In Ito's calculus one often comes $dW^2=dt$. How does this come about? What is it's relation to the Milstein method?


2 Answers 2


In general $dX^2$ is an ad-hoc or heuristic form of $d\langle X, X\rangle_t$, where $\langle X, X\rangle_t$ is the quadratic variation, which is defined by \begin{align*} \langle X, X\rangle_t = \lim_{\pi\rightarrow 0} \sum_{i=1}^n (X_{t_i}-X_{t_{i-1}})^2. \end{align*} Here, $0=t_0 < \cdots < t_n = t$, and $\pi = \max\{ t_i-t_{i-1}, i=1,\ldots, n\}$.

For a Brownian motion $\{W_t, t \ge 0\}$, it can be shown that $\langle W, W\rangle_t = t$. Therefore \begin{align*} dW^2 = d\langle W, W\rangle_t = dt. \end{align*}

  • $\begingroup$ How would you should $\langle W, W \rangle_t=t$? My first guess would be to treat each term in the summation as an independent $\chi^2$ distribution and then show that in the limit that $dt \to 0$ the standard deviation squeezes to 0 proportional to $dt$ for the biggest term in the summation. If so then this is the method I write in my answer. Is there an alternate way to show this? $\endgroup$ May 24, 2016 at 15:46
  • $\begingroup$ @BorunChowdhury: Check Section 3.4.2, in particular Theorem 3.4.3, of the book amazon.ca/Stochastic-Calculus-Finance-II-Continuous-Time/dp/…. $\endgroup$
    – Gordon
    May 24, 2016 at 16:04
  • 2
    $\begingroup$ @Borun Chowdhury. In your answer you are not computing the same quantity. Above is quadratic variation of a stochastic process over a finite horizon $[0,t]$. You sort of compute qv over a single time step only. You could apply your reasoning + CLT to derive a similar result but it will presumably amount to assuming a uniform partition of $[0,t]$ while the true result holds for any kind of partition as long as the limit is taken according to what is mentioned in Gordon's answer. $\endgroup$
    – Quantuple
    May 24, 2016 at 16:58
  • $\begingroup$ @Quantuple I checked theorem 3.4.3 in the book Gordon mentioned and it is indeed doing what I am doing. In fact I think my proof is a bit shorter because I use the fact that each increment is normally distributed and hence its square is a chi-squared distribution and therefore I directly know the mean and variance. $\endgroup$ May 25, 2016 at 8:51
  • 1
    $\begingroup$ Yes Gordon I do see your point now. I think Shreve is saying that interpreting $dW^2= dt$ outside an integral doesn't make sense as the distribution of $dW^2/dt$ does not depend on the size of $dt$. The convergence can only be understood with the summation/integration and this is written informally as $dW^2= dt$. Very nice. Thanks! $\endgroup$ May 25, 2016 at 13:17

