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Meaning of $H_{mathbb{R}}(W)$

For $X_{t}=expleft{left(mu-r-frac{sigma^{2}}{2}right)t+sigma W_{t}right}$, do we have $mathbb{E}[int_{0}^{tau_{b}}X_{s}dW_{s}]=0$?How do you make dependent Brownian motions independent?Sum of two Markov processes another Markov process?Ornstein-Uhlenbeck process written explicitlyOrnstein-Uhlenbeck process: incrementsBrownian motion / ito's formulaGeometric Brownian Motion PropertiesProof that time integral of OU process is not Markov while together with the integrand it is.Product of Two Correlated Brownian Motionsderivative of integral of Brownian motion

0

$begingroup$

Let $W_{t}$ be Brownian motion. It is said that if $X_{t}$ is an Ornstein–Uhlenbec process of form $X_{t} = sqrt{2alpha} int^{t}_{-infty}e^{-alpha(t-s)}dW_{s}, X_{t}$ then belongs to $H_{mathbb{R}}(W)$. What is $H_{mathbb{R}}(W)$?

real-analysis stochastic-calculus

share|cite|improve this question

asked Mar 14 at 20:10

BayesIsBaeBayesIsBae

697

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Without any context your question is very difficult to answer. Where did you encounter this statement? (In which book/article/...?)
  $endgroup$
  – saz
  Mar 14 at 20:50
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It was a statement by a professor. Some research leads me to believe it could be the "Hilbert space generated by $W_{t}$" but I am unsure of what this means.
  $endgroup$
  – BayesIsBae
  Mar 14 at 20:59
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $W_{t}$ be Brownian motion. It is said that if $X_{t}$ is an Ornstein–Uhlenbec process of form $X_{t} = sqrt{2alpha} int^{t}_{-infty}e^{-alpha(t-s)}dW_{s}, X_{t}$ then belongs to $H_{mathbb{R}}(W)$. What is $H_{mathbb{R}}(W)$?

real-analysis stochastic-calculus

share|cite|improve this question

asked Mar 14 at 20:10

BayesIsBaeBayesIsBae

697

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Without any context your question is very difficult to answer. Where did you encounter this statement? (In which book/article/...?)
  $endgroup$
  – saz
  Mar 14 at 20:50
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It was a statement by a professor. Some research leads me to believe it could be the "Hilbert space generated by $W_{t}$" but I am unsure of what this means.
  $endgroup$
  – BayesIsBae
  Mar 14 at 20:59
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $W_{t}$ be Brownian motion. It is said that if $X_{t}$ is an Ornstein–Uhlenbec process of form $X_{t} = sqrt{2alpha} int^{t}_{-infty}e^{-alpha(t-s)}dW_{s}, X_{t}$ then belongs to $H_{mathbb{R}}(W)$. What is $H_{mathbb{R}}(W)$?

real-analysis stochastic-calculus

share|cite|improve this question

asked Mar 14 at 20:10

BayesIsBaeBayesIsBae

697

$endgroup$

share|cite|improve this question

asked Mar 14 at 20:10

BayesIsBaeBayesIsBae

697

share|cite|improve this question

asked Mar 14 at 20:10

BayesIsBaeBayesIsBae

697

asked Mar 14 at 20:10

BayesIsBaeBayesIsBae

697

asked Mar 14 at 20:10

BayesIsBaeBayesIsBae

697