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Let $W_t$ be a standard Brownian Motion. Solve the stochastic differential equation: $$d X_t = -\frac{1}{2} e^{-2 X_t} dt + e^{-X_t} dW_t$$ Does the solution exist for all times t?

GOAT Messi
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  • What are your ideas? Did you test the assumptions of the existence theorem? You could also try to apply the Ito theorem to $Y_t=e^{2X_t}$ to see if the transformed equation has a more simple structure. – Lutz Lehmann Jan 25 '19 at 22:16
  • @LutzL I know this is quite a late comment, but is there a reason you suggested this particular transformation? I tried to solve this question too, but the transformation did not seem to help according to my workings. But if I'm wrong, I would like to understand why this particular transformation works. – PLY Jun 01 '19 at 13:17
  • @PLY : No, it does not work, it was a guess based on the drift term. Disregarding the stochastic term sometimes works, but more often not. See the answer for a better motivated guess. – Lutz Lehmann Jun 01 '19 at 13:53
  • @LutzL I have looked at your answer and computed it too, which works indeed. But I still wonder if there is a general approach for finding such a transformation (without too much effort), since I wasn't able to deduce it from your method. – PLY Jun 01 '19 at 14:13

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Set $Y=e^X-W$, then by the Ito theorem $$ dY=e^XdX+\frac12e^Xd[X]-dW=0\implies Y=e^{X_0} $$ and the equation $e^{X_t}=e^{X_0}+W_t$ can only be true as long as the right side is positive. As there is a non-zero probability that it becomes non-negative, the solution will not exist for all times.

How to get to this idea with a little hand-waving

On an atomic level one has $(dW)^2=dt$. Then the expression $e^{-X}dW-\frac12e^{-2X}(dW)^2$ is the start of the Taylor series of $\ln(1+e^{-X}dW)$. Taking the exponential gives thus $$ e^{X_{t+dt}-X_t}=1+e^{-X_t}dW_t\implies e^{X_{t+dt}}=e^{X_t}+dW_t\implies e^{X_t}=e^{X_0}+W_t $$ To confirm this heuristic computation, apply as above the Ito theorem to $Y=e^X-W$ to confirm that it is a constant process.

Lutz Lehmann
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