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Sensitivity of path integral w.r.t. initial state
The Next CEO of Stack OverflowCompute partial derivative w.r.t initial conditionDerivative of terminal state w.r.t. the inital conditions.Inseparable terminal and running cost in optimal control problems?Time-dependent inequalities in optimal controllerOptimal control with state dependent action spaceHow to numerically solve an optimal control problem with an indirect method (in MATLAB)?Help with an optimal control problemCalculating the control energy of a non-optimal prespecified trajectory of an LTIInitial guess generation for solving Nonlinear programming problemsTime-Discretize a Linear Quadratic OCP (Bolza Function)Optimal control problem with fixed endpoint; what I am doing wrong?Optimal control problem with a path constraint which involves controls at two distinct time points
$begingroup$
I'm trying to figure out how the value of a path integral depends on the initial condition (starting point) of the path.
Part1:
Consider the state variable $x(t)$ that is governed by an ODE
$$dotx(t) = f(x(t))$$
with some initial condition. This defines our path in time.
Let's say I have a path integral, for example a cost-to-go $C(x_t,t)$ that evaluates how much cost will be accumulated until a final time $t_f$ for a path starting at $x_t$ at time $t$ and governed by the ODE above:
$$ C(x_t,t) = int_t^t_f g(x(tau)) ; dtau qquad textgiven: x(t)=x_t, ; dotx(t) = f(x(t))$$
Now my question: What is $partial C(x_t,t) over partial x_t $, i.e., the sensitivity of the cost-to-go w.r.t. the starting state?
I'm having troubles figuring this out because I don't know how to formalize the dependence of future states on $x_t$.
Part2:
How does the answer change when we consider a stochastic path (random walk) instead
$$dx = f(x,t) dt + sigma(x,t) dw$$
where $dw$ is standard Brownian motion and the cost-to-go is now an expectation over all paths that start at $x_t$ at time $t$:
$$ C(x_t,t) = mathbb E_x_t left[ int_t^t_f g(x(tau)) ; dtau right] qquad textgiven: x(t)=x_t $$
I've tried to apply some Ito calculus rules here but I'm getting stuck again...
Possibly related questions:
Derivative of terminal state w.r.t. the inital conditions
Compute partial derivative w.r.t initial condition
partial-derivative stochastic-integrals optimal-control
asked 2 days ago
Jan CaJan Ca
111
New contributor
Jan Ca is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 1 more comment
$begingroup$
I'm trying to figure out how the value of a path integral depends on the initial condition (starting point) of the path.
Part1:
Consider the state variable $x(t)$ that is governed by an ODE
$$dotx(t) = f(x(t))$$
with some initial condition. This defines our path in time.
Let's say I have a path integral, for example a cost-to-go $C(x_t,t)$ that evaluates how much cost will be accumulated until a final time $t_f$ for a path starting at $x_t$ at time $t$ and governed by the ODE above:
$$ C(x_t,t) = int_t^t_f g(x(tau)) ; dtau qquad textgiven: x(t)=x_t, ; dotx(t) = f(x(t))$$
Now my question: What is $partial C(x_t,t) over partial x_t $, i.e., the sensitivity of the cost-to-go w.r.t. the starting state?
I'm having troubles figuring this out because I don't know how to formalize the dependence of future states on $x_t$.
Part2:
How does the answer change when we consider a stochastic path (random walk) instead
$$dx = f(x,t) dt + sigma(x,t) dw$$
where $dw$ is standard Brownian motion and the cost-to-go is now an expectation over all paths that start at $x_t$ at time $t$:
$$ C(x_t,t) = mathbb E_x_t left[ int_t^t_f g(x(tau)) ; dtau right] qquad textgiven: x(t)=x_t $$
I've tried to apply some Ito calculus rules here but I'm getting stuck again...
