Why Squared Brownian Increments Contribute a Drift Term in Itô’s Lemma

Summary

The document asks why Itô’s multiplication rules set products involving dt to zero while treating the square of a stochastic increment as dt. The answer offers a discrete branching intuition: consider the possible paths over a small step, square their changes, and observe that the squared outcomes share the same leading contribution. In the continuous limit, this common contribution is represented by dt, so the quadratic variation enters the drift-like part of Itô’s formula.

The response points to a paper for further explanation but supplies no derivation or worked example itself. The intuition is therefore only sketched, and should not be read as saying every squared increment is literally nonrandom at finite step size. The deterministic dt term describes the limiting quadratic variation behavior of Brownian motion, while finite increments remain random.

Key ideas

• Itô calculus assigns a nonzero contribution to the square of a Brownian increment because its size is on the order of elapsed time.
• A discrete branching picture can help explain why squared path changes share a common leading contribution.
• In the continuous limit, accumulated squared increments converge to quadratic variation proportional to time.
• The answer sketches this intuition but relies on an external paper for details.

Tags

# Ito's Lemma: Multiplication Rule


# Ito's Lemma: Multiplication Rule












I have a conceptual question about Ito's lemma, in particular, the multiplication. Ito's multiplication rule states, that multiplying dt by itself or by dx (the stochastic differential) equals zero. That I understand. However, it also states, that multiplying dx by itself yields dt. I understand that dx is proportional to the square root of time. Nevertheless, dx^2 is still a stochastic process. Nevertheless, in Ito's Lemma it is then treated as if it were part of the deterministic part of the formula. This I do not understand since the result is still a (albeit different) random process. Thanks very much for any enlightenment!

## Answer by vonjd (score 5, accepted)

https://quant.stackexchange.com/a/37107

May I point your attention to my following paper, where I address this question in an intuitive manner on page 12:

von Jouanne-Diedrich, Holger, Ito, Stratonovich and Friends (May 18, 2017). Available at SSRN: https://ssrn.com/abstract=2956257

The basic idea in the discrete case is to go through all possible branches and then square the differences. In each case you will see that it results in the same number - which renders the stochastic case, for all intents and purposes, deterministic!

Details can be found in the paper (and I would love to get your feedback on the paper - Thank you :-)

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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