Constructing Random Diffusion












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Is it possible to define the following in a sensefull way?



Let $ F:mathbb R to [-1,1]$ a random function so that the family $left(F(x)right)_{xinmathbb R}$ is i.i.d. Now let $$ dX_t = dB_t + F(X_t) , dt, quad X_0=0$$ a diffusion process with standard Brownian motion $B_t$. Is it then true, that $X_t$ never gets trapped? I think of $X_t$ being the continuous limit of a 1d random walk in random environment. Does that make sense at all?










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    Is it possible to define the following in a sensefull way?



    Let $ F:mathbb R to [-1,1]$ a random function so that the family $left(F(x)right)_{xinmathbb R}$ is i.i.d. Now let $$ dX_t = dB_t + F(X_t) , dt, quad X_0=0$$ a diffusion process with standard Brownian motion $B_t$. Is it then true, that $X_t$ never gets trapped? I think of $X_t$ being the continuous limit of a 1d random walk in random environment. Does that make sense at all?










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      Is it possible to define the following in a sensefull way?



      Let $ F:mathbb R to [-1,1]$ a random function so that the family $left(F(x)right)_{xinmathbb R}$ is i.i.d. Now let $$ dX_t = dB_t + F(X_t) , dt, quad X_0=0$$ a diffusion process with standard Brownian motion $B_t$. Is it then true, that $X_t$ never gets trapped? I think of $X_t$ being the continuous limit of a 1d random walk in random environment. Does that make sense at all?










      share|cite|improve this question













      Is it possible to define the following in a sensefull way?



      Let $ F:mathbb R to [-1,1]$ a random function so that the family $left(F(x)right)_{xinmathbb R}$ is i.i.d. Now let $$ dX_t = dB_t + F(X_t) , dt, quad X_0=0$$ a diffusion process with standard Brownian motion $B_t$. Is it then true, that $X_t$ never gets trapped? I think of $X_t$ being the continuous limit of a 1d random walk in random environment. Does that make sense at all?







      stochastic-processes






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      asked Nov 29 '18 at 1:30









      maliesen

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