MHB What is the probability density for a given exponential functional integral?

AI Thread Summary
The discussion focuses on determining the probability density of the integral $$\int_0^t h(s) e^{2\beta(\mu(s) + W_s)}$$ where $\beta > 0$, and $h, \mu$ are continuous functions with $h \geq 0$, while $W$ represents standard Brownian motion. The original poster seeks clarification on the term "law," which refers to the probability density of this integral. They reference a known case where $\mu(s) = -\nu s$ and $h(s) = 1$, citing a result from Marc Yor's book on exponential functionals of Brownian motion. The thread emphasizes the need for further insights into the probability density for the general case presented.
gnob
Messages
11
Reaction score
0
Good day!
I have a question regarding the law of the ff:
$$
\int_0^t h(s) e^{2\beta(\mu(s) + W_s)}
$$
where $\beta >0;$ $h,\mu$ are continuous functions on $\mathbb{R}_+$ with $h\geq 0;$
and $W=\{W_s,s\geq 0\}$ is a standard Brownian motion.

Thanks for any help.:D
 
Physics news on Phys.org
gnob said:
Good day!
I have a question regarding the law of the ff:
$$
\int_0^t h(s) e^{2\beta(\mu(s) + W_s)}
$$
where $\beta >0;$ $h,\mu$ are continuous functions on $\mathbb{R}_+$ with $h\geq 0;$
and $W=\{W_s,s\geq 0\}$ is a standard Brownian motion.

Thanks for any help.:D

Please, can You better explain what is the question You have?...

Kind regards

$\chi$ $\sigma$
 
gnob said:
Good day!
I have a question regarding the law of the ff:
$$
\int_0^t h(s) e^{2\beta(\mu(s) + W_s)}
$$
where $\beta >0;$ $h,\mu$ are continuous functions on $\mathbb{R}_+$ with $h\geq 0;$
and $W=\{W_s,s\geq 0\}$ is a standard Brownian motion.

Thanks for any help.:D

Many thanks for the reply. What I meant of "law" is the probability density of the given integral. For the case $\mu(s) = -\nu s$ where $\nu$ is a positive constant and $h(s)=1,$ the law was already known (Corollary 1.2, p95) from Mar Yor's book given here Exponential Functionals of Brownian Motion and Related Processes - Marc Yor - Google Books .

Thanks again for any insights. :D
 
Back
Top