If $\sum_{n=1}^\infty E(X_n-X_{n-1})^2/b_n^2 < \infty$, then $X_n/b_n \rightarrow 0$ a.s.

Question

This is Durrett Exercise 5.4.9.

I'm trying to show that if $X_n$ is a martingale, and $b_m \uparrow \infty$, $\sum_{n=1}^\infty E(X_n-X_{n-1})^2/b_n^2 < \infty$, then $X_n/b_n \rightarrow 0$ a.s..

If $X_n \in L^2$, I can show that this reduces to Durrett Exercise 5.4.8, which I have solved, since the above also gives us that the the sum of $\sum_{n=1}^\infty E\left(\frac{X_n}{b_n}-\frac{X_{n-1}}{b_{n-1}}\right)^2<\infty$, and then I can use orthogonality of martingale increments, and $L^p$ convergence theorem to conclude that it converges in almost surely and in $L^2$, although I'm not sure that I can show it converges to 0.

But, in this case, I am stuck and unable to proceed

Answer 1

Hints: Since $b_n \uparrow \infty$ we may assume without loss of generality that $b_n>1$ for all $n \geq 1$.

Let's first assume additionally that $(X_n)_{n \in \mathbb{N}}$ is square integrable and $X_0 = 0$.

1. Set $$Y_n := \sum_{i=1}^n \frac{X_i-X_{i-1}}{b_i}.$$ Show that $(Y_n)_{n \in \mathbb{N}}$ is a squre-integrable martingale with $$\mathbb{E}(Y_n^2) = \sum_{i=1}^n \frac{\mathbb{E}((X_i-X_{i-1})^2)}{b_i^2}.$$
2. It follows from Step 1 and the assumptions on $(X_n)_{n \in \mathbb{N}}$ that $(Y_n)_{n \in \mathbb{N}}$ is an $L^2$-bounded martingale. The martingale convergence thorem yields that $Y_n$ converges almost surely, i.e. $$\sum_{i=1}^{\infty} \frac{X_i-X_{i-1}}{b_i}$$ converges with probability $1$.
3. Apply Kronecker's lemma to deduce that $$\lim_{n \to \infty} \frac{X_n}{b_n} = 0 \quad \text{a.s.}$$

For the general case: Define $\tilde{X}_n := X_n-X_0$.

1. Check that $\sum_n \mathbb{E}((X_n-X_{n-1})^2)/b_n^2 < \infty$ implies $\sum_n \mathbb{E}((\tilde{X}_n-\tilde{X}_{n-1})^2)/b_n^2 < \infty$.
2. Deduce from $\mathbb{E}((\tilde{X}_n-\tilde{X}_{n-1})^2)< \infty$, $n \in \mathbb{N}$, and $\tilde{X}_0 =0 \in L^2(\mathbb{P})$ that $\tilde{X}_n \in L^2(\mathbb{P})$ for all $n \geq 1$.
3. Apply the first part of this answer to $\tilde{X}_n$.