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I've been tasked to solve the following SDE:

$dX_t = dt + 2\sqrt{X_t}dW_t, \ \ \ t \in \mathbb{R}_+$

where $W_t$ is a standard Brownian motion and $X_0=x$.

I need a closed form solution.

What is the best approach?

Is it to write in in "integral" form: $X_t = x + t + 2\int_0^t\sqrt{X_u}dW_u$ and solve the stochastic integral term? If so, I can't find a function $f$ to use in Ito's lemma so that I get a good-enough solution.

Any advice would be appreciated as I am pretty much a bigger to stochastic calculus. Thanks

Edit: Note: I am allowed to assume X stays non-negative all the time

saz
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Vladimir Nabokov
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1 Answers1

6

Hints:

  1. Suppose that $(X_t)_{t \geq 0}$ is a solution to the SDE. Using Itô's formula show that $Y_t := \sqrt{X_t}$ satisfies $$dY_t = dW_t.$$ (See the second part of this answer to get a better understanding how to choose a suitable transform.)
  2. Conclude from step 1 that $X_t := (\sqrt{x}+W_t)^2$ is a candidate for the solution.
  3. Use Itô's formula to show that $(X_t)_{t \geq 0}$ from step 2 is indeed a solution to the given SDE.
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saz
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  • @VladimirNabokov You are welcome. – saz Apr 10 '17 at 14:04
  • Is it too naive to ask why $dY_t = dW_t$ in step 1? I am a complete beginner in stochastic calculus. Although I know Ito's formula, I don't understand how to apply it to get the above SDE. – Mathews Boban Dec 25 '23 at 01:39
  • If we naively take $dY_t = \frac{1}{2\sqrt{X_t}}dX_t$, we get that $dY_t = dW_t + \frac{1}{2Y_t}dt$. I suspect $dY_t = \frac{1}{2\sqrt{X_t}}dX_t$ is not a valid inference. – Mathews Boban Dec 25 '23 at 03:21
  • I just learned that if $f$ is a smooth real-valued function over the reals, then $df(X_t) = f'(X_t)dX_t + \frac{1}{2}f''(X_t)d[X]_t$ (source). I also learned that if $dX_t = \sigma(X_t,t) dW_t + \mu(X_t,t) dt$, then $d[X]_t = \sigma(X_t,t)^2 dt$ (source). I think that it's best to not delete my naive thoughts above because it might help other confused beginning learners of stochastic calculus in the future. – Mathews Boban Dec 25 '23 at 07:27