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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)

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  • $\begingroup$ Given that $u(t)$ is random, then your expectation is incorrect. $\endgroup$ – Gordon Apr 23 at 14:00
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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.

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  • $\begingroup$ 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. $\endgroup$ – Ranu Castaneda Apr 21 at 23:15
  • $\begingroup$ 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. $\endgroup$ – Magic is in the chain Apr 21 at 23:21
  • $\begingroup$ 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? $\endgroup$ – Ranu Castaneda Apr 22 at 16:59
  • $\begingroup$ Hi Ranu, sorry don’t follow. Do you mind editing the question to make it clearer. $\endgroup$ – Magic is in the chain Apr 22 at 17:33
  • $\begingroup$ Hi Magic, I've made some changes in the original question, I hope it is clearer now. $\endgroup$ – Ranu Castaneda Apr 24 at 22:21

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