Controlled Diffusions and Admissible Controls

How a Control Enters a Diffusion

A controlled diffusion is a state process whose instantaneous drift, diffusion coefficient, or both may be selected through a time-dependent control. The control may depend on information observed up to the current time, but it cannot use future information. This requirement is encoded by progressive measurability. Before discussing optimization, it is necessary to specify which controls produce a well-defined state process.

Fix a finite horizon $T>0$ and a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to that filtration. Let $U$ be a nonempty Borel subset of a Euclidean space. The state is real-valued here; the same definitions extend to vector states.

Definition: A control $u=(u_t)_{0\le t\le T}$ is progressively measurable if, for each $t\le T$, the map $(s,\omega)\mapsto u_s(\omega)$ restricted to $[0,t]\times\Omega$ is measurable with respect to $\mathcal{B}([0,t])\otimes\mathcal{F}_t$. A $U$-valued control takes values in $U$ almost everywhere with respect to $dt\otimes d\mathbb{P}$. Progressive measurability implies adaptedness, but adaptedness alone is not the definition used here.

Given an initial state $x\in\mathbb{R}$, the controlled stochastic differential equation (SDE) is $dX_t=\mu(t,X_t,u_t)\,dt+\sigma(t,X_t,u_t)\,dB_t,\quad X_0=x.$ The coefficient $\mu$ is the drift and $\sigma$ is the diffusion coefficient. The ordinary integral is a Lebesgue integral in time. Every Ito integral below is defined for a progressively measurable integrand $H$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; this is the class $\mathcal{L}^2$. If an equation is first considered only up to a stopping time, its stopped integrand is required to satisfy this condition on each bounded time interval.

Definition: A strong solution for a specified control $u$ is a continuous, adapted process $X$ on the given probability space, driven by the given Brownian motion $B$, such that $\int_0^T|\mu(t,X_t,u_t)|\,dt<\infty$ almost surely, $\mathbb{E}\int_0^T\sigma(t,X_t,u_t)^2\,dt<\infty$, and the integral form of the SDE holds almost surely for every $t\le T$. Pathwise uniqueness means that two strong solutions on this same space, with the same $B$, $u$, and initial state, are indistinguishable. A weak solution, in contrast, may use a different probability space and Brownian motion. Uniqueness in law means that solutions have the same distribution; it is not the same assertion as pathwise uniqueness.

Admissibility and a Well-Posedness Result

Admissible-control classes depend on the application. Here is a useful finite-horizon convention that makes the state equation part of the definition.

Definition: For initial state $x$, call a control admissible if it is progressively measurable, $U$-valued almost everywhere, and its state equation has a unique strong solution satisfying $\mathbb{E}\sup_{0\le t\le T}|X_t|^2<\infty$. The latter condition is called membership of the solution in $S^2$. This definition imposes no cost or optimality condition.

The following sufficient conditions ensure that every control in the specified class is admissible. Uniformity in the control variable is essential: the bounds must hold for all $a\in U$, including when $u_t$ is random.

Theorem: Suppose $\mu(t,x,a)$ and $\sigma(t,x,a)$ are measurable in their arguments and, for some finite constants $L,K$, satisfy for all $t\in[0,T]$, $x,y\in\mathbb{R}$, and $a\in U$, $$ |\mu(t,x,a)-\mu(t,y,a)|+|\sigma(t,x,a)-\sigma(t,y,a)|\le L|x-y|, $$ $$ |\mu(t,x,a)|+|\sigma(t,x,a)|\le K(1+|x|). $$ Then, for every progressively measurable $U$-valued control $u$, the controlled SDE has a unique strong solution in $S^2$. Moreover, any continuous adapted solution satisfying the integral equation is in $S^2$, so pathwise uniqueness holds among all such solutions, not just among solutions assumed in advance to have finite second moments.

Proof. We first construct a solution. Set $X^0_t=x$, and recursively define $X^{k+1}_t=x+\int_0^t\mu(s,X^k_s,u_s)\,ds+\int_0^t\sigma(s,X^k_s,u_s)\,dB_s.$ The coefficient integrands are progressively measurable. The growth bound and the $S^2$ bounds at each finite iteration ensure that the Ito integrals belong to $\mathcal{L}^2$. We use the Ito isometry and Doob's $L^2$ maximal inequality; the latter is the standard martingale maximal inequality. The Ito integral has finite second moment by the isometry and conditional expectation of each future integral increment zero, so it is a square-integrable martingale and Doob's inequality applies.

