The following question is more math than quant, but since it arises from a mathematical finance textbook, I've figured the good people in this sub might be able to help me. So here goes.

In the 3rd edition of his textbook "Introduction to Stochastic Calculus with Applications" (Imperial College Press 2012), Fima Klebaner defines quadratic variation as follows (p. 8):

If $g$ is a function of real variable, define its quadratic variation over the interval $[0, t]$ as the limit (when it exits) $$ [g](t) = \lim_{\delta_n \rightarrow 0}\sum_{i = 1}^n\left(g(t^n_i) - g(t^n_{i - 1})\right)^2, $$ where the limit is taken over partitions: $0 = t^n_0 < t^n_1 < \cdots < t^n_n = t$, with $\delta_n = \max_{1 \leq i \leq n} \left(t^n_i - t^n_{i - 1}\right)$.

Klebaner goes on to remark that this definition is not the same as $$ \sup \sum_{i = 1}^n \left(g(t^n_i) - g(t^n_{i - 1})\right)^2 $$ where supremum is taken over all partitions.

I don't understand the highlighted definition. I feel that it is lacking in precision. If anyone understands it, as well as the distinction between it and the other definition, I would appreciate if you could explain it to me. Thanks.

  • 3
    $\begingroup$ If you understand the definition of Riemann integral, you should be able to understand this definition. For first-order variation, they are equal because of the triangle inequality. However, for quadratic case, no such triangle inequality can be applied, and then they are generally different. $\endgroup$ – Gordon May 17 '16 at 17:20
  • $\begingroup$ @Gordon: Thanks. This answers my question. $\endgroup$ – Evan Aad May 17 '16 at 17:50

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