Mixing
Existence of a unique solution of $dX_t=X_tB_tdB_t+X_tB_tdt, X_0=1$
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Does this SDE have a unique solution:
$$dX_t=X_tB_tdB_t+X_tB_tdt, quad X_0=1?$$
I have to check if the Lipschitz condition holds and if the growth condition holds.
I presume not, as the Brownian motion can take any value (it is unbounded, so the growth and the Lipschitz condition are not satisfied. But how can I formally prove this, if my intuition is indeed correct?
How can one generally prove/disprove the conditions for existence and uniqueness, besides showing if it is differentiable?
brownian-motion stochastic-pde
asked Jan 28 at 18:27
RavonripRavonrip
898
$endgroup$
add a comment |
$begingroup$
Does this SDE have a unique solution:
$$dX_t=X_tB_tdB_t+X_tB_tdt, quad X_0=1?$$
I have to check if the Lipschitz condition holds and if the growth condition holds.
I presume not, as the Brownian motion can take any value (it is unbounded, so the growth and the Lipschitz condition are not satisfied. But how can I formally prove this, if my intuition is indeed correct?
How can one generally prove/disprove the conditions for existence and uniqueness, besides showing if it is differentiable?
brownian-motion stochastic-pde
asked Jan 28 at 18:27
RavonripRavonrip
898
$endgroup$
$begingroup$
The classical Lipschitz conditions are for SDEs of the form $$dX_t = b(t,X_t) ,dt + sigma(t,X_t) , dB_t$$ where $b$ and $sigma$ are deterministic functions. Note that your coefficients are not deterministic and, hence, the classical results do not apply.
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– saz
Jan 28 at 18:44
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Ah, I see, thank you. And how can I then prove whether or not a unique solution exists?
$endgroup$
– Ravonrip
Jan 28 at 18:56
$begingroup$
Well, there are general uniqueness results for SDEs with random coefficients; I'm just saying that you should be careful which kind of results are applicable.
$endgroup$
– saz
Jan 28 at 19:22
add a comment |
$begingroup$
Does this SDE have a unique solution:
$$dX_t=X_tB_tdB_t+X_tB_tdt, quad X_0=1?$$
I have to check if the Lipschitz condition holds and if the growth condition holds.
I presume not, as the Brownian motion can take any value (it is unbounded, so the growth and the Lipschitz condition are not satisfied. But how can I formally prove this, if my intuition is indeed correct?
How can one generally prove/disprove the conditions for existence and uniqueness, besides showing if it is differentiable?
brownian-motion stochastic-pde
asked Jan 28 at 18:27
RavonripRavonrip
898
$endgroup$
Does this SDE have a unique solution:
$$dX_t=X_tB_tdB_t+X_tB_tdt, quad X_0=1?$$
I have to check if the Lipschitz condition holds and if the growth condition holds.
I presume not, as the Brownian motion can take any value (it is unbounded, so the growth and the Lipschitz condition are not satisfied. But how can I formally prove this, if my intuition is indeed correct?
How can one generally prove/disprove the conditions for existence and uniqueness, besides showing if it is differentiable?
brownian-motion stochastic-pde
brownian-motion stochastic-pde
asked Jan 28 at 18:27
RavonripRavonrip
898
asked Jan 28 at 18:27
RavonripRavonrip
898
asked Jan 28 at 18:27
RavonripRavonrip
898
asked Jan 28 at 18:27
RavonripRavonrip
898
asked Jan 28 at 18:27
RavonripRavonrip
898
898
$begingroup$
The classical Lipschitz conditions are for SDEs of the form $$dX_t = b(t,X_t) ,dt + sigma(t,X_t) , dB_t$$ where $b$ and $sigma$ are deterministic functions. Note that your coefficients are not deterministic and, hence, the classical results do not apply.
$endgroup$
– saz
Jan 28 at 18:44
$begingroup$
Ah, I see, thank you. And how can I then prove whether or not a unique solution exists?
$endgroup$
– Ravonrip
Jan 28 at 18:56
$begingroup$
Well, there are general uniqueness results for SDEs with random coefficients; I'm just saying that you should be careful which kind of results are applicable.
$endgroup$
– saz
Jan 28 at 19:22
add a comment |
$begingroup$
The classical Lipschitz conditions are for SDEs of the form $$dX_t = b(t,X_t) ,dt + sigma(t,X_t) , dB_t$$ where $b$ and $sigma$ are deterministic functions. Note that your coefficients are not deterministic and, hence, the classical results do not apply.
$endgroup$
– saz
Jan 28 at 18:44
$begingroup$
Ah, I see, thank you. And how can I then prove whether or not a unique solution exists?
$endgroup$
– Ravonrip
Jan 28 at 18:56
$begingroup$
Well, there are general uniqueness results for SDEs with random coefficients; I'm just saying that you should be careful which kind of results are applicable.
$endgroup$
– saz
Jan 28 at 19:22
$begingroup$
The classical Lipschitz conditions are for SDEs of the form $$dX_t = b(t,X_t) ,dt + sigma(t,X_t) , dB_t$$ where $b$ and $sigma$ are deterministic functions. Note that your coefficients are not deterministic and, hence, the classical results do not apply.
$endgroup$
– saz
Jan 28 at 18:44
$begingroup$
The classical Lipschitz conditions are for SDEs of the form $$dX_t = b(t,X_t) ,dt + sigma(t,X_t) , dB_t$$ where $b$ and $sigma$ are deterministic functions. Note that your coefficients are not deterministic and, hence, the classical results do not apply.
$endgroup$
– saz
Jan 28 at 18:44
$begingroup$
Ah, I see, thank you. And how can I then prove whether or not a unique solution exists?
$endgroup$
– Ravonrip
Jan 28 at 18:56
$begingroup$
Ah, I see, thank you. And how can I then prove whether or not a unique solution exists?
$endgroup$
– Ravonrip
Jan 28 at 18:56
$begingroup$
Well, there are general uniqueness results for SDEs with random coefficients; I'm just saying that you should be careful which kind of results are applicable.
$endgroup$
– saz
Jan 28 at 19:22
$begingroup$
Well, there are general uniqueness results for SDEs with random coefficients; I'm just saying that you should be careful which kind of results are applicable.
$endgroup$
– saz
Jan 28 at 19:22
add a comment |
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$begingroup$
The classical Lipschitz conditions are for SDEs of the form $$dX_t = b(t,X_t) ,dt + sigma(t,X_t) , dB_t$$ where $b$ and $sigma$ are deterministic functions. Note that your coefficients are not deterministic and, hence, the classical results do not apply.
$endgroup$
– saz
Jan 28 at 18:44
$begingroup$
Ah, I see, thank you. And how can I then prove whether or not a unique solution exists?
$endgroup$
– Ravonrip
Jan 28 at 18:56
$begingroup$
Well, there are general uniqueness results for SDEs with random coefficients; I'm just saying that you should be careful which kind of results are applicable.
$endgroup$
– saz
Jan 28 at 19:22