I am wondering, polynomials like

$S_n^4-6n S_n^2+3n^2+2n$ for $$S_n=\sum_{i=1}^n{X_i}$$ where $$\mathbb{P}(X_i=1)=\mathbb{P}(X_i=-1)=\frac{1}{2}$$ is a martingale (under the conventional filtration). While $$B_t^4-6t B_t^2+3t^2$$ for Brownian motion $B_t$ is also a martingale.

Note the difference between the two, and the similarity!

What's the general conclusion about the polynomials of $S_n$ and $n$, also about $B_t$ and $t$, to make them into martingales?



1 Answer 1


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 $u$.

  • 2
    $\begingroup$ Just to add, it is widely known that Hermite polynomials and Brownian Motions are deeply connected with regards to (local) martingale property (and the chaos decomposition property) . math.ucsd.edu/~pfitz/downloads/hermite.pdf $\endgroup$
    – The Bridge
    Feb 12, 2011 at 0:05
  • $\begingroup$ @Didier: this is great! I just have a few comments. 1. For functions not in $\mathbb{C}^1$ (continuous differentiable to the first order) in $t$ and $\mathbb{C}^2$ in $B_t$, are there any examples of martingale in the form of $Q(B_t, t)$? 2. Does the condition $P(s+1,n+1)+P(s-1,n+1)=2P(s,n) P(S_n, n)$ truly enumerate all martingales in the form of $P(S_n, n)$? $\endgroup$
    – Qiang Li
    Feb 13, 2011 at 2:02
  • $\begingroup$ @Qiang Li: ("comments" in the sense of "questions".) Well, after an interesting initial question, you once again fall back on some standard textbook stuff. This is not MO purpose. Please get yourself some lecture notes (this time, on Brownian martingales), as was already suggested to you about other MO basic probability questions, and study them. Specific references were provided to you, did you go and read them? Considering the rythm of your questions on MO and elsewhere, this is doubtful. .../... $\endgroup$
    – Did
    Feb 13, 2011 at 11:19
  • $\begingroup$ .../... Anyway, about Brownian martingales, Durrett is excellent, Tsirelson is kind enough to put tau.ac.il/~tsirel/Courses/Brown/syllabus.html on the web, and many other good introductory texts exist. Re 2.: Does it, indeed? Either you did not spend one minute thinking about the answer, or you have no clue of what you are talking about. $\endgroup$
    – Did
    Feb 13, 2011 at 11:21
  • 5
    $\begingroup$ @Didier Piau : Sometimes I feel frustrated on MO 'cause I can't vote twice for a nice answer. $\endgroup$
    – The Bridge
    Feb 14, 2011 at 10:15

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