A càdlàg function is a function $f:[a,b]\rightarrow\mathbb R$ such that i) the left limit $\lim_{t\uparrow x}f(t)$ exists, and ii) the right limit $\lim_{t\downarrow x}f(t)$ exists and equals $f(x)$.
A càdlàd function is a function $g:[a,b]\rightarrow\mathbb R$ such that i) the left limit $\lim_{t\uparrow x}f(t)$ exists and equals $f(x)$, and ii) the right limit $\lim_{t\downarrow x}f(t)$ exists.
In my application the càdlàg and the càdàlad functions $f$ and $g$ both are of bounded variation, i.e. both $$V^1(f) = \sup_{P\in\mathcal P}\sum_{i=1}^{n_P}\vert f(x_i) – f(x_{i-1})\vert\qquad\text{and}\qquad V^1(g) = \sup_{P\in\mathcal P}\sum_{i=1}^{n_P}\vert g(x_i) – g(x_{i-1})\vert$$ are finite. Here $\mathcal P$ is the set of all partitions $P = \{x_0, x_1, \dots, x_{n_P}\}$ of $[a,b]$ such that $x_0 = a$, $x_{n_P} = b$ and $x_{i-1}\leq x_i$ for all $i=1,2,\dots, n_P$.
Is it true that their quadratic variations are zero?
If a continuous function $h:[0,1]\rightarrow\mathbb R$ has bounded variation, $V^1(h) < \infty$, it holds that $V^2(h) = 0$, where $$V^2(h) = \sup_{P\in\mathcal P}\sum_{i=1}^{n_P}\vert h(x_i) – h(x_{i-1})\vert^2.$$ This is because for any $P\in\mathcal P$ it holds that \begin{align*} \sum_{i=1}^{n_P}\vert h(x_i) – h(x_{i-1})\vert^2 &\leq\left(\sum_{i=1}^{n_P}\vert h(x_i) – h(x_{i-1})\vert\right)\left(\sum_{i=1}^{n_P}\vert h(x_i) – h(x_{i-1})\vert\right) \\ &\leq \sup_{i\in\{1,2,\dots,n_P\}}\vert h(x_i) – h(x_{i-1})\vert V^1(h), \end{align*}
and since $h$ is continuous, it is uniformly continuous on compact sets. Hence $\sup_{i\in\{,1,2,\dots,n_P\}}\vert h(x_i) – h(x_{i-1})\vert$ converges to zero.
I see that for non-continuous functions, e. g. functions with a jump, there will always be a distance on some interval $[x_i,x_{i-1}]$ that can not converge to zero no matter what partition is chosen. However, I was wondering whether there still exists an approach for controlling the quadratic variation for functions that are at least "close" to being continous, such as càdlàg and càdlàd functions.
Best Answer
The quadratic variation of a discontinuous function with bounded variation is not zero. The prime example are the paths of a Poisson process $N_t$ ($\mathbb N$-valued, increasing by jumps of size one and càdlàg). It holds that $V^2(N)$ is $N$ itself which is written as $$V^2(N)_t=[N,N]_t=N_t.$$ This holds because the quadratic variation of $N$ is the sum of its jumps and $N$ is nothing else than that, too, so $$ [N,N]_t=\sum_{s\le t}(\Delta N_s)^2=\sum_{s\le t}\Delta N_s=N_t\,. $$