Hi,
Regarding Martingales you can see them as fair games
This means that if the (martingale) process represents your (random) wealth, you should not be able to design a strategy to increase your current wealth, no matter what the outcome of the sample space is.
Brownian Motion can be seen as a limit of rather simple random walks but I'm sure that you know about this.
Markov processes "disconnect" Future and Past of the process conditionnally on the present value of the process. Where "disconnect" means that functions of past and of future values of the process are independent conditionnally on the present value of the process.
Does it make things more clear ?
It is true that $\langle M\rangle^n_1 \to \langle M\rangle_1$ in $L^1$ when $M$ is a continuous square-integrable martingale. In fact, the result holds even if $M$ is cadlag, as long as $\langle M\rangle$ is continuous.
Indeed, $M^2$ is then a submartingale of class D and so, since
$$E[(M_{t_{i+1}} - M_{t_{i}})^2 |\mathcal{F}_{t_i}]=E[M_{t_{i+1}}^2 - M_{t_{i}}^2 |\mathcal{F}_{t_i}],$$
the result follows from the analogous results for the predictable component of the Doob decomposition of a submartingale of class D, which was proved in
Doléans, Catherine. "Existence du processus croissant naturel associé à un potentiel de la classe (D)." Zeitschrift für Wahrscheinlichkeitstheorie und verwandte Gebiete 9, no. 4 (1968): 309-314.
and which constitutes Theorem 31.2, chapter 6 of
Rogers, L. Chris G., and David Williams. Diffusions, Markov processes and martingales: Volume 2, Itô calculus. Vol. 2. Cambridge university press, 2000.
When $M$ is cadlag (and $\langle M\rangle$ is not continuous), $\langle M\rangle^n_1$ converges to $\langle M\rangle_1$, but only in the $\sigma\left(L^{1}, L^{\infty}\right)$-topology, see for example
Rao, K. Murali. "On decomposition theorems of Meyer." Mathematica Scandinavica 24, no. 1 (1969): 66-78.
In this case, it follows from Mazur's lemma that one can obtain (strong) convergence in $L^1$ by
replacing $(\langle M\rangle^n_1)_n$ with a forward convex combinations thereof. In fact, taking forward convex combinations, one can obtain convergence in $L^1$ simultaneously at all times $t\in [0,1]$, as I showed in
Siorpaes, Pietro. "On a dyadic approximation of predictable processes of finite variation." Electronic Communications in Probability 19 (2014): 1-12.
Best Answer
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 $u$.