For $k\ge1$, write $D_k(t)=\mathbb{E}\sup_{r\le t}|X^k_r-X^{k-1}_r|^2$. The inequalities $|a+b|^2\le2|a|^2+2|b|^2$, Cauchy–Schwarz for the time integral, Doob's inequality for the stochastic integral, and the Lipschitz bound give a constant $C$, independent of $k$, such that $D_{k+1}(t)\le C\int_0^tD_k(s)\,ds.$ The growth condition gives $D_1(t)\le A$ for a finite constant $A$ on $[0,T]$. Iteration therefore gives $D_{k+1}(T)\le A C^k T^k/k!$. The sum of the square roots of these bounds is finite. Thus the Picard iterates are Cauchy in the norm $(\mathbb{E}\sup_{t\le T}|Y_t|^2)^{1/2}$. Their limit has a continuous adapted version: convergence in this norm gives a subsequence converging uniformly almost surely, and a uniform limit of continuous paths is continuous. The Lipschitz bound, the time-integral estimate, and the Ito isometry show that both integrals in the recursion converge to the corresponding integrals of the limit. Hence the limit solves the SDE and belongs to $S^2$.

We now justify that any solution is in $S^2$; this is needed before applying global second-moment estimates to an arbitrary second solution. Let $X$ be any continuous adapted solution and put $\tau_n=\inf\{t\ge0:|X_t|\ge n\}$, with the infimum of the empty set defined as infinity. For $t\le T$, set $F_n(t)=\mathbb{E}\sup_{r\le t}|X_{r\wedge\tau_n}|^2$. The stopped equation, Cauchy–Schwarz, Doob's inequality, and the linear-growth bound yield constants $C_0,C_1$, independent of $n$, such that $F_n(t)\le C_0+C_1\int_0^t(1+F_n(s))\,ds.$ Indeed, the drift contribution is bounded by $t\mathbb{E}\int_0^{t\wedge\tau_n}|\mu(s,X_s,u_s)|^2ds$, and the stochastic contribution by $4\mathbb{E}\int_0^{t\wedge\tau_n}|\sigma(s,X_s,u_s)|^2ds$; the growth bound controls both by a constant times $\int_0^t(1+F_n(s))ds$. Gronwall's inequality gives a bound on $F_n(T)$ independent of $n$. By continuity, $\tau_n$ eventually exceeds $T$ on each path, so Fatou's lemma gives $\mathbb{E}\sup_{t\le T}|X_t|^2<\infty$.

Finally, for two solutions $X,Y$ with the same data, define $D(t)=\mathbb{E}\sup_{r\le t}|X_r-Y_r|^2$, which is finite by the preceding argument. The same integral estimates and Lipschitz condition give $D(t)\le C\int_0^tD(s)\,ds$. Gronwall's inequality implies $D(t)=0$ for every $t\le T$. Continuity then gives indistinguishability. This proves existence, the $S^2$ claim, and pathwise uniqueness. $\square$

The linear-growth and Lipschitz hypotheses are sufficient conditions, not a definition of all possible controlled models. If coefficients are discontinuous, degenerate, or constrained by state boundaries, existence and uniqueness require separate arguments. Also, the theorem is a strong-solution result on the given filtered space; it does not identify a risk-neutral measure $\mathbb{Q}$ or change the underlying probability measure $\mathbb{P}$.

Worked Controlled-Diffusion Examples

Worked Example: A Bounded Drift Control

On $0\le t\le1$, take $U=[-1,1]$, $X_0=0$, $\mu(t,x,a)=a$, and $\sigma(t,x,a)=1$. Use the deterministic control $u_t=1$ for $0\le t\le\tfrac12$ and $u_t=-1$ for $\tfrac12<t\le1$. It is progressive and $U$-valued. Both coefficients are globally Lipschitz in $x$, and satisfy a uniform linear-growth bound.

Integrating the equation gives $X_t=B_t+\int_0^t u_s\,ds$. The drift integral equals $t$ for $t\le\tfrac12$, and equals $1-t$ for $\tfrac12<t\le1$. In particular, $X_1=B_1$. The deterministic drift accumulated over the full interval is $\tfrac12-\tfrac12=0$, so $\mathbb{E}X_1=0$ and $\mathbb{E}X_1^2=1$. For every $t\le1$, $|\int_0^t u_sds|\le1$, hence $\sup_{t\le1}|X_t|^2\le2\sup_{t\le1}|B_t|^2+2$. Doob's inequality gives $\mathbb{E}\sup_{t\le1}|B_t|^2\le4\mathbb{E}|B_1|^2=4$, proving the required $S^2$ integrability.

Check: substituting $t=1$ into the integrated equation gives $X_1=B_1+0$, whose mean and second moment are indeed $0$ and $1$. The control changes the intermediate drift but cancels its total time integral.

Worked Example: A Bounded Drift Depending on Current Information

Let $T=1$, $U=\{0,1\}$, $X_0=0$, and $dX_t=u_tdt+dB_t$, with $u_t=\mathbf{1}_{\{B_t\ge0\}}$. The process $B$ is continuous and adapted, so $(t,\omega)\mapsto\mathbf{1}_{\{B_t(\omega)\ge0\}}$ is progressively measurable. Thus the control uses the current state of the noise, not its future values. The coefficients are bounded in $u$, constant in $x$, and satisfy the theorem's hypotheses.

