# Example of a martingale with non-independent increments and fixed variance

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.

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 $$\mathbb{E}[\Delta_n^2 \Delta_m^2] = \mathbb{E}[Z_0^4]\mathbb{E}[Z_n^2]\mathbb{E}[Z_m^2] \ne \mathbb{E}[Z_0^2]^2 \mathbb{E}[Z_n^2]\mathbb{E}[Z_m^2] = \mathbb{E}[\Delta_n^2] \mathbb{E}[\Delta_m^2].$$
• @George 1) This shows non-independence by showing the covariance is non-zero. 2) I think that $\mathbb{E}[\Delta_n \Delta_m] = 0 = \mathbb{E}[\Delta_n]\mathbb{E}[\Delta_m]$. Mar 1 at 16:26
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 $$n\ge 3$$.