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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.

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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