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Why Log Returns and Proportional Price Changes Differ in Itô Calculus

Article Quant Q&A · Author: Slade
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Summary

The document examines why the differential of a log price is not the same as the proportional price change for an Itô process, even though their squared differentials agree in quadratic-variation calculations. It applies Itô’s lemma to a price with drift and Brownian volatility: the log differential includes a negative half-variance drift correction, while both squared expressions reduce to the same instantaneous variance term.

The central lesson is that stochastic differentials cannot be treated like ordinary numbers when manipulating their squares or taking square roots. Equality after retaining only quadratic-variation terms does not imply equality of the original differentials or a simple plus-or-minus relationship. The example is a local continuous-time calculation; it does not provide a general catalog of invalid stochastic operations or discuss discrete realized-variance estimators.

Key ideas

  • Itô’s lemma adds a drift correction to the differential of a log price.
  • The log differential and proportional price change are generally unequal for an Itô process.
  • Their squares agree in the quadratic-variation calculation because drift terms vanish at that order.
  • Equality of squared stochastic differentials does not justify taking ordinary square roots.

Tags

StatisticsVolatility
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# 44786


# For an Ito Process, $d\ln{X} \neq \frac{dX}{X}$ and $(d\ln{X})^2 = (\frac{dX}{X})^2$, but $d\ln{X} \neq \pm \frac{dX}{X}$












In normal calculus we can write $d\ln{x} = \frac{dx}{x}$ since there is no quadratic variation to deal with. This isn't true for stochastic processes, and Ito's Lemma is used to calculate $d\ln{X}$. So when I was reading about volatility/realized volatility, I saw that there were two expressions used for realized variance $\sigma ^2 dt$, one is $\frac{dX}{X}^2$, while the other is $d\ln{X}^2$. So although these are equal I was wondering how come taking the square root doesn't lead to a valid relationship. I know that certain terms are 'negligible' thus leading the square operation to give the same results, but I don't see how to 'reverse' this using the square root.

Here's my work (it's just using Ito's Lemma and simplifying quadratic terms):

For an Ito Process of form: $\frac{dX_t}{X_t} = \mu(t,X_t) dt + \sigma(t,X_t) dW_t$

\begin{equation} (\frac{dX}{X})^2 = (\mu dt + \sigma dW)^2 = \sigma^2dt \end{equation} while \begin{align} d\ln{X} = \frac{1}{X}dX - \frac{1}{2X^2}*dX^2 = \frac{dX}{X} - \frac{1}{2} \sigma ^2dt = (\mu - \frac{1}{2} \sigma ^2 ) dt + \sigma dW \\ (d\ln{X})^2 = ((\mu - \frac{1}{2} \sigma ^2 ) dt + \sigma dW)^2 = (\sigma dW) ^2 = \sigma ^2 dt \end{align}

so $d\ln{X} \neq \frac{dX}{X}$ and $(d\ln{X})^2 = (\frac{dX}{X})^2$.

In general, if $x^2 = y^2$, then $x = \pm y$, but $d\ln{X} \neq \pm \frac{dX}{X}$ since $d\ln{X}$ has the $- \frac{1}{2} \sigma ^2 dt $ term due to the quadratic variation of a Brownian motion.

So basically I am wondering why this is invalid, and what other sorts of operations on stochastic processes are invalid. Thanks!

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.