Brownian motion on the n-sphere
From course notes on SDE's. We consider a Stratonovich equation.
$dX_t=left(I-frac{1}{|X_t|^2}X_tX_t^T right)circ dB_t$
With $X_tin mathbb{R}^n$ and ${B_t}$ being n-dimensional brownian motion. We wish to show that $|X_t|^2$ is constant along trajectories, so that there cannot exist a unique stationary trajectory.
Now, does the forward Kolmogorov equation work in the same way for a Stratonovich equation as it does the Ito case? If that is the case, how would I look at trajectories in general? Ergodic theory is not a part of the notes.
stochastic-processes stochastic-calculus brownian-motion stochastic-integrals sde
asked Nov 28 at 23:32
thaumoctopus
12818
add a comment |
From course notes on SDE's. We consider a Stratonovich equation.
$dX_t=left(I-frac{1}{|X_t|^2}X_tX_t^T right)circ dB_t$
With $X_tin mathbb{R}^n$ and ${B_t}$ being n-dimensional brownian motion. We wish to show that $|X_t|^2$ is constant along trajectories, so that there cannot exist a unique stationary trajectory.
Now, does the forward Kolmogorov equation work in the same way for a Stratonovich equation as it does the Ito case? If that is the case, how would I look at trajectories in general? Ergodic theory is not a part of the notes.
stochastic-processes stochastic-calculus brownian-motion stochastic-integrals sde
asked Nov 28 at 23:32
thaumoctopus
12818
add a comment |
From course notes on SDE's. We consider a Stratonovich equation.
$dX_t=left(I-frac{1}{|X_t|^2}X_tX_t^T right)circ dB_t$
With $X_tin mathbb{R}^n$ and ${B_t}$ being n-dimensional brownian motion. We wish to show that $|X_t|^2$ is constant along trajectories, so that there cannot exist a unique stationary trajectory.
Now, does the forward Kolmogorov equation work in the same way for a Stratonovich equation as it does the Ito case? If that is the case, how would I look at trajectories in general? Ergodic theory is not a part of the notes.
stochastic-processes stochastic-calculus brownian-motion stochastic-integrals sde
asked Nov 28 at 23:32
thaumoctopus
12818
From course notes on SDE's. We consider a Stratonovich equation.
$dX_t=left(I-frac{1}{|X_t|^2}X_tX_t^T right)circ dB_t$
With $X_tin mathbb{R}^n$ and ${B_t}$ being n-dimensional brownian motion. We wish to show that $|X_t|^2$ is constant along trajectories, so that there cannot exist a unique stationary trajectory.
Now, does the forward Kolmogorov equation work in the same way for a Stratonovich equation as it does the Ito case? If that is the case, how would I look at trajectories in general? Ergodic theory is not a part of the notes.
stochastic-processes stochastic-calculus brownian-motion stochastic-integrals sde
stochastic-processes stochastic-calculus brownian-motion stochastic-integrals sde
asked Nov 28 at 23:32
thaumoctopus
12818
asked Nov 28 at 23:32
thaumoctopus
12818
asked Nov 28 at 23:32
thaumoctopus
12818
asked Nov 28 at 23:32
thaumoctopus
12818
asked Nov 28 at 23:32
thaumoctopus
12818
12818
add a comment |
add a comment |
1 Answer
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That $|X_t|^2$ is constant can be seen as follows:
begin{align}
d(|X_t|^2)
& = 2X_t^Tcirc dX_t\
& = 2left(X_t^T-frac{1}{|X_t|^2}X_t^TX_tX_t^Tright)circ dB_t\
& = 0.
end{align}
(I used $P_tcirc (Q_tcirc dB_t)=(P_tQ_t)circ d B_t$.)
answered Nov 29 at 6:42
AddSup
361211
Wow, thank you, should have remembered that it more or less just reduces to normal calculus for the Stratonovich case :)
– thaumoctopus
Nov 29 at 14:17
@thaumoctopus You're welcome ;)
– AddSup
Nov 30 at 6:22
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
That $|X_t|^2$ is constant can be seen as follows:
begin{align}
d(|X_t|^2)
& = 2X_t^Tcirc dX_t\
& = 2left(X_t^T-frac{1}{|X_t|^2}X_t^TX_tX_t^Tright)circ dB_t\
& = 0.
end{align}
(I used $P_tcirc (Q_tcirc dB_t)=(P_tQ_t)circ d B_t$.)
answered Nov 29 at 6:42
AddSup
361211
Wow, thank you, should have remembered that it more or less just reduces to normal calculus for the Stratonovich case :)
– thaumoctopus
Nov 29 at 14:17
@thaumoctopus You're welcome ;)
– AddSup
Nov 30 at 6:22
add a comment |
That $|X_t|^2$ is constant can be seen as follows:
begin{align}
d(|X_t|^2)
& = 2X_t^Tcirc dX_t\
& = 2left(X_t^T-frac{1}{|X_t|^2}X_t^TX_tX_t^Tright)circ dB_t\
& = 0.
end{align}
(I used $P_tcirc (Q_tcirc dB_t)=(P_tQ_t)circ d B_t$.)
answered Nov 29 at 6:42
AddSup
361211
Wow, thank you, should have remembered that it more or less just reduces to normal calculus for the Stratonovich case :)
– thaumoctopus
Nov 29 at 14:17
@thaumoctopus You're welcome ;)
– AddSup
Nov 30 at 6:22
add a comment |
That $|X_t|^2$ is constant can be seen as follows:
begin{align}
d(|X_t|^2)
& = 2X_t^Tcirc dX_t\
& = 2left(X_t^T-frac{1}{|X_t|^2}X_t^TX_tX_t^Tright)circ dB_t\
& = 0.
end{align}
(I used $P_tcirc (Q_tcirc dB_t)=(P_tQ_t)circ d B_t$.)
answered Nov 29 at 6:42
AddSup
361211
That $|X_t|^2$ is constant can be seen as follows:
begin{align}
d(|X_t|^2)
& = 2X_t^Tcirc dX_t\
& = 2left(X_t^T-frac{1}{|X_t|^2}X_t^TX_tX_t^Tright)circ dB_t\
& = 0.
end{align}
(I used $P_tcirc (Q_tcirc dB_t)=(P_tQ_t)circ d B_t$.)
answered Nov 29 at 6:42
AddSup
361211
edited Nov 29 at 6:48
answered Nov 29 at 6:42
AddSup
361211
answered Nov 29 at 6:42
AddSup
361211
answered Nov 29 at 6:42
AddSup
361211
361211
Wow, thank you, should have remembered that it more or less just reduces to normal calculus for the Stratonovich case :)
– thaumoctopus
Nov 29 at 14:17
@thaumoctopus You're welcome ;)
– AddSup
Nov 30 at 6:22
add a comment |
Wow, thank you, should have remembered that it more or less just reduces to normal calculus for the Stratonovich case :)
– thaumoctopus
Nov 29 at 14:17
@thaumoctopus You're welcome ;)
– AddSup
Nov 30 at 6:22
Wow, thank you, should have remembered that it more or less just reduces to normal calculus for the Stratonovich case :)
– thaumoctopus
Nov 29 at 14:17
Wow, thank you, should have remembered that it more or less just reduces to normal calculus for the Stratonovich case :)
– thaumoctopus
Nov 29 at 14:17
@thaumoctopus You're welcome ;)
– AddSup
Nov 30 at 6:22
@thaumoctopus You're welcome ;)
– AddSup
Nov 30 at 6:22
add a comment |
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