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Solve the following stochastic differential equations

$ dX_t = \frac{1}{2 X_t} dt + dB_t$

or equivalently with a transformation $Y_t = X_t^2$

$ dY_t = dt + 2 \sqrt{Y_t} dB_t$ with $Y_0 = y_0 > 0$

where B is one dimentional standard Brownian motion.

The solution is Given by $Y_t = (\sqrt{y_0} + B_t)^2$. You can check this by finding the differential of the solution using Ito's formula. This solution was by guess. The way to solve this problem is by introducing a transformation of $Y_t$ that will give a linear SDE that one can solve easily. Then the solution can be found by the inverse of the used transformation.

user144410
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  • Just to be sure: It's really $\frac{1}{2Y_t}$ and not $\frac{1}{2} Y_t$ (in the 1st SDE), right? And: What have you tried? Do you know any approach to solve SDEs? – saz Dec 18 '14 at 15:34
  • If one try to solve the second SDE, we can let $X_t = \sqrt{Y_t}$. Then $dX_t = 0.5 Y_t^{-0.5} dY_t = \frac{1}{2 X_t} dY_t$ and we get the first one. I'm learning SDEs. I usually use Ito formula. – user144410 Dec 18 '14 at 15:37
  • As far as I see you didn't apply Itô's formula correctly. However; this means that you actually want to solve the second SDE - am I right? – saz Dec 18 '14 at 15:40
  • Here I did not apply any Ito formula, I was answering you regarding "Do you know how to approach SDEs. Both SDEs are the same up to the transformation $X_t = \sqrt{Y_t}$. A solution to any of them will yield the solution to the other. – user144410 Dec 18 '14 at 15:42
  • I think the way to solve this is to find a transformation that gives a linear SDE. – user144410 Dec 18 '14 at 15:44
  • Yeah; sorry, my mistake. – saz Dec 18 '14 at 15:44
  • Your question was closed for "missing context". This is a somewhat unsatisfactory abbreviation that may be difficult to decipher. It may mean that the members who wanted to answer, felt that they couldn't, because they would need more contextual information about the tools you have mastered. It may also mean that the close voters received a vibe that you wanted somebody to do your homework for you. I don't know this topic at all, so I cannot tell. Anyway, well established members felt that it needs improvement. It is a bit unfortunate that none of them left clearer instructions. – Jyrki Lahtonen Dec 25 '14 at 15:41
  • (cont'd) You may not have noticed that the question has started receiving reopening votes. The prudent thing now is to wait and, if possible, make further improvements to the question. – Jyrki Lahtonen Dec 25 '14 at 15:43

1 Answers1

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Hint: For the second SDE use the substitution $$Z_t := \exp(-\sqrt{Y_t}).$$ For the general approach see e.g. (the second part of) this answer.

Remark: As @user144410 pointed out, the solution of the first SDE can be obtained from the second one using the transformation $X_t = \sqrt{Y_t}$.

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