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(just joined, this is my first post),

I've been perusing SDE’s, and a simple one is $$\mathrm{d} Y(t) = Y(t)\hspace{0.1cm}\mathrm{d}W(t)$$ where $W$ is stochastic. The solution is $$ Y = \text{exp}[W(t) - \frac{t}{2}] $$ or $\ln(Y) = W(t) - \frac{t}{2}.$ I don't understand where the $-\frac{t}{2}$ comes from. I saw the mathematical derivation ($\mathrm{d}W$ is sort of the square root of $\mathrm{d}t$), but I would like an intuitive explanation if possible. Thanks.

Air Mike
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Daniel
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  • Do you understand why it "should" be a martingale or why that would imply the correction term to exp(W(t))? ( https://math.stackexchange.com/questions/176196/calculate-the-expected-value-of-y-ex-where-x-sim-n-mu-sigma2 ) – Elle Najt Sep 16 '20 at 23:48
  • Sorry, no I don't (I'm new to this subject). Thanks for the link even though it didn't help me. It just seems to me that, since W bounces up and down then Y should bounce up and down, I don't see where the time trend comes from. – Daniel Sep 17 '20 at 01:02
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    It would help if you explained your background. Do you know what a martingale is? What about the discrete time martingale transform? What that equation is shorthand for? It's actually not a time trend -- it keeps the expectation $1$ at every time, because of the calculation in the link I gave you earlier. – Elle Najt Sep 17 '20 at 04:00

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First the math derivation: by Itô formula for $f(x)=lnx$ we have

$$dZ_{t}=d(ln(Y_{t}))=f'(Y_{t})dY_{t}+\frac{1}{2}f''(Y_{t})d\langle Y\rangle_{t}$$

$$=\frac{1}{Y_{t}}Y_{t}dB_{t}+\frac{1}{2}\frac{-1}{Y_{t}^{2}}Y_{t}^{2}dt$$

$$\Rightarrow Z_{t}=B_{t}+\frac{-1}{2}t.$$

As explained in the comments, the $t/2$ appears because when we Taylor expand in the proof Itô formula, we get a second-order term.

Now for heuristics. Suppose Brownian motion is differentiable. Then we study the ODE

$$\dot{y}_{t}=y_{t}\dot{B}_{t},$$

which is solved by separation of variables by

$$y(t)=cexp\left(\int_{0}^{t} \dot{B}_{s}ds \right)=cexp\left(B_{t}-B_{0} \right).$$

So the main reason we get that extra term $t/2$ is precisely due to the non-differentiability of Brownian motion.

Thomas Kojar
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  • Thanks, this helps. In considering the way it "bounces up and down" randomly, I didn't realize that (a) Brownian motion is non-differentiable and (b) that would affect the integration. – Daniel Sep 28 '23 at 20:52