4.5. Square integrable martingales
In this section, we look into martingales with special property - square integrability. Square integrability gives martingale an upper bound for maximal expectation so that it can further be used to determine the convergence of the sequence.
Square integrable martingales
A martingale $X_n$ is square integrable if $EX_n^2 < \infty$ for all $n.$
In the following discussion, we assume $X_0 = 0.$ Notice that $X_n^2$ is a submartingale and if we let $A_n = A_{n-1} + E(X_n^2 | \mathcal{F}_{n-1}) - X_{n-1}^2,$ $A_0=0,$ which is from Doob’s decomposition, then $EX_n^2 = EA_n$ and
\[\begin{aligned} A_n &= \sum\limits_{m=1}^n \left( E(X_m^2|\mathcal{F}_{m-1}) - X_{m-1}^2 \right) \\ &= \sum\limits_{m-1}^n E\left( (X_m - X_{m-1})^2 | \mathcal{F}_{m-1} \right). \end{aligned}\]
For a square integrable martingale $X_n,$ let $A_\infty = \lim_n A_n.$ The followings hold.
(i) $E \sup_n X_n^2 \le 4 EA_\infty.$
(ii) $E \sup_n |X_n| \le 3 EA_\infty^\frac{1}{2}$
(iii) $\lim_n X_n$ exists and is almost surely finite on $\{A_\infty < \infty\}.$
(iv) If $f:\mathbb{R} \to \mathbb{R}$ is increasing and $\int_0^\infty f^{-2}(t)dt < \infty,$ $f(t) \ge 1,\forall t,$ then $\frac{X_n}{f(A_n)} \to 0$ a.s. on $\{A_\infty = \infty\}.$
(i) $E \sup_n X_n^2 \le 4 EA_\infty.$
(ii) $E \sup_n |X_n| \le 3 EA_\infty^\frac{1}{2}$
(iii) $\lim_n X_n$ exists and is almost surely finite on $\{A_\infty < \infty\}.$
(iv) If $f:\mathbb{R} \to \mathbb{R}$ is increasing and $\int_0^\infty f^{-2}(t)dt < \infty,$ $f(t) \ge 1,\forall t,$ then $\frac{X_n}{f(A_n)} \to 0$ a.s. on $\{A_\infty = \infty\}.$
expand proof
(i) is direct from $L^p$ maximal inequality.
(ii) Let $N_a = \inf\{n: A_{n+1} > a^2\},$ then it is a stopping time. $$\begin{aligned} P(\sup_n|X_n| > a) &= P(\sup_n |X_n| > a, N < \infty) + P(\sup_n |X_n| > a, N = \infty) \\ &\le P(N<\infty) + P(\sup_n |X_{n \wedge N} > a) \\ &= P(N<\infty) + \lim_n P(\sup_{m\le n} |X_{m \wedge N}| > a) \\ &\le P(N<\infty) + \frac{1}{a^2} \lim_n E|X_{n \wedge N}|^2 \\ &= P(N<\infty) + \frac{1}{a^2} \lim_n E A_{n\wedge N} \\ &\le P(N<\infty) + \frac{1}{a^2} E(A_\infty \wedge a^2) \\ &= P(A_\infty > a^2) + \frac{1}{a^2} E(A_\infty \wedge a^2). \end{aligned}$$ The last inequality is from the fact that $$EA_{n\wedge N} \le EA_N \le a^2 \text{ on } \{N<\infty\}, \\ EA_n \le EA_\infty \le a^2 \text{ on } \{N=\infty\}.$$ Using this, Fubini's theorem and integration by substitution, we get $$\begin{aligned} E \sup_n |X_n| &= \int P(\sup_n |X_n| \ge a) da \\ &\le \int_0^\infty P(A_\infty^{1/2} > a) da + \int_0^\infty \frac{1}{a^2} E(A_\infty \wedge a^2) da \\ &= EA_\infty^{1/2} + \int_0^\infty \frac{1}{a^2} \int_0^{a^2} P(A_\infty > b) db da \\ &= 3EA_\infty^{1/2}. \end{aligned}$$ (iii) Given $a>0,$ by (i), $E \sup_n X_{n \wedge N_a} \le 4a^2 < \infty.$ By submartingale convergence, $X_{n\wedge N_a}$ converges a.s. and in $L^2.$ Now let $C_k = \{X_{n\wedge N_k} \text{ converges}\},$ then $P(C_k) = 1$ and $P(\cap_k C_k) = 1$ as well. For an arbitrary $\omega \in (\cap_k C_k) \cap (A_\infty < \infty),$ $N_k(\omega) = \inf\{n: A_n(\omega)\ge k\} = \infty$ for large enough $k$ since $A_\infty(\omega) < \infty.$ Hence $X_{n\wedge N_k}(\omega) = X_n(\omega)$ converges.
