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Assume $\frac{dX_t}{X_t} = \mu dt + \sigma d B_t$ where $\mu$ is a constant and $B_t$ is a Brownian motion, and let $Y_t = \ln X_t$. I understand that $B_t$ is nowhere differentiable and both $X_t$ and $Y_t$ are functions of $B_t$, so neither of them are differentiable. Thus, $d(\ln X_t) \neq \frac{d X_t}{X_t}$.

But can someone please explain why:

  1. $\frac{\partial Y_t}{\partial X_t} = \frac{1}{X_t}$
  2. $\int d Y_t = Y_t$

Assume the entire question is under Ito's Lemma if necessary.

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    $\begingroup$ I guess you define 1 in the Malliavin sense? Then it makes sense. Otherwise, what is the definition of 1? $\endgroup$ – Yair Daon Nov 21 '17 at 15:55
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(1) Forget about random variables for a second. You have a function $Y_t=Y(t,X_t):=\ln(X_t)$. Then as a function it's partial derivative is defined in the usual sense: $\frac{\partial Y_t}{\partial X_t}=\frac{1}{X_t}$. The fact that $Y_t$ is a random variable makes no difference: your $X_t=X(t,\omega)$, but you're still asking what the change in $Y_t$ is if you keep $t$ constant, and vary $X_t$. Whenever you see such partials involved, they are always in the classical multivariable calculus sense.

Don't confuse this with terms like $dY_t$ which are formally defined as stochastic differentials:

(2) This can be taken as the definition of $dY_t$. In fact:

$$Y_t-Y_0=\int_0^tdY_s,$$

where the integral is an Ito Integral, and not Lebesgue or Riemannian/Stieltjes.

If now $Y_t=Y(t,X_t)$, then Ito's lemma in it's full form says that:

$$dY_t=\frac{\partial Y}{\partial t}dt+\frac{\partial Y}{\partial X_t}dX_t+\frac{1}{2}\frac{\partial^2 Y}{\partial X_t^2}dX_t^2+...$$

This should look really familiar: It's the Taylor series of $Y_t$ expanded around $t,dX_t$, with higher order terms of $dt$ suppressed. If you now substitute $dX_t=\mu dt+\sigma dB_t$, you'll recover Ito's lemma after assuming $dt^2$ and $dtdB_t$ is higher order, and $dB_t^2=dt$. You can formaly prove those last facts using the definition of quadratic variation.

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