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I have come across the following unclear definition:

Consider $dS(t) = S(t)[\mu(t)dt + \sigma(t) dW(t)]$

"Assume that the coefficient $\sigma$ is Markovian. That is, (with abuse of notation) $\sigma(t) = \sigma(S(t))$."

I don't see what we are abusing here. Does anyone know what this actually means?

Thanks

user30201
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  • it seems we require more context to help you. (like what is $S(t)$) – rlartiga Apr 23 '14 at 02:09
  • I put your statement in google books, and well the unique abuse of notation I find is the fact S is not a function of the time is a stochastic process. – rlartiga Apr 23 '14 at 02:20

1 Answers1

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I think the abuse is that the notation $\sigma(t) = \sigma(S(t))$ would literally mean that $\sigma(t)$ and $\sigma(S(t))$ are equal, e.g. if $t=2$ and $S(t)=S(2)=3$, the value of function $\sigma$ evaluated at argument $2$ would be equal to the value of function $\sigma$ at argument $3$. However, this is surely not meant. Instead, I guess the notation is (technically incorrectly, i.e., abusingly) used to mean that $\sigma$ at time $t$ is determined by $S(t)$.

Juho Kokkala
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