I'm trying to come up with a martingale whose increments $\Delta_n = M_n - M_{n-1}$ are non-independent, and have fixed variance, $Var(\Delta_n) = k$.

Attempt 1

Initially I was thinking that $M_N = \sum_{i=1}^n Z_i$ where $Z_i \sim N(0,1)$. This results in increments with fixed variance $Var(\Delta_i)=Var(Z_n)=1$. However, I think that the increments in this case are independent.

Attempt 2

In order to produce dependent increments it feels like something multiplicative might work. For example $M_n = \prod_{i=1}^n Z_i$ where $Z_i \sim N(0,1)$. This approach results in dependent increments, but the variance of the increments seems pretty intractable, and certainly not fixed.


2 Answers 2


Let $Z_i \sim N(0,1)$ be i.i.d. random variables and define $M_n := Z_0 \sum_{i=1}^n Z_i$. Then $\Delta_n = Z_0 Z_n$, so the increments are identically distributed and hence have constant variance, but are not independent because $$\mathbb{E}[\Delta_n^2 \Delta_m^2] = \mathbb{E}[Z_0^4]\mathbb{E}[Z_n^2]\mathbb{E}[Z_m^2] \ne \mathbb{E}[Z_0^2]^2 \mathbb{E}[Z_n^2]\mathbb{E}[Z_m^2] = \mathbb{E}[\Delta_n^2] \mathbb{E}[\Delta_m^2].$$

  • $\begingroup$ This is very helpful, thank you! Two questions: 1) Does this actually show independence, or just zero covariance? 2) Do you actually need to square the increments? Doesn't the conclusion follow just as easily from showing E[delta_n * delta_m] not equal to E[delta_n] * E[delta_m]? $\endgroup$
    – George
    Mar 1 at 15:35
  • 1
    $\begingroup$ @George 1) This shows non-independence by showing the covariance is non-zero. 2) I think that $\mathbb{E}[\Delta_n \Delta_m] = 0 = \mathbb{E}[\Delta_n]\mathbb{E}[\Delta_m]$. $\endgroup$ Mar 1 at 16:26

Here's a construction of many examples.

Let $S_n$ be a simple symmetric random walk: $S_0=0$ and $$ S_n=\sum_{k=1}^n \xi_k, $$ where the $\xi_k$ are iid with $P[\xi_k=1]=P[\xi_k=-1]=1/2$. Let $\mathcal F_n:=\sigma\{\xi_1,\ldots,\xi_n\}$ be the associated filtration ($\mathcal F_0:=\{\emptyset,\Omega\}$). Let $\{H_k\}$ be any sequence of bounded random variables with $H_k$ chosen to be $\mathcal F_{k-1}$ measurable, for $k=1,2,\ldots$. (Thus $\xi_k$ and $H_k$ are independent.) Then $$ M_n:=\sum_{k=1}^n H_k\cdot\xi_k,\qquad n=0,1,2,\ldots,\qquad\qquad(*) $$ is a martingale, with $E[(\Delta M_n)^2]=E[H_n^2]$. Just take care (normalize if necessary) to choose the $H_k$ so that they all have the same mean square.

For example, define inductively $H_1\equiv 1$ and $$ H_n:=\phi_n(\Delta M_{n-1}), $$ where $M_k$ ($k=1,2,\ldots,n-1$) are determined by ($*$) and $\phi_n:\Bbb R\to(0,\infty)$ is continuous (say) and such that $E[(\phi_n(\Delta M_{n-1}))^2]=1$. Notice that $E[(\Delta M_n)^2]=1$ for all $n$.

Fix $n\ge 1$. If $\Delta M_n$ and $\Delta M_{n-1}$ were independent we would have $$ \eqalign{ E[f(\Delta M_{n-1})]=E[(\Delta M_n)^2]\cdot E[f(\Delta M_{n-1}] &=E[(\Delta M_n)^2 f(\Delta M_{n-1}]\cr &=E[H_n^2 f(\Delta M_{n-1}]\cr &=E[\phi_n(\Delta M_{n-1})^2 f(\Delta M_{n-1})]\cr } $$ for all measurable $f$. In particular, taking $f(x) =\phi_n(x)^2$ we would have $$ 1=E[(\phi_n(\Delta M_{n-1}))^2]=E[\phi_n(\Delta M_{n-1})^4]. $$ But by Jensen's nequality $$ E[\phi_n(\Delta M_{n-1})^4]\ge\left\{E[(\phi_n(\Delta M_{n-1}))^2]\right\}^2=1, $$ and the inequality is strict (resulting in a contradiction) unless the random variable $(\phi_n(\Delta M_{n-1}))^2$ is a.s. constant. This degeneracy can be avoided by taking $\phi_n^2$ to be injective (and $n\ge 3$), for example. Things being so, $\Delta M_{n-1}$ and $\Delta M_n$ are not independent if $n\ge 3$.


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