Find an $\mathcal{L}^2$-martingale $(M_n)$ with $M_n \rightarrow M_\infty$ almost surely for a finite real valued $M_\infty$ but $\langle M \rangle_n \rightarrow \infty$ almost surely.
$\langle M \rangle_n$ denotes the square variation of $(M_n)$ as defined here, but there is also an alternative formula, I will use: $$\langle M \rangle_n := \sum_{i=1}^n \mathbf{E}\bigl[(M_i - M_{i-1})^2|\mathcal{F}_{i-1}\bigr]$$
Ok, I tried the following:
Let $Y_n$ be independent random variables, defined the following way: $$\mathbf{P}[Y_n = n] = \mathbf{P}[Y_n = -n] = \frac{1}{2n^3}, \quad \mathbf{P}[Y_n = 0] = 1 - \frac{1}{n^3}\, .$$
Obviously $\mathbf{E}[Y_n] = 0$ and $\mathbf{Var}[Y_n] < \infty$.
We define $M_n := \sum_{i=1}^n Y_i$ adapted to $\mathcal{F}_n = \sigma(Y_1, \ldots, Y_n)$, this is a martingale, because
$$\mathbf{E}[M_t|\mathcal{F}_s]=\mathbf{E}\Biggl[\sum_{i=1}^t Y_i\Bigg\vert\mathcal{F}_s\Biggr] = \sum_{i=1}^t\mathbf{E}[Y_i |\mathcal{F}_s] = \sum_{i=s+1}^t\mathbf{E}[Y_i] + \sum_{i=1}^s Y_i = M_s \quad \text{ for } t> s\, ,$$ and because $\mathbf{Var}[M_n] = \mathbf{Var}[Y_1] + \cdots + \mathbf{Var}[Y_n]$ it has finite variance for all $n\in\mathbb{N}$. By the Borel-Cantelli lemma: $$\sum_{n=1}^\infty \mathbf{P}[Y_n \neq 0] = \sum_{n=1}^\infty \frac{1}{n^3} < \infty $$ we see that $\mathbf{P}\bigl[Y_n \neq 0 \text{ i. o.}\bigr] = 0$, so $M_n$ converges almost surely. Similarly $\mathbf{P}[M_\infty = +\infty] < \mathbf{P}\bigl[Y_n = +1 \text{ i. o.}\bigr] = 0$, same for "$-\infty$", so $M_\infty$ is finite real valued.
$\langle M \rangle_n = \sum_{i=1}^n \mathbf{E}\bigl[(M_i - M_{i-1})^2|\mathcal{F}_{i-1}\bigr] = \sum_{i=1}^n\mathbf{E}\bigl[Y_i^2|\mathcal{F}_{i-1}\bigr] = \sum_{i=1}^n\mathbf{E}\bigl[Y_i^2\bigr] = \sum_{i=1}^n\frac{1}{i}\, . $
So we see that $\langle M \rangle_n \rightarrow \infty$ almost surely. $\square$
Is my example correct?
Is there some insight to be taken from this exercise, because I didn't...
Thanks for any help!
You example is correct. For brevity, you could have used Doob's Martingale convergence theorem.
$$ E[|M_N|] \leq \sum_{i=1}^N E[|Y_i|] = \sum_{i=1}^N \frac{1}{i^2} < \frac{\pi^2}{6} $$ therefore $M_n \rightarrow M_\infty\, a.s.$ and $M_\infty \in L^1$.
It does not imply convergence in $L^1$. For that you need stronger conditions such as uniform integrability. A sufficient condition for UI is: $\sup \|M_n\|_p < \infty$, for $p>1$. In your case, you satisfy this property for up to $p<3/2$, so you can say that $M_t \rightarrow M_{\infty}$ in $L_p$ for $p\in [1,3/2)$, (of course $M_\infty \in L^p$). You also showed that $\langle M\rangle_t\rightarrow \infty$, therefore $M_\infty \notin L^2$.
I also believe there's an interesting remark if you make the connection with continuous martingales. If $\langle M \rangle_{\infty} = \infty, a.s.$ then $\limsup_{t\rightarrow \infty} M_t = +\infty$ and $\liminf_{t\rightarrow \infty} M_t = -\infty$ so $(M)_t$ does not converge.