Expected number of leaf nodes resulting from branching process

Suppose I have two different branching processes, which can be visualized as a tree.

Process A starts with a single node containing value 1, then at each recursive step, it splits into two nodes. Each node flips a coin, and with probability 1/2 the node's value is set to 0, otherwise it preserves the value of its parent.

Process B is similar, except that instead of two child nodes being created independently, only one node flips a coin, and the second node copies its value from the first.

At recursive step $k$ there are $2^k$ leaf nodes of this tree, and I am trying to figure out the expected number of those nodes which have value 1 for each process.

I figure that for Process A, the expected value should be 1, since the probability that a leaf node still has value 1 is simply $(1/2)^{k}$.

Intuitively, I think the expected value for Process B should be smaller, and in addition the assignment that I am doing right now asks me to show that it is smaller. But I have calculated the expected value via brute force for $k = 2, 3, 4$ and am getting the same result (1).

e.g., when $k = 2$, with probability 1/2 there will be 0 leaf nodes with a 1, and with probability 1/2 there will be 2 leaf nodes with a 1, leading to an expected value of 1. When $k = 3$, with probability 1/2 there will be 0 leaf nodes with a 1, and with probability 1/2 there will be 2 child nodes which branch to produce an expected 1 leaf still containing 1, for a total expected value of 1.

Am I doing something wrong here? Or am I being asked to prove something that is wrong?

• I would expect the two processes to have the same expected value, but process B to have a greater variance for the number of leaf nodes with value $1$ Sep 16, 2015 at 7:12

The expected number of children of each individual is 1 in each case. The expected size of the $n$th generation is $\mu^n$ where $\mu$ is the expected number of children of each individual. Therefore, the expected size is 1 at each generation.
The variance is $n \sigma^2$ where $\sigma^2$ is the variance of the offspring distribution - that is, $(1/2 + 1) - 1 = 1/2$ in the first case, or $2-1$ = 1 in the second case. Therefore the variance is $\frac{1}{2} n$ vs $n$ in the two cases.