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.
 A: 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 $$\mathbb{E}[\Delta_n^2 \Delta_m^2] = \mathbb{E}[Z_0^4]\mathbb{E}[Z_n^2]\mathbb{E}[Z_m^2] \ne \mathbb{E}[Z_0^2]^2 \mathbb{E}[Z_n^2]\mathbb{E}[Z_m^2] = \mathbb{E}[\Delta_n^2] \mathbb{E}[\Delta_m^2].$$
A: 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 $n\ge 3$.
