Integral Notation of an exponential Brownian motion

In summary, the conversation discusses the unfamiliarity of equation (1.a) in Marc Yor's paper, which has the notation of $ds$ coming before the exponential part. The person asks for clarification on whether there is a difference compared to writing it the other way around. They also ask for references on this notation and how the scaling property applies to (1.a). The response confirms that the two notations mean the same thing and provides some book references where it is used.
  • #1
gnob
11
0
Good day!

I am reading the paper of Marc Yor (www.jstor.org/stable/1427477). equation (1.a) seems unfamiliar to me since the $ds$ comes first before the exponential part;
$$
\int_0^t ds \exp(aB_s + bs).
$$
Can you please help me clarify if there is a difference with the above notation as compared to if I write it this way:
$$
\int_0^t \exp(aB_s + bs) ds.
$$
Please give me some reference (books) on this. thanks

Secondly, how does the scaling property applied to (1.a) to become
$$
\int_0^t ds \exp 2(B_s + vs).
$$
Thanks a lot for your response. I know that the Brownian scaling states that if $B_s$ is a standard Brownian motion, then $\sqrt{c}B_{cs}$ is also a standard Brownian motion.
 
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  • #2
gnob said:
Good day!

I am reading the paper of Marc Yor (www.jstor.org/stable/1427477). equation (1.a) seems unfamiliar to me since the $ds$ comes first before the exponential part;
$$
\int_0^t ds \exp(aB_s + bs).
$$
Can you please help me clarify if there is a difference with the above notation as compared to if I write it this way:
$$
\int_0^t \exp(aB_s + bs) ds.
$$
Please give me some reference (books) on this. thanks

Hi gnob, :)

Yes they do mean the same thing. I have seen this notation used in quantum mechanics books such as,

1) Modern Quantum Mechanics by J. Sakurai

2) Quantum Physics by S. Gasiorowicz

3) Quantum Mechanics by C.C. Tannoudji

Also a brief description about the two notations can be found >>here<<.

Kind Regards,
Sudharaka.
 

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