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In my experience $\wedge$ has something to do with the outer product, but I am not sure what it means when $t$ and $s$ are not vectors and the book I am reading does not explain it. I thought maybe it means $min\{s,t\}$ but before the book used it it had already specified that $s<t$, so we already know a priori what the minimum is.

MathStudent
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  • There's no universally agreed on meaning to that. I don't recall ever having seen $\wedge$ used with vectors. – Henrik supports the community Sep 01 '15 at 15:24
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    Like you said, usually (at least in probability theory from what I've seen), between two numbers it has meant their minimum. What source are you using where you found this? – layman Sep 01 '15 at 15:24
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    Could you give some more context? Usually $\wedge$ denotes the minimum. – Dominik Sep 01 '15 at 15:24
  • Well I have also seen it refer to the smallest sigma algebra containing both of those sigma algebras, which is why I thought it meant the minimum – MathStudent Sep 01 '15 at 15:24
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    I would imagine it means $\min(s,t)$ regardless. Does everything else make sense on that interpretation? – David C. Ullrich Sep 01 '15 at 15:24
  • It is in a book on Fractional Brownian Motion. It doesn't... not make sense, except that it already specified $s<t$, which confused me. If that is the generally accepted definition than I guess the book may have just had a tad bit of inconsistency there. – MathStudent Sep 01 '15 at 15:27
  • @MathStudent If you want a more definitive answer, maybe give us the title of the book and which page the symbol is used. One of us might have the book and can check out the context. – layman Sep 01 '15 at 15:33
  • Stochastic Calculus for Fractional Brownian Motion and Applications pg. 25 – MathStudent Sep 01 '15 at 15:40

1 Answers1

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I was able to find a free (legal) preview of the book by searching for it on Amazon -- the preview includes page 25. I see why you are confused. I don't know if this is the right answer, but here is how I am understanding it.

They assumed $t > s$ when they defined the kernel $K_{H}(t,s)$. And then they defined another function afterward:

$R_{H}(t,s) = \displaystyle \int_{0}^{t \land s} K_{H}(t,u)K_{H}(s,u) \,du$.

I think that the formula they gave for $K_{H}(t,s)$ only holds if, when $t$ is fixed, the input $s$ is smaller than $t$. And you can see in the definition of $R_{H}(t,s)$ that it makes sense. In that definition, for fixed $t, s$, since we are integrating with respect to $u$ over $[0, t \land s]$, then each $u$ is already smaller than $t$, so the first factor in the product, $K_{H}(t,u)$ makes sense by the formula they gave. Also, each $u$ is smaller than $s$, so the second factor in the product, $K_{H}(s,u)$, make sense by the formula they gave.

So, if $t$ is fixed, then for each $s < t$, $K_{H}(t,s) = $ that formula they gave. Then for any $s, t$ (no matter which is bigger), $R_{H}(t,s)$ is the function that sends $(t,s)$ to the integral (w.r.t. $u$) over $[0, t\land s]$ of the product $K_{H}(t,u)K_{H}(s,u)$, and each of the factors in this product are defined by the $K_{H}$ formula since $u$ is always smaller than $t$ and $s$ (since it's smaller than $t \land s$ if $\land$ is the min symbol).

layman
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