# 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)

• Given that $u(t)$ is random, then your expectation is incorrect. – Gordon Apr 23 '19 at 14:00

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.

• Thanks for your answer. How can obtain the expected value of $X(t)$ if the relation between $u(t)$ and it, is $u(t).X(t)= X(t)+v(t)$?, where $v(t)$ is a deterministic function. – Ranu Castaneda Apr 21 '19 at 23:15
• Very easy if I understood the question correctly. Expected value of the sum of two variables is equal to the sum of their expected values, E[x+y]= E[x]+E[y], and then the expected value of a deterministic quantity is just that quantity. – Magic is in the chain Apr 21 '19 at 23:21
• What I mean is if $X(t)$ has a Brownian component and there is a relation between the control $u(t)$ and it, the third component of expected value of the $X(t)$ can't be eliminated, is it right? – Ranu Castaneda Apr 22 '19 at 16:59
• Hi Ranu, sorry donâ€™t follow. Do you mind editing the question to make it clearer. – Magic is in the chain Apr 22 '19 at 17:33
• Hi Magic, I've made some changes in the original question, I hope it is clearer now. – Ranu Castaneda Apr 24 '19 at 22:21