# Is the Brownian motion multiplication rule a definition or is it a theorem?

Is the Brownian motion multiplication rule a definition or is it a theorem?

Refer to the highlight part of http://i.stack.imgur.com/doQuT.png where $dw_1(t)dw_1(t)=dt$

It's a lemma! Ito's Lemma gives the change of coordinates rule for stochastic calculus. The multiplication rule is a shorthand way of expressing it.

• Why is it known as a lemma? If you go through the original paper from 1944, you will note that is actually a theorem. – Olorun Jun 2 '15 at 2:30
• I don't know what it's named as a lemma, but everybody believes it's a definition. – SmallChess Jun 2 '15 at 3:14
• most books call it "Ito's Lemma"! – Mark Joshi Jun 2 '15 at 3:56
• Then I would argue that most books are mistaken, since it is a theorem in the original paper. Not that it really matter though. – Olorun Jun 2 '15 at 13:01
• @Olorun: There is no technical difference, only an organizational one. An axiom is a logical sentence with truth value set to true, with no questions asked. A logical framework is a collection of axioms + a collection of derivation rules (rules that tell you how new ture sentences are produced from other true ones). A theorem is a logical sentence that is derivable within a particular logical framework. Now a corollary is what you call a theorem if it follows immediately from other theorems. And a lemma is what you call a theorem when it's just an intermediate result towards another theorem. – Nikolaj-K Jun 30 '15 at 12:39

Formally, this is a shorthand for the quadratic variation. For a more rudimentary definition, $\langle W, W\rangle$ is a process such that $W^2-\langle W, W\rangle$ is a martingale. Moreover, $\langle W, W\rangle_t$ is a limit, in probability, of the variation \begin{align*} \sum_{i=1}^n|W_{t_{i}}-W_{t_{i-1}}|^2, \end{align*} over the partition \begin{align*} 0=t_0 < t_1 < \cdots < t_n = t. \end{align*} From the property of a standard Brownian motion, it can be shown that the above limit equals $t$. That is, $\langle W, W\rangle_t = t$.

In general, if $X_t = \int_0^t \xi_s dW_s$ and $Y_t = \int_0^t \eta_s dW_s$, then \begin{align*} \langle X, Y \rangle_t &= \int_0^t \xi_s \eta_s ds, \end{align*} which we also write as \begin{align*} \langle dX_t, dY_t\rangle = \xi_t\eta_tdt. \end{align*} For $\xi_t=\eta_t = 1$, we then have \begin{align*} \langle dW_t, dW_t\rangle = dt, \end{align*} which is usually write as $dW_t dW_t = dt$.

• Can I add that it means the Brownian motion accumulates t units of quadratic variation. – SmallChess Jun 2 '15 at 15:41
• @StudentT: That is correct. – Gordon Jun 2 '15 at 17:06