# Square of Wiener process

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

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*}

• 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? – Borun Chowdhury May 24 '16 at 15:46
• @BorunChowdhury: Check Section 3.4.2, in particular Theorem 3.4.3, of the book amazon.ca/Stochastic-Calculus-Finance-II-Continuous-Time/dp/…. – Gordon May 24 '16 at 16:04
• @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. – Quantuple May 24 '16 at 16:58
• @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. – Borun Chowdhury May 25 '16 at 8:51
• 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! – Borun Chowdhury May 25 '16 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 $$S_{t+\Delta t}=S_t+a(t,S_t)\Delta t+b(t,S_t)\sqrt{\Delta t}\,Z+\frac{1}{2}b(t,S_t)\frac{\partial b(t,S_t)}{\partial S}\Delta t(Z^2-1)$$

• 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. – Borun Chowdhury May 25 '16 at 9:02