7 added 15 characters in body

One knows that $P(S_n,n)$ is a martingale iff 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 iff 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 $\frac{u^i}{i!}$ 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$\frac{u^i}{i!}$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 homogenous 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 homogenous 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 $u$.

6 deleted 36 characters in body

One knows that $P(S_n,n)$ is a martingale iff $P(s+1,n+1)+P(s-1,n+1)=2P(s,n)$ and $Q(B_t,t)$ is a martingale iff $2\partial_tQ(x,t)+\partial^2_{xx}Q(x,t)=0$.

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 along increasing powers as a series of multiples of $u$ \frac{u^i}{i!}$for nonnegative integers$i$is also a martingale. This yields the well known fact that $$1,\ B_t,\ \frac12B^2_t-\frac12t,\ B^2_t-t,\ frac16B^3_t-\frac12tB_t,\ B^3_t-3tB_t,\ frac1{24}B^4_t-\frac14tB_t^2+\frac18t^2, B^4_t-6tB_t^2+3t^2,$$ etc., are all martingales. One recognizes the sequence of Hermite polynomials$H_n(B_t,t)$, a fact which is not very surprising fact since these polynomials may be defined precisely through the expansion of$\exp(ux-u^2t/2)$. So far, so good. What 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 $$\exp(-(\ln\cosh(u))n)=1-\frac12u^2n+\frac1{12}u^4n+\frac18u^4n^2+O(u^6). exp(-(\ln\cosh(u))n)=1-\frac12u^2n+\frac1{24}u^4(2n+3n^2)+O(u^6).$$ 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 powers of terms$u$\frac{u^i}{i!}$ in this expansion yields the martingales $$1,\ S_n,\ \frac12S_n^2-\frac12n,\ S_n^2-n,\ frac16S_n^3-\frac12nS_n, S_n^3-3nS_n,$$ and $$\frac1{24}S_n^4-\frac14nS_n^2+\frac1{12}n+\frac18n^2,\ S_n^4-6nS_n^2+2n+3n^2,\ frac1{120}S_n^5-\frac1{12}nS_n^3+\frac18n^2S_n+\frac1{12}nS_n. S_n^5-10nS_n^3+5(2n+3n^2)S_n.$$ Thus, in $M_n^u$, 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 homogenous 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 homogenous 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 noted that$\mathrm{e}^u-1$is not a power of$u$. 5 added 3 characters in body One knows that$P(S_n,n)$is a martingale iff$P(s+1,n+1)+P(s-1,n+1)=2P(s,n)$and$Q(B_t,t)$is a martingale iff$2\partial_tQ(x,t)+\partial^2_{xx}Q(x,t)=0$. 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 along increasing powers of$u$is also a martingale. This yields the well known fact that $$1,\ B_t,\ \frac12B^2_t-\frac12t,\ \frac16B^3_t-\frac12tB_t,\ \frac1{24}B^4_t-\frac14tB_t^2+\frac18t^2,$$ etc., are all martingales. One recognizes the sequence of Hermite polynomials$H_n(B_t,t)$, a not very surprising fact since these may be defined precisely through the expansion of$\exp(ux-u^2t/2)$. So far so good. 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 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 $$\exp(-(\ln\cosh(u))n)=1-\frac12u^2n+\frac1{12}u^4n+\frac18u^4n^2+O(u^6).$$ 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 powers of$u$yields the martingales $$1,\ S_n,\ \frac12S_n^2-\frac12n,\ \frac16S_n^3-\frac12nS_n,$$ and $$\frac1{24}S_n^4-\frac14nS_n^2+\frac1{12}n+\frac18n^2,\ \frac1{120}S_n^5-\frac1{12}nS_n^3+\frac18n^2S_n+\frac1{12}nS_n.$$ Thus, in$M_n^u$,$B_t$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 homogenous 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 homogenous in$u$.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 noted that $\mathrm{e}^u-1$ is not a power of $u$.