# 4.3. Application of martingales

For applications of martingales, I would like to cover the case of martingales with bounded increments and the branching process.

## Martingales with bounded increments

Before getting to the topic, I would like to state a very useful theorem when constructing a (sub)martingale.

The uniqueness in the statement is in almost sure sense.

Let $A_n = A_{n+1}+(E(X_n|\mathcal{F}_{n-1})-X_{n-1},$ $A_0=0.$ It is clear that $A_n$ is an increasing predictable sequence. Let $M_n = X_n - A_n$ accordingly, then it is a martingale.

Now suppose $X_n = M_n+A_n = M_n'+A_n'.$ Then $M_n-M_n' = A_n'-A_n \in \mathcal{F}_{n-1}$ and $$\begin{aligned} &M_n-M_n' \\ &= E(M_n-M_n'|\mathcal{F}_{n-1}) \\ &= M_{n-1} - M_{n-1}'. \end{aligned}$$ Thus $M_n - M_n' = A_0'-A_0 = 0$ for all $n$ and the uniqueness follows.

The theorem insists that every submartingales can be decomposed into an increasing sequence and a martingale. The important part is where we constructed $A_n.$ Since $A_0=0,$

\[\begin{aligned} A_n &= \sum_{m=1}^n \left( E(X_m|\mathcal{F}_{m-1})-X_{m-1} \right) \\ &= \sum_{m=1}^n E(X_m-X_{m-1}|\mathcal{F}_{m-1}). \end{aligned}\]This gives us a form of *conditional increment*. In quite a lot of situations constructing a sequence like this leads to a (sub)martingale with bounded increments.

The main theorem of this subsection is a dichotomy that applies to martingales with bounded increments.

Without loss of generality, let $X_0=0.$ For $k>0,$ let $N_k = \inf\{n:X_n\le-k\}$ be a stopping time so that $X_{n\wedge N_k}$ also be a martingale. If $N_k = \infty,$ $X_{n\wedge N_k} = X_n > -k$ for all $n.$ If $N_k < \infty,$ $X_{N_k} \le -k$ and $X_t > -k$ for $t=1,2,\cdots,N_k-1,$ thus $X_{N_k} = X_{N_k-1} + (X_{N_k} - X_{N_k-1}) \ge -k-M.$ Since $X_{n\wedge N_k}+k+m$ is a non-negative martingale, by supermartingale convergence $X_{n\wedge N_k}$ converges a.s.

This implies $X_n$ converges on $\{N_k=\infty\}.$ Since $\liminf_n X_n > -\infty$ implies $X_n \ge -k'$ for all but finite $n$'s, for some $k'$ and so $N_{k'+1}=\infty,$ we get $$\{\liminf_n X_n > -\infty\} \subset \bigcup_{k=1}^\infty \{N_k = \infty\}.$$ Apply the same to $(-X_n)$ and we get $$\{\limsup_n X_n < \infty\} \subset \bigcup_{k=1}^\infty \{N_k = \infty\}.$$ Hence $D^c \subset C$ and it follows that $P(C\cup D) = 1.$

As a corollary we get an extension of the second Borel-Cantelli lemma for dependent sequence.

Let $X_n = \sum_{m=1}^n (\mathbf{1}_{A_m} - P(A_m|\mathcal{F}_{m-1})),$ $X_0=0.$ Then it is easy to check that $X_n$ is a martingale with bounded increment.By the dichotomy, we get $C$ or $D$ almost surely.

On $C,$ in order to make $X_n$ convergent, $$\sum_n \mathbf{1}_{A_n} = \infty \iff \sum_n P(A_n|\mathcal{F}_{n-1}) = \infty.$$ On $D,$ $$\begin{gathered} \sum_n \mathbf{1}_{A_n} \ge \limsup_n X_n = \infty, \\ \sum_n P(A_n|\mathcal{F}_{n-1}) \ge \limsup_n (-X_n) = \infty. \end{gathered}$$ Thus in any case, the desired result follows.

Notice that $X_n$ in the proof is in the form of $A_n$ from Doob’s decomposition.

## Braching process

Think of $\xi_i^n$ as the number of offsprings that $n$th individual produce in $i$th generation. $Z_n$ naturally be the total number of offsprings in $n$th generation. By construction, $Z_n$’s are independent.

It is clear that $Z_n/\mu^n \in \mathcal{F}_n$ and is integrable for all $n.$ $$\begin{aligned} E(Z_{n+1}|\mathcal{F}_n) &= E(Z_{n+1} \sum_{k=0}^\infty \mathbf{1}_{Z_n=k}|\mathcal{F}_n) \\ &= \sum_{k=0}^\infty E(Z_{n+1} \mathbf{1}_{Z_n=k}|\mathcal{F}_n) \\ &= \sum_{k=0}^\infty E(\sum_{i=1}^{k}\xi_i^{n+1} \mathbf{1}_{Z_n=k}|\mathcal{F}_n) \\ &= \sum_{k=0}^\infty \mathbf{1}_{Z_n=k}k\mu \\ &= \sum_{k=0}^\infty \mathbf{1}_{Z_n=k} Z_n \mu = Z_n \mu. \end{aligned}$$

Using this, we can confirm our naturale guess that the population will be extinct if the average number of offsprings per individual is below one.

$P(Z_n > 0) = E\mathbf{1}_{Z_n > 0} \le EZ_n\mathbf{1}_{Z_n>0} = EZ_n.$ By the lemma, $E(\frac{Z_n}{\mu^n}) = E(\frac{Z_0}{\mu^0})=1$ thus $EZ_n = \mu^n.$ $$\sum_{n=1}^\infty P(Z_n>0) \le \sum_{n=1}^\infty \mu^n < \infty.$$ By the first Borel-Cantelli lemma, $P(Z_n = 0 \text{ eventually})=1.$

*Acknowledgement*

This post series is based on the textbook *Probability: Theory and Examples*, 5th edition (Durrett, 2019) and the lecture at Seoul National University, Republic of Korea (instructor: Prof. Sangyeol Lee).