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Solve the following stochastic differential equation: $$ dX_t=X_t\,dt+dW_t. $$ Thank you very much for help! I even don't know where to start...

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(Shreve: Stochastic Calculus for Finance Volume II, Exercise 4.8). This is a specific case of the Vasicek SDE, $$ dR_t = (\alpha - \beta R_t)dt + \sigma dW_t. $$

Use Ito's formula to compute $$d(e^{\beta t} R_t)= \beta e^{\beta t} R_tdt + e^{\beta t} dR_t = \cdots $$ the right side will not involve $R_t$. Integrate this answer to obtain the solution.

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    $\begingroup$ Outside of finance the Vasicek model is also known as an Ornstein-Uhlenbeck process. I believe the analytical solution is first due to Doob 1946. $\endgroup$ – horchler Jun 3 '13 at 21:42
  • $\begingroup$ Hi, what is $R_t$ here? Is it just any random variable? $\endgroup$ – Matt Phillips Mar 4 '15 at 16:11
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    $\begingroup$ @MattPhillips, $R_t$ is an unknown stochastic process that you are trying to solve for. $\endgroup$ – nullUser Mar 4 '15 at 22:16
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The usual tool used here is Ito's Formula. To use it, you'll want to propose a possible solution, say $$X_t = e^{t}\left(X_0 + \int_0^t e^{-s} \, dW_s\right),$$ and use Ito, by defining a function, say, $g(t,y)$ that is a function of deterministic $t$ and stochastic $y$. For a nice introduction to SDEs I half-heartedly recommend Oksendal.

Indeed, I do not respond directly to your question. Readily available is not my intuition about generating this solution (I happened to have it "in my back pocket"), but I think it derives from intuition you might have developed in ODE (notice the exponential term, a common term in the solution to ODEs).

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  • $\begingroup$ are you sure? Let's calculate $d\,X_t$ for your $X_t$, assuming that $X_0=0$: $$d\,X_t=-e^{-t}\cdot\int_{0}^{t}e^s\,dW_s+e^{-t}e^t\,dW_t$$ and this is not $X_t$ but $-X_t$. Am I right? $\endgroup$ – Almost sure Jun 3 '13 at 15:17
  • $\begingroup$ So finally the solution is $-X_t$, not $X_t$, am I right? But I am not sure my calculations are correct: what about the equality $$\frac{d}{dt} \int_{0}^{t}e^s \,dW_s=e^t\,dW_t?$$ $\endgroup$ – Almost sure Jun 3 '13 at 15:25
  • $\begingroup$ Indeed, the notation you give is just shorthand. The right hand side is shorthand for the left. $\endgroup$ – soren Jun 3 '13 at 16:34
  • $\begingroup$ Soren, thank you for assuring me about correctness of my calculation - now I understand it. You have also to edit $e^s$ into $e^{-s}$ in your first answer. $\endgroup$ – Almost sure Jun 3 '13 at 16:58

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