Brownian Motion - Total/Quadratic Variation

[KingWithoutACrown]

What are the significance of these concepts?

[Captain Nemo]

Originally Posted by KingWithoutACrown

What are the significance of these concepts?

I'm not sure there's much the SOA can ask about this, except as a set of "conceptual" multiple-choice or true/false.

There are some interesting theorems in functional analysis regarding the variation of functions (for example, if a function has finite total variation over any finite interval, then it can be expressed as the difference of two monotone nondecreasing functions over that interval) but other than referencing "total variation", "quadratic variation", and "infinite crossing property", I don't think it's all that testable.

"Which of the following statements is true:

I. If Z(t) is Brownian motion, then Z(t) has finite expected quadratic variation.

II. If Z(t) is Brownian motion, then Z(t) has finite expected total variation.

III. If Z(t) is Brownian motion, and Z(3) = 2, then the expected number of values of t between 3 and 5 such that Z(t) = 2 is equal to 2^(0.5).

(A) I
(B) II
(C) I and III
(D) II and III
(E) None of the above.
"

(A)

[Londondrug]

anyone please explain a bit about II and III above?

[AAABBBCCC]

Hes talking about the expected measure of the zero set of brownian motion over a compact interval.

LOL

[Londondrug]

Originally Posted by AAABBBCCC

Hes talking about the expected measure of the zero set of brownian motion over a compact interval.

LOL

I know. But I have no clue how to judge II and III, especially III.

[AAABBBCCC]

which is zero, but I dont know what the expected Hausdorff dimension is.

[Londondrug]

For II, I guess is wrong, because the first order of variation of Z(t) is infinite.

[AAABBBCCC]

Originally Posted by Londondrug

I know. But I have no clue how to judge II and III, especially III.

statement III makes no sense (as it stands). it says how many values hit 2 and claims that it's the square root of 2. That doesnt make any sense.

Hes talking about the expected measure of the set of points {t: z(t)=2, 3<=t<=5}, which I believe is zero. It is just fubini's theorem.

[AAABBBCCC]

Originally Posted by Londondrug

For II, I guess is wrong, because the first order of variation of Z(t) is infinite.

Yes, it is wrong. Brownian paths are differentiable nowhere with probability one while functions of bounded variation are differentiable almost everywhere.

[Londondrug]

Many thanks, AAABBBCCC.
probably III is trying to say the variance of Z(5)-Z(3) is 2, then the sigma is 2^0.5, however, it has nothing to do with Quadratic Variation.