Is continuous, uniformly integrable martingale indistinguishable from BMO martingale?

Question

Definition:

BMO: Let $M$ be a martingale in $\mathcal{H}^2$. $M$ is said to be in BMO if there exists a constant $c$ such that for any stopping time $T$ we have $$ E\{(M_\infty-M_{T_-})^2 \mid\mathcal{F}_t\}\leq c^2 \quad a.s. $$ The smallest such $c$ is defined to be the BMO norm of $M$.

$\mathcal{H}^2$: Let $M$ be a local martingale with cadlag paths. We define the $\mathcal{H}^2$ norm of M to be $\|M\|_2 = E\{[M,M]_{\infty}\}^{\frac{1}{2}}$.

Integrable variation: The stochastic process $X$ is called a process of integrable variation if $E[V_t] < \infty$ for each $t \geq 0$, where $V_t = \sup \Sigma_{k = 1}^{n} \mid X_{t_k} - X_{t_{k-1}} \mid$.

Natural process: A stochastic process $A = \{A_t, t \geq 0\}$ of integrable variation with $A_0 = 0$ a.s. is called natural if the equality $$\mathbb{E} \int_0^t Z_s d A_s=\mathbb{E} \int_0^t Z_{s-} d A_s$$ holds for each $t > 0$ and for each bounded, right-regular $\mathbb{F}$-martingale $Z = \{Z_t, t \geq 0\}$.

Some quotes:

Let $M$ be a uniformly integrable martingale. From Exercise 21, Chapter IV, 2005 Stochastic Integration and Differential Equations(Second Edition, Version 2.1 Springer-Verlag, Heidelberg), $M = U + V$, where $U$ is of integrable variation and $V$ is in BMO.

And it is shown in Theorem 2(Kruglov, Victor M., On natural and predictable processes, Sankhyā, Ser. A 78, No. 1, 43-51 (2016). ZBL1338.60104.) that any natural martingale of integrable variation is indistinguishable from zero.

Here is my question: Do those two propositions together suggest that any continuous, uniformly integrable martingale is indistinguishable from BMO martingale?

Answer 1

We follow the online note A remark on H1 martingales (also see Revuz-Yor on BMO) where they give a recipe for constructing martingales $X_{t}$ that are in uniformly integrable but not in in $\mathcal{H}^{1}$ i.e. $X_{t}\in \mathcal{M}\setminus \mathcal{H}^{1}$. This will give a counterexample for BMO too because we have the inclusion

$$BMO\subset \mathcal{H}^{p},\forall p\geq 1.$$

For this inequality (see pg.262) and another counterexample (see example 2) in "On the transformation of some classes of martingales by a change of law"., they give a bounded martingale that is not in BMO.

By BDG inequalities we have that $M_{t}\in \mathcal{H}^1$ iff

$$\sum_{n\geq 1} P\left[\sup_{t\geq 0}|M_{t}|\geq n\right]<\infty.$$

We set $M_{t}:=exp(B_{t}-\frac{t}{2})$. We know from Prove $\mathbb{P}(\sup_{t \geq 0} M_t > x \mid \mathcal{F}_0)= 1 \wedge \frac{M_0}{x}$ for a martingale $(M_t)_{t \geq 0}$ or $P(\sup X_t\ge x)=1/x$ if $\lim_{t\to \infty}X_t=0$ a.s. and $X_0 = 1$ that

$$P\left[\sup_{t\geq 0}|M_{t}|\geq n\right]=1\wedge \frac{1}{n}.$$

Consider an independent random variable $Y$ with

$$P[Y>n]=\frac{1}{c_{n}},$$

for $c_{n}:=\ln(e+\sum_{k=1}^{n}\frac{1}{k})$ (so about $\log(e+\log n)$). We finally define the martingale

$$X_{t}:=M_{t\wedge \sigma},$$

for $\sigma:=\inf\{s\geq 0: M_s> Y\}$. We claim that $X_{t}\in \mathcal{M}\setminus \mathcal{H}^{1}$.

We have that

$$\sum_{n\geq 1} P[\sup_{t\geq 0}|X_{t}|\geq n]\geq \lim_{m\to +\infty}\frac{1}{c_{m}}\sum_{n=1}^{m} P[\sup_{t\geq 0}|M_{t}|\geq n]=\lim_{m\to +\infty}\frac{e^{c_{m}}-e}{c_{m}}=+\infty.$$

So $X_{t}\notin \mathcal{H}^{1}$. On the other hand, since $\sigma$ is a stopping time, we can apply the optional stopping theorem

$$E[X_{t}]=E[M_{t\wedge \sigma}]=E[M_{0\wedge \sigma}]=1$$

and so $X_{t}\in \mathcal{M}$.

When are they related? See (3.16) Exercise (BMO-martingales) in Revuz Yor:

For $Y$ continuous uniformly integrable martingale, if $Y_t = E [Y_{\infty}|\mathcal{F}_{t}]$ for a bounded r.v. $Y_{\infty}$, then $Y\in BMO$ and $\|Y\|_{BMO}\leq 2\|Y_{\infty}\|_{\infty}$.