Convergence of the quadratic variation process

Question

Suppose we are given a sequence of stochastic processes $X^n, n\in\mathbb{N},$ with finite quadratic variations and a stochastic process $X$ such that for every $t\geq0$ $$ \lim_{n\to\infty}\mathbb{E}(|X^n_t-X_t|^2) =0. $$ Is it then possible to infer convergence of the quadratic variations, i.e. for every $t\geq0$, $\langle X^n\rangle_t\to\langle X\rangle_t$ in some sense, perhaps $L^2$?

Answer 1

The answer is no - I will construct processes $X^n$ on $[0, 1]$ such that the limit in question holds for $X$ the zero process, but $\limsup_n \langle X^n \rangle_t =\infty$ almost surely for every $t \in (0, 1]$.

Indeed, for $n \in \mathbb Z_+$, and $1 \leq k \leq 2^n$, let $Z_{n, k}$ be identically distributed jointly independent random variables equal to $0$ or $1$ with probability $\frac{1}{2}$ each.

Define

$$X^n = \sum_{k = 1}^{2^{n}} Z_{n, k} \, \mathbf 1_{(\frac{k-1}{2^n}, \frac{k-1}{2^n} + \frac{1}{2^{2n}})}.$$

That the limit in question holds is obvious, since for every $0 \leq t \leq 1$, $X_n^t$ is almost surely $0$ for all large enough $n$.

We clearly have

$$\langle X^n \rangle_t = 2 \, |\{1 \leq k \leq 2^n t \, | \, Z_{n, k} = 1\}|.$$

Since the $Z_{n, k}$ are independent of each other, $B_{n, t} := |\{1 \leq k \leq 2^n t \, | \, Z_{n, k} = 1\}|$ is binomially distributed with success rate $\frac{1}{2}$ and number of trials equal to $[2^n t]$. Thus by symmetry of the binomial distribution,

$$\mathbb P \left (\langle X^n \rangle_t \geq [2^n t] \right ) \geq \frac{1}{2}.$$

By independence of the $X^n$ from each other, the events $E_n := \{ \langle X^n \rangle_t \geq [2^n t] \}$ are independent.

Thus by the second Borel-Cantelli lemma, $\langle X^n \rangle_t \geq [2^n t]$ infinitely often, almost surely; and so $\limsup_n \langle X^n \rangle_t = \infty$ almost surely, as claimed.