(iv) Let $H_m = \frac{1}{f(A_m)}$ be a bounded predictable sequence. Then $Y_n := (H \cdot X)_n = \sum\limits_{m=1}^n \frac{X_m - X_{m-1}}{f(A_m)}$ is a square integrable martingale. Let $B_n = \sum\limits_{m=1}^n E\left( (Y_m - Y_{m-1})^2 | \mathcal{F}_{m-1} \right),$ then $$\begin{aligned} B_\infty &= \sum\limits_{m=0}^\infty \frac{A_{m+1} - A_m}{f(A_{m+1})^2} \\ &\le \sum\limits_{m=0}^\infty \int_{A_m}^{A_{m+1}} f^{-2}(t) dt \\ &\le \int_0^\infty f^{-2}(t) dt < \infty \text{ a.s.} \end{aligned}$$ By (iii), $\lim_n Y_n$ exists and is finite almost surely. By Kronecker's lemma, it suffices to show that $f(A_n) \uparrow \infty.$ Since $\int_0^\infty f^{-2}(t) dt < \infty,$ $\lim_t f(t)$ should be $\infty$ otherwise it gives contradiction. Since $A_n, f$ is increasing and $f(A_\infty) = \infty$ on $(A_\infty = \infty),$ this is true.
From the facts, we get another form of conditional Borel-Cantelli lemma.
Let $B_n \in \mathcal{F}_n$ for all $n \ge 0$ and $p_n = P(B_n | \mathcal{F}_{n-1}), n\ge 1.$ Then
$$\frac{\sum_{n=1}^\infty\mathbf{1}{B_n}}{\sum_{n=1}^\infty p_n} \to 1 \text{ a.s. on } \left\{\sum\limits_{n=1}^\infty p_n = \infty \right\}.$$
Let $X_n = X_{n-1} + \mathbb{1}_{B_n} - P(B_n | \mathcal{F}_{n-1}),$ $X_0=0$ be a square integrable martingale. Then $A_n$ from Doob's decomposition yields $A_m - A_{m-1} = p_m - p_m^2$ and $A_n = \sum\limits_{m=1}^n p_m - p_m^2 \le \sum\limits_{m=1}^n p_n.$
On $(A_\infty < \infty),$ $X_n$ converges a.s. $$\frac{X_n}{\sum_{m=1}^n p_m} = \frac{\sum_{m=1}^n \mathbf{1}_{B_m}}{\sum_{m=1}^n p_m} - 1 \to 0 \text{ a.s. on } (\sum_{n=1}^\infty p_n = \infty).$$ On $(A_\infty = \infty),$ let $f(t) = 1 \vee t$ so that such $f$ satisfies conditions in (iv) of the previous theorem. Then $\frac{X_n}{f(A_n)} = \frac{X_n}{A_n \vee 1} \to 0$ a.s. on $(A_\infty = \infty).$ Since $A_n \le \sum_{m=1}^n p_m,$ we get $\frac{X_n}{\sum_{m=1}^n p_m} \to 0$ a.s. on $(A_\infty = \infty).$
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).