I don't think just knowing $dW_t\,dW_t=dt$ is enough.We assume $$d{{X}_{t}}=\mu (t,{{X}_{t}})dt+\sigma (t,{{X}_{t}})d{{W}_{t}}\,\,\,\,\,(1)$$ The idea behind the Milstein scheme is that the accuracy of the discretization can be increased by expanding the coefficients$\mu =\mu (t,{{X}_{t}})$ and $\sigma =\sigma (t,{{X}_{t}})$ via Ito’s lemma. This is sensible since the coefficients are also functions of $X_t$. Indeed, we can apply Ito’s Lemma to the functions $\mu_t$ and $\sigma_t$ as we would for any differentiable function of $X_t$. By Ito’s lemma, then, the coefficients follow the SDEs \begin{align} & d\mu =({{\mu }_{t}}+\mu {{\mu }_{x}}+\frac{1}{2}{{\sigma }^{2}}{{\mu }_{xx}})dt+{{\mu }_{x}}\sigma d{{W}_{t}} \\ & d\sigma =({{\sigma }_{t}}+\mu {{\sigma }_{x}}+\frac{1}{2}{{\sigma }^{2}}{{\sigma }_{xx}})dt+{{\sigma }_{x}}\sigma d{{W}_{t}} \\ \end{align} then \begin{align} & \mu (s,{{X}_{s}})=\mu (t,{{X}_{t}})+\int_{t}^{s}{({{\mu }_{u}}+\mu {{\mu }_{x}}+\frac{1}{2}{{\sigma }^{2}}{{\mu }_{xx}})du}+\int_{t}^{s}{{{\mu }_{x}}\sigma d{{W}_{u}}} \\ & \sigma (s,{{X}_{s}})=\sigma (t,{{X}_{t}})+\int_{t}^{s}{({{\sigma }_{u}}+\mu {{\sigma }_{x}}+\frac{1}{2}{{\sigma }^{2}}{{\sigma }_{xx}})du}+\int_{t}^{s}{{{\sigma }_{x}}\sigma d{{W}_{u}}} \\ \end{align} Substitute for $\mu_t$ and $\sigma_t$ inside the integrals of Equation (1), we have \begin{align} {{X}_{t+\Delta t}}={{X}_{t}}+\int_{t}^{t+\Delta t}{[\mu (t,{{X}_{t}})+\int_{t}^{s}{({{\mu }_{u}}+\mu {{\mu }_{x}}+\frac{1}{2}{{\sigma }^{2}}{{\mu }_{xx}})du}+\int_{t}^{s}{{{\mu }_{x}}\sigma d{{W}_{u}}}]}\,ds \\ +\int_{t}^{t+\Delta t}{[\sigma (t,{{X}_{t}})+\int_{t}^{s}{({{\sigma }_{u}}+\mu {{\sigma }_{x}}+\frac{1}{2}{{\sigma }^{2}}{{\sigma }_{xx}})du}+\int_{t}^{s}{{{\sigma }_{x}}\sigma d{{W}_{u}}}]\,d{{W}_{s}}} \\ \end{align} The differentials higher than order one are $dsdu=\mathcal O(dt^2)$ and $dsdW_u=\mathcal O(dt^{\frac{3}{2}})$ are ignored. The term involving $dW_u dW_s$ is retained since it is $\mathcal O(dt)$, of order one. This implies that $${{X}_{t+\Delta t}}={{X}_{t}}+\mu (t,{{X}_{t}})\Delta t+\sigma (t,{{X}_{t}})({{W}_{t+\Delta t}}-{{W}_{t}})+\int_{t}^{t+\Delta t}{\int_{t}^{s}{{{\sigma }_{x}}\sigma d{{W}_{u}}}}d{{W}_{s}}\,\,(2)$$ Apply Euler discretization to the last term in (2) to obtain \begin{align} & \int_{t}^{t+\Delta t}{\int_{t}^{s}{{{\sigma }_{x}}\sigma d{{W}_{u}}}}d{{W}_{s}}\approx {{\sigma }_{x}}\sigma \int_{t}^{t+\Delta t}{\int_{t}^{s}{d{{W}_{u}}}}d{{W}_{s}}={{\sigma }_{x}}\sigma \int_{t}^{t+\Delta t}{({{W}_{s}}-{{W}_{t}})}d{{W}_{s}} \\ & \quad \quad \quad \quad \quad \quad \quad \quad ={{\sigma }_{x}}\sigma \int_{t}^{t+\Delta t}{{{W}_{s}}\,}d{{W}_{s}}-{{\sigma }_{x}}\sigma ({{W}_{t+\Delta t}}{{W}_{t}}-{{W}_{t}}^{2}) \\ & \quad \quad \quad \quad \quad \quad \quad \quad =\frac{1}{2}{{\sigma }_{x}}\sigma (W_{t+\Delta t}^{2}-W_{t}^{2}-\Delta t)-{{\sigma }_{x}}\sigma ({{W}_{t+\Delta t}}{{W}_{t}}-{{W}_{t}}^{2}) \\ & \quad \quad \quad \quad \quad \quad \quad \quad =\frac{1}{2}{{\sigma }_{x}}\sigma [{{({{W}_{t+\Delta t}}-{{W}_{t}})}^{2}}-\Delta t]\,\,\,(3)\\ \end{align} we know ${{W}_{t+\Delta t}}-{{W}_{t}}\overset{d}{\mathop{=}}\,\sqrt{\Delta t}Z$, by substitute (3) in (2) we have $${{\widehat{X}}_{t+\Delta t}}={{\widehat{X}}_{t}}+\mu \,\Delta t+\sigma \sqrt{\Delta t}Z+\frac{1}{2}{{\sigma }_{x}}\sigma \Delta t\,({{Z}^{2}}-1)$$ as a result $$S_{t+\Delta t}=S_t+a(t,S_t)\Delta t+b(t,S_t)\sqrt{\Delta t}\,Z+\frac{1}{2}b(t,S_t)\frac{\partial b(t,S_t)}{\partial S}\Delta t(Z^2-1)$$

  • $\begingroup$ I am not saying knowing $dW^2= dt$ is enough. I apologise if I gave that impression. I am saying that the additional term in the Milstein method involves a $(\Delta W^2- \Delta t)$ and while the mean of this term is zero so it may look odd on first sight, its variance is $2 \Delta t$. Anyway, I like your detailed answer. My comment about the Milstein method was about a confusion I had related what I say in this comment. $\endgroup$ May 25, 2016 at 9:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.