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Integral Notation of an exponential Brownian motion

gnob

New member
Dec 11, 2012
11
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.
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.