The expectation of Yule process

A type of bacteria cell divides at a constant rate $\lambda$ over time (that is, the probability that a cell divides in a small interval of time $t$ is approximately $\lambda t$).

Given that a population starts out at time zero with $k$ cells of this bacteria and that cell divisions are independent of one another, the size of the population at time $t$, say $X(t)$, has the probability distribution

$\displaystyle P(X(t) = n) = \binom{n - 1}{k - 1} e^{-\lambda kt} (1 - e^{-\lambda t})^{n-k}, \quad n = k, k+1, \ldots$

It is called the Yule process.

The expected value of the population at time $t$, $E(X(t))$, is given by:

$\displaystyle E(X(t)) = \sum_{n=k}^{\infty} n \binom{n-1}{k-1} e^{-\lambda kt} (1 - e^{-\lambda t})^{n-k}$

Simplifying, we can write:

$\displaystyle = k e^{-\lambda kt} \sum_{n=k}^{\infty} \frac{n!}{(n-k)!k!} (1 - e^{-\lambda t})^{n-k}$

$\displaystyle = k e^{-\lambda kt} \sum_{i=0}^{\infty} \binom{k+i}{i} (1 - e^{-\lambda t})^i$

Using $\displaystyle \sum_{i=0}^{\infty} \binom{i+k}{i} x^i = \frac{1}{(1-x)^{k+1}}$ when $\left| x \right| \lt 1$, this further simplifies to:

$\displaystyle= k e^{-\lambda kt} \frac{1}{(1 - (1 - e^{-\lambda t}))^{k+1}}$

Finally, simplifying the expression we get:

$\displaystyle = k e^{-\lambda kt} e^{(k+1)\lambda t} = k e^{\lambda t}$

Thus, the expected number of bacteria at time $t$ is $k e^{\lambda t}$.