One knows that $P(S_n,n)$ is a martingale if and only if $P(s+1,n+1)+P(s-1,n+1)=2P(s,n)$ and
that $Q(B_t,t)$ is a martingale if and only if $2\partial_tQ(x,t)+\partial^2_{xx}Q(x,t)=0$.
Assume that $P(S_n,n)$ is a martingale and, for a given $d$ and for every $h>0$, let
$$
Q_h(x,t)=h^{d}P(x/\sqrt{h},t/h),
$$
in the sense that one evaluates $P(s,n)$ at the integer parts $s$ and $n$ of $x/\sqrt{h}$ and $t/h$.
If $Q_h\to Q$ when $h\to0$, writing $\partial_t$ and $\partial^2_{xx}$ as limits of finite differences of orders $1$ and $2$, one sees that $2\partial_tQ+\partial^2_{xx}Q=0$, hence $Q(B_t,t)$ is a martingale.
Example: $P(s,n)=s^2-n$. For $d=1$, $Q_h(x,t)=x^2-t$ hence $Q(x,t)=x^2-t$.
Other example:
$P(s,n)=s^4-6ns^2+3n^2+2n$. For $d=2$, $Q_h(x,t)=x^4-6tx^2+3t^2+2ht$ hence $Q(x,t)=x^4-6tx^2+3t^2$.
In the other direction, to deduce a martingale in $S_n$ and $n$ from a martingale in $B_t$ and $t$, one should probably replace each monomial by a sum of its first derivative. This means something like replacing $q(t)=3t^2$ by $\displaystyle\sum_{k=1}^n(\partial_tq)(k)=3n^2+3n$ but I did not look into the details.
Edit (Thanks to The Bridge for a comment on the part of this answer above this line)
Recall that a natural way to build in one strike a full family of martingales that are polynomial functions of $(B_t,t)$ is to consider so-called exponential martingales. For every parameter $u$,
$$
M^u_t=\exp(uB_t-u^2t/2)
$$
is a martingale hence every "coefficient" of its expansion as a series of multiples of $u^i$ for nonnegative integers $i$ is also a martingale. This yields the well known fact that
$$1,\ B_t,\ B^2_t-t,\ B^3_t-3tB_t,\ B^4_t-6tB_t^2+3t^2,
$$
etc., are all martingales. One recognizes the sequence of Hermite polynomials $H_n(B_t,t)$, a fact which is not very surprising since these polynomials may be defined precisely through the expansion of $\exp(ux-u^2t/2)$.
So far, so good. But what could be an analogue of this for standard random walks? The exponential martingale becomes
$$
D^u_n=\exp(uS_n-(\ln\cosh(u))n)
$$
and the rest is simultaneously straightforward (in theory) and somewhat messy (in practice): one should expand $\ln\cosh(u)$ along increasing powers of $u$ (warning, here comes the family of Bernoulli numbers), then deduce from this the expansion of $D^u_n$ along increasing powers of $u$, and finally collect the resulting sequence of martingales polynomial in $(S_n,n)$.
Let us see what happens in practice. Keeping only two terms in the expansion of $\ln\cosh(u)$ yields $\ln\cosh(u)=\frac12u^2-\frac1{12}u^4+O(u^6)$ hence
$$
\exp(-(\ln\cosh(u))n)=1-\frac12u^2n+\frac1{24}u^4(2n+3n^2)+O(u^6).
$$
Multiplying this by
$$
\exp(uS_n)=1+uS_n+\frac12u^2S_n^2+\frac16u^3S_n^3+\frac1{24}u^4S_n^4+\frac1{120}u^5S_n^5+O(u^6),
$$
and looking for the coefficients of the terms $u^i$ in this expansion yields the martingales
$$
1,\ S_n,\ S_n^2-n,\ S_n^3-3nS_n,
$$
and
$$
S_n^4-6nS_n^2+2n+3n^2,\
S_n^5-10nS_n^3+5(2n+3n^2)S_n.
$$
Thus, in $M_t^u$, $B_t$ scales like $1/u$ and $t$ like $1/u^2$ hence Hermite polynomials are homogeneous when one replaces $t$ by $B_t^2$. The analogues of Hermite polynomials for $(S_n,n)$, from degree $4$ on, are not homogeneous in the sense of this dimensional analysis where $n$ is like $S_n^2$. Ultimately, this is simply because in $D_n^u$ one has to compensate $uS_n$ by $(\ln\cosh(u))n$, which is not homogeneous in $u^2n$.
Note that this argument of non homogeneity carries through to continuous time processes. For instance, the exponential martingales for the standard Poisson process $(N_t)_t$ are
$$
\exp(uN_t-(\mathrm{e}^u-1)t),
$$
and the rest of the argument is valid once one has noted that $\mathrm{e}^u-1$ is not a power of $u$.
No, that is not true. Consider the following, defined on a filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}\_{t\in[0,T]},\mathbb{P})$.
- $W$ is a standard Brownian motion.
- $U$ is an $\mathcal{F}_0$-measurable Bernoulli random variable independent of $W$, with $\mathbb{P}(U=0)=\mathbb{P}(U=1)=1/2$.
Then, set $M_t=UW_t$. This is a continuous martingale. If $\mathcal{F}^M_t$ is its completed natural filtration then $U$ is $\mathcal{F}^M_t$-measurable for all $t > 0$. Then, $U$ is $\mathcal{F}^M_{0+}$-measurable but is not measurable with respect to $\mathcal{F}^M_0$ (which only contains sets with probability 0 and 1). So $\mathcal{F}^M_{0+}\not=\mathcal{F}^M_0$.
Also, this is essentially the same as the example I gave in a previous answer of a Markov process which is not strong Markov.
As another example to show that there is not really any simple way you can modify the question to get an affirmative answer, consider the following; a Brownian motion $W$ and left-continuous, positive, and locally bounded adapted process $H$. Then, $M=H_0+\int H\\,dW$ is a local martingale. Also, $M$ has quadratic variation $[M]=\int H^2_t\\,dt$ which has left-derivative $H^2$ for all $t > 0$. So, $H_t$ is $\mathcal{F}^M_t$-measurable, as is $W_t=\int H^{-1}\\,dM$. In fact, $\mathbb{F}^M$ is the completed natural filtration generated by $W$ and $H$. If $H$ is taken to be independent of $W$, then $\mathbb{F}^M$ will only be right-continuous if $\mathbb{F}^H$ is, and it easy to pick left-continuous processes whose completed natural filtration fails to be right-continuous.
Best Answer
Charles Fefferman proved that the space of functions of bounded mean oscillation $BMO$ is the dual of the Hardy space $H^1$. The key was the observation that the classical Hardy space $H^1$ is a natural substitution for $L^1$, and $BMO$ is a natural substitution for $L^{\infty}$.
One result of comparing classical Hardy and martingale Hardy spaces is that we can establish that the dual of the martingale Hardy space $\mathcal{H}^1$ is martingale $BMO$.
Check out "Martingale Inequalities: Seminar Notes on Recent Progress (Mathematics Lecture Note Series)" by Adriano Garsia