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

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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
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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
 
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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
 
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