Expected Values of Stochastic Differential Equation Solutions

Summary

The document considers a controlled stochastic differential equation with deterministic drift components and a Brownian noise term. It presents an integrating-factor solution for the state variable and asks whether taking its expectation removes the stochastic integral. The cited answer says that this expectation is zero when the integrand multiplying Brownian motion is deterministic, leaving the non-random terms as the expected value of the solution.

That result relies on the stated determinism condition and suitable integrability assumptions for the stochastic integral. It does not establish the separate goal of proving the control is bounded: the control is expressed using the state in its denominator, so bounds on the state and on the deterministic functions would also matter. The exchange offers a basic expectation property, not a full boundedness argument or an optimization analysis.

Key ideas

• An integrating factor can express the state variable as deterministic terms plus a stochastic integral.
• A Brownian stochastic integral with a deterministic integrand has expectation zero under standard integrability conditions.
• The proposed expectation follows only when the noise integrand is deterministic as assumed in the answer.
• Taking expectations alone does not prove that a state-dependent control is bounded.

Tags

# Expected value of stochastic optimization


# Expected value of stochastic optimization












I have a optimization problem where the SDE is:

$$ dX(t) = [X(t)(u(t)-\beta(t))+\theta(t)]dt+X(t)u(t)\sigma dW(t), t \in [0,T], X(0) = X_0 $$ where $\beta(t)$ and $\theta(t)$ are deterministic functions. I found the solution of the SDE is the following: $$ X(t)=e^{\int_{0}^{t}(u(s)-\beta(s))ds}.[X_0+\int_{0}^{t}\theta(s).e^{-\int_{0}^{s}(u(z)-\beta(z))dz}ds+\sigma\int_{0}^{t}u(s).e^{-\int_{0}^{s}(u(z)-\beta(z))dz}dW_s] $$ I found a relation between the control $u(t)$ and $X(t)$, which is the following: $$ u(t)=k.\left(1+\frac{\rho(t)}{X(t)}\right) $$ where $\rho(t)$ is a deterministic function and $k$ is a constant. I want to prove that $u(t)$ is bounded. For this reason I was trying to make a relation of $u(t)$ with the expected value of $X(t)$. One of my tries was to determinate if this expresion is correct: $$ E[X(t)]=e^{\int_{0}^{t}(u(s)-\beta(s))ds}.[X_0+\int_{0}^{t}\theta(s).e^{-\int_{0}^{s}(u(z)-\beta(z))dz}ds] $$ any idea? (I hope it is clearer now)

## Answer by Magic is in the chain (score 3)

https://quant.stackexchange.com/a/45205

The expectation looks correct, assuming the function in front of the Brownian is deterministic. It is a standard result in stochastic calculus that the expected value of the integral of a deterministic function with respect to the Brownian motion is zero. You may want to check the properties of the stochastic integral, one of which is the property that I just mentioned.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.