The solution is $X_t=B_t+\int_0^t\mathbf{1}_{\{B_s\ge0\}}ds$. For each $s>0$, $B_s$ has the centered normal distribution with positive variance $s$, so $\mathbb{P}(B_s\ge0)=\tfrac12$. Tonelli's theorem applies to the nonnegative indicator and gives $\mathbb{E}\int_0^1\mathbf{1}_{\{B_s\ge0\}}ds=\int_0^1\tfrac12ds=\tfrac12.$ Since $\mathbb{E}B_1=0$, it follows that $\mathbb{E}X_1=\tfrac12$. Also $0\le\int_0^t\mathbf{1}_{\{B_s\ge0\}}ds\le1$, which gives $\mathbb{E}\sup_{t\le1}|X_t|^2\le2\mathbb{E}\sup_{t\le1}|B_t|^2+2\le10$.

Check: the time integral is between $0$ and $1$, and its expected value is $\tfrac12$; the resulting mean $\tfrac12$ has the correct sign because the drift is never negative. The state equation holds directly by substituting the displayed expression for $X_t$.

Worked Example: A Controlled Volatility

Take $T=2$, $U=[0,2]$, $X_0=1$, zero drift, and $\sigma(t,x,a)=ax$. Use $u_t=1$ on $[0,1]$ and $u_t=2$ on $(1,2]$. The control is deterministic and progressive. The coefficient is uniformly Lipschitz in $x$, since $|ax-ay|\le2|x-y|$, and has linear growth, since $|ax|\le2|x|$.

Define $A_t=\int_0^t u_s^2ds$ and $Y_t=\int_0^t u_s\,dB_s$. The Ito integrand $u$ is progressive and $\mathbb{E}\int_0^2u_s^2ds=A_2=1+4=5<\infty$. Consider $X_t=\exp(Y_t-\tfrac12A_t)$. For $f(t,y)=\exp(y-\tfrac12A_t)$, the derivatives are $f_t=-\tfrac12u_t^2f$, $f_y=f$, and $f_{yy}=f$ at times where $A'_t=u_t^2$. Ito's formula, including its second-derivative correction, gives $$ dX_t=f_tdt+f_yu_t\,dB_t+\tfrac12f_{yy}u_t^2dt =u_tX_t\,dB_t. $$ Thus the candidate satisfies the SDE and $X_0=1$.

Here $Y_2=B_1+2(B_2-B_1)$ is centered normal with variance $1+4=5$. Therefore $\mathbb{E}X_2=e^{-5/2}\mathbb{E}e^{Y_2}=e^{-5/2}e^{5/2}=1$, and $\mathbb{E}X_2^2=e^{-5}\mathbb{E}e^{2Y_2}=e^{-5}e^{10}=e^5$. These finite moments also verify $\mathbb{E}\int_0^2u_t^2X_t^2dt<\infty$: for $t\le2$, $\mathbb{E}X_t^2=e^{A_t}\le e^5$, and $u_t^2\le4$, so the integral expectation is at most $8e^5$.

Check: setting $u_t=c$ constant in the same formula yields $X_t=\exp(cB_t-\tfrac12c^2t)$, and the displayed Ito calculation again gives $dX_t=cX_t\,dB_t$. For the specified control, $A_2=5$, matching both the Gaussian variance and the exponent in the moment calculations.

Common Pitfall: An Adapted-Looking Rule Can Use the Future

A control must be progressive, not merely specified by a formula involving Brownian motion. For example, on the time interval $[0,1]$, the rule $u_t=\mathbf{1}_{\{B_1\ge0\}}$ uses the terminal value even when $t<1$. It is generally not $\mathcal{F}_t$-measurable at those times and is therefore not an admissible adapted control under the definition above. Replacing $B_1$ with $B_t$, as in the preceding example, makes the rule depend only on information available at time $t$.

Another important distinction is that boundedness of the control alone does not establish well-posedness. The coefficients must also be sufficiently regular in the state, or another existence and uniqueness argument must be supplied. Under the uniform Lipschitz and linear-growth conditions, the theorem verifies both the state equation and its finite-second-moment requirement; without those conditions, admissibility must be checked for the particular model.

Feature	What it specifies	What it does not imply by itself
Progressive control	Uses information available through each time	Existence of a state solution
Drift coefficient $\mu$	Finite-variation part of the state dynamics	Noise intensity
Diffusion coefficient $\sigma$	Coefficient multiplying $dB_t$	Boundedness of the state
Strong solution	State constructed on the given space with the given Brownian motion	Uniqueness in law alone

Check Your Understanding

Use the definitions and theorem to decide which conditions are being tested in each question.

For a deterministic measurable control $u_t\in U$, explain why it is progressively measurable.
In the bounded-drift example, calculate the value of $\int_0^1u_sds$ and use it to recover $X_1$.
Why is $\mathbf{1}_{\{B_t\ge0\}}$ an allowable current-information control, while $\mathbf{1}_{\{B_1\ge0\}}$ need not be allowable for $t<1$?
For $\sigma(t,x,a)=ax$ with $a\in[0,2]$, give a uniform Lipschitz constant in $x$ and a linear-growth bound.
Why must a uniqueness proof first justify finite second moments for an arbitrary solution before applying global $L^2$ estimates?

Controlled Diffusions and Admissible Controls

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