Two questions that perhaps I shouldn't be asking...I read the definition of the local time but I didn't get why the name. Can anyone give me some intuition of why the local time is actually called "local time"? As far as I understand for a Brownian motion L_t = int_0^t \delta(B_t) dt, with \delta Dirac's function, or L_t = |B_t|-\int_0^t sgn(B_t) dB_tAlso, I think that I remember that a compact set is one that is bounded and closed. I presume that when we talk of a function with compact support we mean that the support is a compact set. Now I read somewhere that functions on the real line (which is unbounded) that vanish at infinity have compact support. Is that so? If yes why is that?Thanks for your help,Kyriakos

local time is kind of occupation time with random clock d[X, X]_t, so I guess that's tells you a bit why, but I am not sure about the naming history though

You might want to have a look at the article in Wikipedia, which explains it wellhttp://en.wikipedia.org/wiki/Support_(mathematics)

Thanks antonio. Actually it was this article that caused some of the initial confusion.It says "the support of f is defined as the smallest closed subset of X outside of which f is zero", therefore for a continuous function over the real line, as R is clopen it will be its support, whether it vanishes at infinity or not. Right?Now it says for the compact support that "Functions with compact support in X are those with support that is a compact subset of X. For example, if X is the real line, they are examples of functions that vanish at infinity (and negative infinity)."A compact set is one that is closed and bounded isn't it? From Wikipedia again "In mathematics, a subset of Euclidean space Rn is called compact if it is closed and bounded. For example, in R, the closed unit interval [0, 1] is compact, but the set of integers Z is not (it is not bounded) and neither is the half-open interval [0, 1) (it is not closed)."The real line is not bounded, therefore it is not compact. Why then functions that vanish at infinity have compact support?Cheers,K

QuoteOriginally posted by: RezThanks antonio. Actually it was this article that caused some of the initial confusion.It says "the support of f is defined as the smallest closed subset of X outside of which f is zero", therefore for a continuous function over the real line, as R is clopen it will be its support, whether it vanishes at infinity or not. Right?Now it says for the compact support that "Functions with compact support in X are those with support that is a compact subset of X. For example, if X is the real line, they are examples of functions that vanish at infinity (and negative infinity)."The real line is not bounded, therefore it is not compact. Why then functions that vanish at infinity have compact support?Cheers,KYeah, but the next sentence there is "Indeed, they are special cases of such functions that must vanish at finite bounds".It's just awkwardly written. It's literally true that a function that vanishes outside of (a,b) is an "example of a function thatvanishes at infinity". So, yes, the functions vanish at infinity, but that does not imply that every function that vanishes atinfinity has compact support. regards,

QuoteOriginally posted by: RezTwo questions that perhaps I shouldn't be asking...I read the definition of the local time but I didn't get why the name. Can anyone give me some intuition of why the local time is actually called "local time"? As far as I understand for a Brownian motion L_t = int_0^t \delta(B_t) dt, with \delta Dirac's function, or L_t = |B_t|-\int_0^t sgn(B_t) dB_tAlso, I think that I remember that a compact set is one that is bounded and closed. I presume that when we talk of a function with compact support we mean that the support is a compact set. Now I read somewhere that functions on the real line (which is unbounded) that vanish at infinity have compact support. Is that so? If yes why is that?Thanks for your help,KyriakosIt seems to me that there is a bit of confusion in the definition of local time.In the book by Chung and Williams (1983 I think) the local time is defined as the limit:L(T,x) = lim(e->0):1/2e*int(t from 0 to T)(1{X(t) in [-e+x,e+x]})*dt.Obviously it is the density of occupancy time at some level x. In other words, the occupancy time of process X(t) in some spacial interval [a,b] is the integral of the local time in that interval:occupancy time in interval [a,b] = int(x from a to b)L(T,x)*dx.The concept of local time is closely related to the generalised Ito's formula which tackles functions with discontinuous derivatives. Harrison (1985) provides a complete discussion but I dont think you can buy it anywhere...ofc it is classic.

QuoteOriginally posted by: AlanYeah, but the next sentence there is "Indeed, they are special cases of such functions that must vanish at finite bounds".It's just awkwardly written. It's literally true that a function that vanishes outside of (a,b) is an "example of a function thatvanishes at infinity". So, yes, the functions vanish at infinity, but that does not imply that every function that vanishes atinfinity has compact support. That's great. I read it as 'all functions that vanish at infinity have compact support' but now I can see the truth ~). Then for example, the normal density function does not have compact support.K

I've been searching back in the books. If the support is closed, then we don't have any problem. If the support is the real line, apparently, you consider the closure by adding the two cemetery states + and - infinity, so that your support becomes compact. When the function vanishes at these two states.But I'm not 100% about this. I'll check further.

Concerning the normal distribution, it is definitely not compact on the real line, in the sense that R is unbounded. But you can somehow add the boundaries to R so as to make it compact, as said in the previous message. But brutely speaking, the first assertion is not true. In a different topology, it might indeed be right.Hope it helps