research36 min studyLesson 681 in Mathematics

Controlled diffusions and admissible controls

Specify the information, integrability, and well-posedness needed before optimizing a noisy state equation.

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What you’ll understand

  • Specify a nonanticipating control class on a filtered probability space.
  • Check assumptions ensuring the controlled state equation and its cost are well defined.
  • Optimize a constant-control subclass without confusing it with the full feedback problem.

A control problem begins before its objective is minimized. We must decide what the controller can observe, which actions are permitted, and whether each permitted action produces a well-defined state and finite cost. Allowing knowledge of future noise changes the problem, while omitting an integrability condition can make an apparent optimum meaningless. A scalar tracking example will separate these modeling questions from the easier task of optimizing a deterministic constant action.

The state equation includes an information structure

Fix a finite horizon, a complete filtered probability space with a right-continuous filtration, and a Brownian motion relative to that filtration. A control takes values in a specified Borel action set and is progressively measurable. Progressive measurability ensures that its restriction up to each time is measurable using only information then available; together with suitable moment bounds it permits the time and stochastic integrals. A controller cannot select an action at time zero from the realized terminal Brownian increment.

A feedback rule can use the current observed state, while an open-loop admissible control may depend on the available history. Neither label alone permits anticipation. Under partial observation, admissibility is relative to the observation filtration rather than the full state filtration. Enlarging information may lower a minimizing value, so the filtration is a mathematical component of the objective. A comparison of two policies is meaningful only when both belong to the same specified information class.

dXs=b(s,Xs,us) ds+σ(s,Xs,us) dWs,Xt=x.dX_s=b(s,X_s,u_s)\,ds+\sigma(s,X_s,u_s)\,dW_s,\qquad X_t=x.
The state is driven by a progressively measurable control and Brownian noise in the stated filtration.↔ Scroll wide equations horizontally.

Admissibility includes existence and finite moments

A useful sufficient setting has coefficients measurable in time and action, uniformly Lipschitz in the state, and at most linear growth in the state and action. For square-integrable initial data and controls with finite expected integrated squared norm, standard SDE estimates give a unique strong state solution and a finite second moment of its time supremum. Compact bounded action sets simplify the growth condition, but compactness alone cannot repair a nonmeasurable coefficient or a non-Lipschitz state equation.

Define the expected running-plus-terminal cost only after verifying its integrability. Quadratic-growth costs are controlled by the second-moment estimates in this setting. More singular costs or exponential criteria need stronger assumptions, and state constraints may remove controls whose otherwise well-defined states leave the permitted region. One may instead optimize over weak formulations where the probability space also varies, but that is a different admissible class. Here we keep the strong formulation and its information fixed.

E∫tT∣us∣2 ds<∞,J(t,x;u)=E[g(XT)+∫tTℓ(s,Xs,us) ds].\mathbb E\int_t^T|u_s|^2\,ds<\infty,\qquad J(t,x;u)=\mathbb E\left[g(X_T)+\int_t^T\ell(s,X_s,u_s)\,ds\right].
The moment requirement is a sufficient baseline for the linear-growth dynamics and quadratic costs considered here.↔ Scroll wide equations horizontally.

A restricted optimization gives a valid comparison

The value is the infimum over all admissible controls, so optimizing a smaller subclass produces an upper bound on a minimizing value. It need not find the globally optimal feedback policy. For additive noise and a deterministic constant action, the terminal mean and variance can be calculated separately. This makes a transparent benchmark: the control shifts the mean, the diffusion contributes an irreducible variance within that subclass, and an effort penalty discourages overly large shifts.

For an adapted action that changes with the observed noise, the integral of the action and the terminal noise need not be independent. Reusing the constant-control variance formula for every feedback policy would therefore be wrong. The stochastic integral remains nonanticipating, but dependence is precisely what allows a feedback controller to respond to disturbances. Later lessons derive dynamic programming equations that account for that response rather than assuming it away in a static mean calculation.

V(t,x)=inf⁡u∈At,TJ(t,x;u)≤inf⁡u∈AconstantJ(t,x;u).V(t,x)=\inf_{u\in\mathcal A_{t,T}}J(t,x;u)\le\inf_{u\in\mathcal A_{\mathrm{constant}}}J(t,x;u).
Restricting the admissible class can only increase a minimizing infimum.↔ Scroll wide equations horizontally.

Find the best constant tracking action

On [0,1][0,1], let dXs=u ds+12dWsdX_s=u\,ds+\tfrac12dW_s, X0=1X_0=1, and restrict to deterministic constants u∈[−1,1]u\in[-1,1]. Minimize E[X12]+12∫01u2 ds\mathbb E[X_1^2]+\tfrac12\int_0^1u^2\,ds. State exactly what this calculation establishes.

  1. The terminal state has mean 1+u1+u and variance one quarter. Its second moment is therefore (1+u)2+1/4(1+u)^2+1/4. The deterministic effort contribution is u2/2u^2/2.

  2. Combine the terms to get 5/4+2u+3u2/25/4+2u+3u^2/2. Its derivative vanishes at u=−2/3u=-2/3, inside the allowed interval; its positive second derivative proves this is the unique best constant.

  3. Substitution gives 5/4−2/3=7/125/4-2/3=7/12. Every constant in the interval is bounded and progressively measurable, so it is admissible. The calculation does not compare this policy with all state-dependent admissible controls.

J(u)=32(u+23)2+712.J(u)=\frac32\left(u+\frac23\right)^2+\frac7{12}.
Completing the square verifies both the minimizer and the restricted optimal cost.↔ Scroll wide equations horizontally.
The result

The best constant is −2/3 and its cost is 7/12. It supplies an upper bound on the value for the larger admissible feedback class.

Explore the model

Interactive lab

Interactive lab · probability

Separate a mean shift from a finite noise variance

Pause and Reset, then set success probability to 0.5 and trials to 4. Let K be the displayed binomial count and define a separate noise variable Z=(K−2)/2 on paper. Predict mean zero and variance one quarter; these match the terminal noise variance in the constant-action example. Adding 1+u shifts the mean without changing this variance. The bars are an exact finite binomial model, not Brownian paths or a controlled SDE. For each fresh comparison Reset and set both controls; parameter edits preserve elapsed time.

Separate a mean shift from a finite noise varianceThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA finite noise model can verify a moment calculation while leaving the admissible stochastic-control problem to separate analysis.

Check your understanding

Which proposed control violates the nonanticipation requirement on [0,1] with the natural Brownian filtration?

Try it yourself

Use the same cost convention with X0=2X_0=2, horizon T=1/2T=1/2, diffusion coefficient one, and deterministic constant control minus one. Compute the terminal mean, variance, effort cost, and total cost.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The Brownian increment variance is the elapsed horizon. The effort integral also scales by that horizon.

Compare with the solution

The mean is 2−1/2=3/22-1/2=3/2 and the variance is 1/21/2. Hence the terminal second moment is 9/4+1/2=11/49/4+1/2=11/4. The effort contribution is (1/2)(1/2)=1/4(1/2)(1/2)=1/4, so the total cost is three. This is the cost of the specified admissible policy, without any claim that it minimizes the enlarged control problem.

Take it with you

  • The admissible filtration specifies what information actions may use.
  • State well-posedness and cost integrability must be checked before optimization.
  • A best constant policy bounds the full minimizing value but need not achieve it.

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