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Suppose a tree is growing. It begins with a root, and every day I may have some branches sprout. The number of branches is a random variable, for instance an evenly chosen finite ordinal $i \in \mathbb{N_{n-1}} \cup {0} = N$.

So, on day 1 I may expect to have on average $\frac{n - 1}{2}$ branches, on day 2 there would be $\frac{(n-1)^2}{4}$, and so on, so for day $t$ I will have $\left(\frac{n}{2}\right)^t$ branches. But there is also a slight chance that there will be 0 twigs sprouting from every existing branch at some point, which means the tree has died.

  • Should that skew my computation?
  • The more time passes, the larger the accumulated chance that the tree is dead. But does the fraction of the chances of the tree being dead to the total number of outcomes tend to 0, or 1, or to some specific value?
  • How can I predict the chances that the tree will grow exactly that large?

 

One way I can approach this problem is by considering that the only thing that matters is, how many branches do I have at a given point in time. This will be the sequence $b_t$. Then, the number of ways to get to that state $w_{b, t}$ is proportional to the probability of the tree being at that state at that point in time.

For example, how can the tree be dead on the second day?

  1. Observe that there is only one way for a tree to die at a given day: all of its branches have to sprout nothing. So, $w_{0, 2} = 2$.

  2. And how many total ways there are for a tree to grow for 2 days?

    • At day 0, there is only 1 possibility: $\{1^1\}$.
    • At day 1, there are $n$ possibilities: $\{0^1 \dots (n - 1)^1\}$.

    • At day 2... If there were $i$ branches previously, and each of them sprouts some new twigs, then there is some function $f_i: x \mapsto y$ that determines the number $y$ of ways $x$ twigs may sprout out of those $i$ branches. This is similar to rolling $i$ $n$-sided dice, except the dice are labeled from 0. Then, I may add these functions together to obtain the likelihood of getting a certain number of twigs from any situation.

      $$ \begin{align} f_i(x) &= \sum_{x' \in \mathbb{N}} f_{i - 1}(x - x') f_1(x') \\ f_1(x) &= \begin{cases} 1 \text{ if $x \in N$}\\ 0 \text{ otherwise} \end{cases}\\ \end{align} $$

      This wild device is called discrete convolution and I am not sure if it has a closed form. So, the total probability of having $x$ twigs sprout from $i$ branches is:

      $$ f = \sum_{i \in N} f_i $$

      — And the total number of ways the fate may have evolved is:

      $$ u_2 = \sum_{x \in \mathbb{N}} f(x) $$

So, the tree is dead on day 2 with the likelihood of $\frac{2}{u_2}$.

Ignat Insarov
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  • This is a work in progress, I am actually trying to solve at least for some small $t$ by napkin computing. Feel free to join! – Ignat Insarov Oct 06 '19 at 23:36

1 Answers1

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This is known as a Galton-Watson tree, and you're actually getting close to recreating the classical way of dealing with these. The best tool for this is probability generating functions: For a random variable $X$, its generating function is defined as the function $$\phi_X(z) = \mathbb{E}[z^X] = \sum_{k \geq 0} \mathbb{P}(X = k)z^k.$$

Now, let $Z_k$ be the random variable equal to the number of elements there are at time $n$. Then it's not so bad to show that the generating function $f_k$ of $Z_k$ satisfies $$f_k(z) = f(f_{k-1}(z))$$ where $f(z) = \frac{1}{k+1}(1 + z + \cdots + z^{k}).$

Continuing in this way, it's not so bad to show that the expected value of $Z_k$ is $(n/2)^k$, and the probability that it eventually dies---let's call it $q$---satisfies $$q = f(q)\,.$$

Marcus M
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