 2P_2,3, and so P_2,3 = 0.45.

Finally, P_2,10 = (0.5)⁷P_2,3 = (0.5)⁷(0.45) ≈ 0.00351562.

Anyway, if my assumptions about the problem were wrong or you wish to discuss this problem further feel free to comment on this answer.