Population of randomly dividing cell

Question

I was really curious. What if, instead of dividing into two every day, the cell divides into one, two, or three, and the probability of all three cases is $1/3$. Initially, there's one. After n days, what is the average number of cell?

Answer 1

Let $X_0=1$ and, for each $k\geq 1$, take $X_k$ as the number of bacteria on the $k^{\text{th}}$ day.

If $U_1,U_2,U_3\dots$ is a sequece of $\text{iid}$ random variables that possess a discrete uniform distribution on the set $\{1,2,3\}$, then we may express $X_k$ as a random sum: $$X_k=U_1 +\dots+U_{X_{k-1}}$$

From total law of expectation, $$\begin{eqnarray*}\mathbb{E}(X_k) &=& \mathbb{E}\Big(\mathbb{E}(X_k|X_{k-1})\Big) \\ &=& \mathbb{E}\Bigg(\mathbb{E}\Big(U_1 +\dots +U_{X_{k-1}}\Big|X_{k-1}\Big)\Bigg) \\ &=& \mathbb{E}\Big(\mathbb{E}(U_1|X_{k-1})+\dots +\mathbb{E}(U_{X_{k-1}}|X_{k-1})\Big) \\ &=& \mathbb{E}(2X_{k-1}) \\ &=& 2\mathbb{E}(X_{k-1})\\ \end{eqnarray*}$$ Since $\mathbb{E}(X_1)$ clearly equals $2$ we see with recursion that $\mathbb{E}(X_k)=2^k$ for any $k\geq 1$.