Possibly related questions:
Derivative of terminal state w.r.t. the inital conditions
Compute partial derivative w.r.t initial condition
partial-derivative stochastic-integrals optimal-control
asked 2 days ago
Jan CaJan Ca
111
New contributor
Jan Ca is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
The first part looks more like calculus of variations than stochastic stuff. Do you know the form of $g?$
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
yep, the first part is not stochastic, but the ODE for $x(t)$ is already known, so I think this is not a variational problem. $g$ can be an arbitrary continuous and differentiable function in my problem
$endgroup$
– Jan Ca
2 days ago
$begingroup$
If $x$ is already determined by the ODE and therefore fixed, then I'm not sure I see the point in calculating that sensitivity. It's really only going to be sensitive to changes in $g,$ since that's all you're allowing to vary. You can probably use the chain rule and the Euler-Lagrange equations to get at a solution to the first problem.
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
I'm varying the initial state $x_t$. $g$ is fixed.
$endgroup$
– Jan Ca
2 days ago
$begingroup$
For part 1, you might be forced to determine the sensitivities numerically. Since $C = int g dt$, if you can't solve the integral analytically then it'd be easiest to evaluate $C$ by quadrature over $x(t)$. Then you can determine $dC/dx(t_0)$ by finite difference wrapped around the quadrature. This isn't variational calculus since $g$ is only dependent on $x$ and $t$. That is, there is no coordinate $h$ such that $h' = g(x,h)$. I have no idea how to extend this for part 2, however. If this is a suitable answer, let me know and I can write it out in an actual answer with more detail.
$endgroup$
– Michael Sparapany
2 days ago
|
show 1 more comment
$begingroup$
I'm trying to figure out how the value of a path integral depends on the initial condition (starting point) of the path.
Part1:
Consider the state variable $x(t)$ that is governed by an ODE
$$dotx(t) = f(x(t))$$
with some initial condition. This defines our path in time.
Let's say I have a path integral, for example a cost-to-go $C(x_t,t)$ that evaluates how much cost will be accumulated until a final time $t_f$ for a path starting at $x_t$ at time $t$ and governed by the ODE above:
$$ C(x_t,t) = int_t^t_f g(x(tau)) ; dtau qquad textgiven: x(t)=x_t, ; dotx(t) = f(x(t))$$
Now my question: What is $partial C(x_t,t) over partial x_t $, i.e., the sensitivity of the cost-to-go w.r.t. the starting state?
I'm having troubles figuring this out because I don't know how to formalize the dependence of future states on $x_t$.
Part2:
How does the answer change when we consider a stochastic path (random walk) instead
$$dx = f(x,t) dt + sigma(x,t) dw$$
where $dw$ is standard Brownian motion and the cost-to-go is now an expectation over all paths that start at $x_t$ at time $t$:
$$ C(x_t,t) = mathbb E_x_t left[ int_t^t_f g(x(tau)) ; dtau right] qquad textgiven: x(t)=x_t $$
I've tried to apply some Ito calculus rules here but I'm getting stuck again...
Possibly related questions:
Derivative of terminal state w.r.t. the inital conditions
Compute partial derivative w.r.t initial condition
partial-derivative stochastic-integrals optimal-control
asked 2 days ago
Jan CaJan Ca
111
New contributor
Jan Ca is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm trying to figure out how the value of a path integral depends on the initial condition (starting point) of the path.
Part1:
Consider the state variable $x(t)$ that is governed by an ODE
$$dotx(t) = f(x(t))$$
with some initial condition. This defines our path in time.
Let's say I have a path integral, for example a cost-to-go $C(x_t,t)$ that evaluates how much cost will be accumulated until a final time $t_f$ for a path starting at $x_t$ at time $t$ and governed by the ODE above:
$$ C(x_t,t) = int_t^t_f g(x(tau)) ; dtau qquad textgiven: x(t)=x_t, ; dotx(t) = f(x(t))$$
Now my question: What is $partial C(x_t,t) over partial x_t $, i.e., the sensitivity of the cost-to-go w.r.t. the starting state?
I'm having troubles figuring this out because I don't know how to formalize the dependence of future states on $x_t$.
