Expression for quadratic variation I read a book and don't understand the following:
Let $X$ be a continuous local martingale and is uniformly bounded. Let $\langle X \rangle^{(n)}_t = \sum_{k \in \mathbb{N}} (X_{t \wedge t^n_k}- X_{t \wedge t^n_{k-1}})^2,$ where $t^n_k = k2^{-n}$. 
We then define a process $M^{(n)}$ by  $M^{(n)}_t = \frac{1}{2} ( X^2_t - \langle X \rangle^{(n)}_t )$ and let $M^{(n)}_{\infty} = \lim_{t \to \infty} M^{(n)}_t$. By telescoping the sum, we get $$M^{(n)}_t = \sum_{k \in \mathbb{N}} X_{t^{n}_{k-1}} (X_{t \wedge t^n_k}- X_{t \wedge t^n_{k-1}}).$$ Then it claims that

$$ \forall n >m, \quad M^{(n)}_{\infty} - M^{(m)}_{\infty} = \sum_{j=1}^{\infty} \big( X_{j 2^{-n}} - X_{{\lfloor{j2^{m-n}}\rfloor}2^{-m}} \big) \big( X_{(j+1) 2^{-n}} -  X_{j 2^{-n}} \big). $$

Can someone please explain this to me?
I know how to show that 
$$ \quad M^{(n)}_{\infty}= \sum_{j=1}^{\infty}  X_{j 2^{-n}} \big( X_{(j+1) 2^{-n}} -  X_{j 2^{-n}} \big). $$ The expression involving $m$ is what bothers me.
 A: It is really helpful to draw a picture. For fixed $m<n$, the partition $$\Pi^m := \left\{\frac{1}{2^m} < \frac{2}{2^m} < \ldots < \frac{k}{2^m} < \ldots \right\}$$ is coarser than the partition
$$\Pi^n := \left\{\frac{1}{2^n} < \frac{2}{2^n} < \ldots < \frac{k}{2^n} < \ldots \right\}.$$ Now, since we are interested in the difference $M^{(n)}-M^{(m)}$, we would like to rewrite the definition of $M_t^{(m)}$ using the partition $\Pi^n$, i.e. to write
$$M_t^{(m)}  =\sum_j Y_j (X_{t \wedge t_{j}^n}-X_{t \wedge t_{j-1}^n})$$
for some random variables $Y_j$. It remains to determine $Y_j$. To this end, we note that for any $k \in \mathbb{N}$ we can find a (unique) $j=j(k) \in \mathbb{N}$ such that
$$t_k^m = k \frac{1}{2^m} = j \frac{1}{2^n} = t_j^n,$$
namely,
$$j(k) = k 2^{n-m}. \tag{1}$$
Now
$$\begin{align*} M_t^{(m)} &= \sum_{j \geq 1} Y_j (X_{t \wedge t_{j}^n}-X_{t \wedge t_{j-1}^n}) \\ &= \sum_{k \geq 1} \sum_{j=j(k-1)}^{j(k)-1} Y_j (X_{t \wedge t_{j+1}^n}-X_{t \wedge t_{j}^n}) \\ &= \sum_{k \geq 1} \tilde{Y}_k \underbrace{\sum_{j=j(k-1)}^{j(k)-1} (X_{t \wedge t_{j+1}^n}-X_{t \wedge t_{j}^n})}_{X_{t \wedge t_{j(k)}^n}-X_{t \wedge t_{j(k-1)}^n} = X_{t \wedge t_{k}^m}-X_{t \wedge t_{k-1}^m}} \end{align*}$$
if we choose $Y_j = \tilde{Y}_k$ for all $j(k-1) \leq j \leq j(k)-1$. Comparing this expression with the definition of $M_t^{(m)}$, we find
$$\tilde{Y}_k = X_{t \wedge t_{k-1}^m}.$$
Note that $Y_j = \tilde{Y}_{k(j)}$ with $k=k(j) $ such that $j(k-1) \leq j \leq j(k)-1$, i.e. for each $j$ we have to find $k(j)$ such that $j \in \left[ (k(j)-1) 2^{n-m}, k(j) 2^{n-m}-1 \right]$. Obviously,
$$k(j) = \lfloor 2^{m-n}j \rfloor.$$
Consequently,
$$Y_j = X_{t \wedge t_{\lfloor 2^{m-n} j \rfloor-1}^m}.$$
