Stochastic Processes – Problem 20: Prove that as , in the sense, and that

Let $\{\xi_{n,i},\, n\geqslant0,\, i\geqslant1\}$ be i.i.d. integer-valued random variables. Define the Galton--Watson branching process $\{Z_{n}: n\geqslant0\}$ by

$$Z_{n}=\sum_{i=1}^{Z_{n-1}}\xi_{n-1,i},\quad Z_{0}=1.$$

Let $\mu = E[\xi_{n,i}] > 1$ and $\mathrm{var}(\xi_{n,i}) = \sigma^{2} > 0$. Prove that as $n\rightarrow\infty$, $\frac{Z_{n}}{\mu^{n}}\rightarrow W$ in the $L^{2}$ sense, and that $E[W]=1$.

(Hint: Consider the $L^{p}$ convergence theorem for $p=2$; one needs to show $\sup_{n}E\left[\left(\frac{Z_{n}}{\mu^{n}}\right)^{2}\right]<\infty$.)

Step 1. Define the natural filtration $$\mathcal F_n:=\sigma\bigl(\xi_{k,i}:0\le k\le n-1,\ i\ge 1\bigr).$$ Set $$X_n:=\frac{Z_n}{\mu^n},\qquad \mu=E[\xi_{n,i}]>1.$$ Since $E[Z_n\mid\mathcal F_{n-1}]=\mu Z_{n-1}$, we obtain $$E[X_n\mid\mathcal F_{n-1}]=\frac{1}{\mu^n}E[Z_n\mid\mathcal F_{n-1}]=\frac{\mu Z_{n-1}}{\mu^n}=X_{n-1}.$$ Therefore $(X_n)$ is a martingale.

Step 2. Compute the second moment recursion. Given $\mathcal F_{n-1}$, write $$Z_n=\sum_{i=1}^{Z_{n-1}}\xi_{n-1,i},$$ with i.i.d. summands of mean $\mu$ and variance $\sigma^2$. Hence $$E[Z_n^2\mid\mathcal F_{n-1}]=\operatorname{Var}(Z_n\mid\mathcal F_{n-1})+\bigl(E[Z_n\mid\mathcal F_{n-1}]\bigr)^2 =\sigma^2 Z_{n-1}+\mu^2 Z_{n-1}^2.$$ Divide by $\mu^{2n}$ and take expectations: $$E[X_n^2]=\frac{\sigma^2}{\mu^{2n}}E[Z_{n-1}]+\frac{\mu^2}{\mu^{2n}}E[Z_{n-1}^2] =\frac{\sigma^2}{\mu^{n+1}}+E[X_{n-1}^2],$$ where we used $E[Z_{n-1}]=\mu^{n-1}$.

Step 3. Sum the recursion from $1$ to $n$: $$E[X_n^2]=E[X_0^2]+\sum_{k=1}^{n}\frac{\sigma^2}{\mu^{k+1}} \le 1+\frac{\sigma^2}{\mu(\mu-1)}<\infty.$$ So $\sup_n E[X_n^2]<\infty$, i.e. $(X_n)$ is bounded in $L^2$.

Step 4. By the martingale convergence theorem in $L^2$, there exists an $L^2$ random variable $W$ such that $$X_n\to W\quad\text{a.s. and in }L^2.$$ Equivalently, $$\frac{Z_n}{\mu^n}\to W\quad\text{in }L^2.$$

Step 5. Since $X_n\to W$ in $L^1$ as well (because $L^2$-convergence implies $L^1$-convergence), we may pass expectations to the limit: $$E[W]=\lim_{n\to\infty}E[X_n].$$ But $(X_n)$ is a martingale with $X_0=1$, so $E[X_n]=E[X_0]=1$ for all $n$. Hence $$E[W]=1.$$

Therefore, for the supercritical Galton--Watson process with finite offspring variance, $$\frac{Z_n}{\mu^n}\xrightarrow[n\to\infty]{L^2}W, \qquad E[W]=1.$$ QED.

QED.