Part2:
How does the answer change when we consider a stochastic path (random walk) instead
$$dx = f(x,t) dt + sigma(x,t) dw$$
where $dw$ is standard Brownian motion and the cost-to-go is now an expectation over all paths that start at $x_t$ at time $t$:
$$ C(x_t,t) = mathbb E_x_t left[ int_t^t_f g(x(tau)) ; dtau right] qquad textgiven: x(t)=x_t $$
I've tried to apply some Ito calculus rules here but I'm getting stuck again...
Possibly related questions:
Derivative of terminal state w.r.t. the inital conditions
Compute partial derivative w.r.t initial condition
partial-derivative stochastic-integrals optimal-control
partial-derivative stochastic-integrals optimal-control
asked 2 days ago
Jan CaJan Ca
111
New contributor
Jan Ca is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Jan CaJan Ca
111
New contributor
Jan Ca is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Jan CaJan Ca
111
New contributor
Jan Ca is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Jan CaJan Ca
111
asked 2 days ago
Jan CaJan Ca
111
111
New contributor
Jan Ca is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jan Ca is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jan Ca is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
The first part looks more like calculus of variations than stochastic stuff. Do you know the form of $g?$
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
yep, the first part is not stochastic, but the ODE for $x(t)$ is already known, so I think this is not a variational problem. $g$ can be an arbitrary continuous and differentiable function in my problem
$endgroup$
– Jan Ca
2 days ago
$begingroup$
If $x$ is already determined by the ODE and therefore fixed, then I'm not sure I see the point in calculating that sensitivity. It's really only going to be sensitive to changes in $g,$ since that's all you're allowing to vary. You can probably use the chain rule and the Euler-Lagrange equations to get at a solution to the first problem.
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
I'm varying the initial state $x_t$. $g$ is fixed.
$endgroup$
– Jan Ca
2 days ago
$begingroup$
For part 1, you might be forced to determine the sensitivities numerically. Since $C = int g dt$, if you can't solve the integral analytically then it'd be easiest to evaluate $C$ by quadrature over $x(t)$. Then you can determine $dC/dx(t_0)$ by finite difference wrapped around the quadrature. This isn't variational calculus since $g$ is only dependent on $x$ and $t$. That is, there is no coordinate $h$ such that $h' = g(x,h)$. I have no idea how to extend this for part 2, however. If this is a suitable answer, let me know and I can write it out in an actual answer with more detail.
$endgroup$
– Michael Sparapany
2 days ago
|
show 1 more comment
$begingroup$
The first part looks more like calculus of variations than stochastic stuff. Do you know the form of $g?$
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
yep, the first part is not stochastic, but the ODE for $x(t)$ is already known, so I think this is not a variational problem. $g$ can be an arbitrary continuous and differentiable function in my problem
$endgroup$
– Jan Ca
2 days ago
$begingroup$
If $x$ is already determined by the ODE and therefore fixed, then I'm not sure I see the point in calculating that sensitivity. It's really only going to be sensitive to changes in $g,$ since that's all you're allowing to vary. You can probably use the chain rule and the Euler-Lagrange equations to get at a solution to the first problem.
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
I'm varying the initial state $x_t$. $g$ is fixed.
$endgroup$
– Jan Ca
2 days ago
$begingroup$
For part 1, you might be forced to determine the sensitivities numerically. Since $C = int g dt$, if you can't solve the integral analytically then it'd be easiest to evaluate $C$ by quadrature over $x(t)$. Then you can determine $dC/dx(t_0)$ by finite difference wrapped around the quadrature. This isn't variational calculus since $g$ is only dependent on $x$ and $t$. That is, there is no coordinate $h$ such that $h' = g(x,h)$. I have no idea how to extend this for part 2, however. If this is a suitable answer, let me know and I can write it out in an actual answer with more detail.
