I have this SDE

$$ dX(t) = [X(t)(u(t)(\delta-r)+r-\beta(t))+\theta(t)(1-\alpha(t))]dt+X(t)u(t)\sigma dW(t), t \in [0,T] \\ X(0) = X_0(1-\alpha(0)) $$

I've checked some books and I find the solution is this: $$ X_t=\Phi_{t,0}.\left( X_0(1-\alpha_0)+\int_{0}^{t}\frac{\theta(s)(1-\alpha(s))}{\Phi_{s,0}}ds + \int_{0}^{t} \frac{\sigma u(s)}{\Phi_{s,0}}dW_s \right)$$ where $ \Phi_{t,0}=\exp \left( \int_{0}^{t}(u(s)(\delta-r) +r-\beta(s))ds \right)\\ $

I need to prove that $X(t)$ is bounded. Some idea?


  • $\begingroup$ Have a look of this question. Your notations are so confusing. Why not merge them, for example, using $\hat{u}(t) \equiv u(t)(\delta-r)+r-\beta(t)$ and $\hat{\theta}(t) \equiv \theta(t)(1-\alpha(t))$? $\endgroup$ – Gordon Feb 21 at 2:20

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