Total bacteria over three generations of splitting

Question

You start with one bacterium (generation $0$ ). Each generation, every bacterium independently splits into $0, 1, 2,$ or $3$ bacteria, each outcome equally likely. Count the total number of bacteria that ever exist across generations $0, 1, 2,$ and $3$ (the original plus all descendants through generation $3$ ).

What is the expected total?

Accepted Answer

First find the expected size of each generation with the tower rule, then add the generations up. Step 1, per-bacterium mean. Each bacterium produces, on average, m = \frac{0 + 1 + 2 + 3}{4} = 1.5 bacteria in the next generation. Step 2, expected size of each generation. Let Xt be the size of generation t. Conditioning on the previous generation, E[Xt \mid X{t-1}] = 1.5\,X{t-1}, so E[Xt] = 1.5\,E[X{t-1}]. Starting from X0 = 1: E[X0] = 1, \quad E[X1] = 1.5, \quad E[X2] = 2.25, \quad E[X3] = 3.375. Step 3, add the generations. Expected total is the sum of the expected generation sizes (linearit…

Total bacteria over three generations of splitting

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