$endgroup$
– Michael Sparapany
2 days ago
$begingroup$
The first part looks more like calculus of variations than stochastic stuff. Do you know the form of $g?$
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
The first part looks more like calculus of variations than stochastic stuff. Do you know the form of $g?$
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
yep, the first part is not stochastic, but the ODE for $x(t)$ is already known, so I think this is not a variational problem. $g$ can be an arbitrary continuous and differentiable function in my problem
$endgroup$
– Jan Ca
2 days ago
$begingroup$
yep, the first part is not stochastic, but the ODE for $x(t)$ is already known, so I think this is not a variational problem. $g$ can be an arbitrary continuous and differentiable function in my problem
$endgroup$
– Jan Ca
2 days ago
$begingroup$
If $x$ is already determined by the ODE and therefore fixed, then I'm not sure I see the point in calculating that sensitivity. It's really only going to be sensitive to changes in $g,$ since that's all you're allowing to vary. You can probably use the chain rule and the Euler-Lagrange equations to get at a solution to the first problem.
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
If $x$ is already determined by the ODE and therefore fixed, then I'm not sure I see the point in calculating that sensitivity. It's really only going to be sensitive to changes in $g,$ since that's all you're allowing to vary. You can probably use the chain rule and the Euler-Lagrange equations to get at a solution to the first problem.
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
I'm varying the initial state $x_t$. $g$ is fixed.
$endgroup$
– Jan Ca
2 days ago
$begingroup$
I'm varying the initial state $x_t$. $g$ is fixed.
$endgroup$
– Jan Ca
2 days ago
$begingroup$
For part 1, you might be forced to determine the sensitivities numerically. Since $C = int g dt$, if you can't solve the integral analytically then it'd be easiest to evaluate $C$ by quadrature over $x(t)$. Then you can determine $dC/dx(t_0)$ by finite difference wrapped around the quadrature. This isn't variational calculus since $g$ is only dependent on $x$ and $t$. That is, there is no coordinate $h$ such that $h' = g(x,h)$. I have no idea how to extend this for part 2, however. If this is a suitable answer, let me know and I can write it out in an actual answer with more detail.
$endgroup$
– Michael Sparapany
2 days ago
$begingroup$
For part 1, you might be forced to determine the sensitivities numerically. Since $C = int g dt$, if you can't solve the integral analytically then it'd be easiest to evaluate $C$ by quadrature over $x(t)$. Then you can determine $dC/dx(t_0)$ by finite difference wrapped around the quadrature. This isn't variational calculus since $g$ is only dependent on $x$ and $t$. That is, there is no coordinate $h$ such that $h' = g(x,h)$. I have no idea how to extend this for part 2, however. If this is a suitable answer, let me know and I can write it out in an actual answer with more detail.
$endgroup$
– Michael Sparapany
2 days ago
|
show 1 more comment
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$begingroup$
The first part looks more like calculus of variations than stochastic stuff. Do you know the form of $g?$
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
yep, the first part is not stochastic, but the ODE for $x(t)$ is already known, so I think this is not a variational problem. $g$ can be an arbitrary continuous and differentiable function in my problem
$endgroup$
– Jan Ca
2 days ago
$begingroup$
If $x$ is already determined by the ODE and therefore fixed, then I'm not sure I see the point in calculating that sensitivity. It's really only going to be sensitive to changes in $g,$ since that's all you're allowing to vary. You can probably use the chain rule and the Euler-Lagrange equations to get at a solution to the first problem.
$endgroup$
– Adrian Keister
2 days ago
$begingroup$
I'm varying the initial state $x_t$. $g$ is fixed.
$endgroup$
– Jan Ca
2 days ago
$begingroup$
For part 1, you might be forced to determine the sensitivities numerically. Since $C = int g dt$, if you can't solve the integral analytically then it'd be easiest to evaluate $C$ by quadrature over $x(t)$. Then you can determine $dC/dx(t_0)$ by finite difference wrapped around the quadrature. This isn't variational calculus since $g$ is only dependent on $x$ and $t$. That is, there is no coordinate $h$ such that $h' = g(x,h)$. I have no idea how to extend this for part 2, however. If this is a suitable answer, let me know and I can write it out in an actual answer with more detail.
$endgroup$
– Michael Sparapany
2 days ago