Branching process expectancy given an initial condition

Question

Given a branching process {$X_n$} with offspring probabilities $p_0 = 1/6, p_1 = 1/3,$ and $p_2 = 1/2$, find ${E[X_2|X_0 = 10]}$. I know that $E[X_n|X_{n-1}] = X_{n-1}\mu$. I tried stating that $E[X_2|X_0 = 10] = \mu E[X_1|X_0 = 10] = \mu^2E[X_0|X_0 = 10] = 10\mu^2$, but I don't know if this is correct, and if it is, I don't know how to state it "elegantly".

Answer 1

You can use "iterated expectations" and I assume there is also some memoryless-ness or Markov property in the system. Iterated expectations gives $$E[X_2|X_0=10]=E[E[X_2|X_1, X_0=10]|X_0=10]$$ and the Markov-type property would imply $$E[X_2|X_1, X_0=10]=E[X_2|X_1]$$ so that information about time $0$ is irrelevant given the state at time $1$.

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In general for random variables $W,Z$ we have the iterated expectations formula: $$E[W]=E[E[W|Z]]$$ and if we want to condition on some event $\mathcal{A}$ then this formula is modified to $$E[W|\mathcal{A}] = E[E[W|Z,\mathcal{A}]|\mathcal{A}]$$ In your case \begin{align} \mathcal{A}&=\{X_0=10\}\\ W &= X_2\\ Z &= X_1 \end{align}