Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The notions of first-order variation and total variation of a function or a stochastic process are equated in this book.

However, I found their definitions different from two other sources:

  1. In Wikipedia the total variation of a function $f$ from time $0$ to time $T$ is defined as $$ \sup_\Pi \sum_{i=0}^{n-1} | f(t_{i+1})-f(t_i) | . $$
  2. In Stochastic Calculus for Finance II: Continuous-time Models by Steve Shreve, the first-order variation of a function or stochastic process $f$ from time $0$ to time $T$ is defined as: $$ \lim_{||\Pi|| \to 0} \sum_{j=0}^{n-1} [f(t_{j+1}) - f(t_j)] $$ where $$\Pi=[t_0, t_1, \dots, t_n] \text{ and } 0 \leq t_0 < t_1 < \cdots < t_n = T$$ and $$||\Pi||= \max_{j=0,\dots, n-1} (t_{j+1} -t_j).$$

Are they two different definitions, or equivalent? Thanks!

share|improve this question
    
Equivalent. Which implication is unclear to you? –  Did Nov 26 '11 at 20:31
    
@Didier: Under what conditions on the function or stochastic process $f$, are they equivalent? How does $\sup$ become $\lim$? –  Ethan Nov 26 '11 at 20:37
    
Forget the stochastic side, this is a deterministic result. How does the total variation $V(f,\Pi)$ varies when $\Pi$ is refined? –  Did Nov 26 '11 at 20:40
1  
@Didier: I guess you expect me to say "non-decreasing"? I don't know why. Plus: can the relation between the partitions not be restricted to refinement? Does it require that $f$ is some kind of continuous? –  Ethan Nov 26 '11 at 20:50
1  
It is indeed just a deterministic result nothing to do with stochastic calculus. W.r.t. your questions: 1. it is non-decreasing because $$|f(t_{i+1})-f(t_{i})|\leq |f(t_{i+1})-f(t_{i+1/2})|+|f(t_{i+1/2})-f(t_i)|$$ where $t_{i+1/2}$ is a new point. $$ $$2. For partitions $\Pi_1$ and $\Pi_2$ there exists $\Pi= \Pi_1\cup\Pi_2$ which is refinement of both, so $$\|\Pi\|\leq\min(\|\Pi_1\|,\|\Pi_2\|)$$ and $$V(f,\Pi)\geq\max(V(f,\Pi_1),V(f,\Pi_2)).$$ –  Ilya Nov 27 '11 at 11:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.