Let $W_t$ be a Wiener process. It is clear to me that $dW_t$ is of size $\sqrt{dt}$. This can be seen because $$ \mathrm{Var}(W_{t+\Delta} - W_{t})=\Delta. $$ But am I allowed to actually write $(dW_t)^2 = dt$? It looks a bit silly... you have the square of a random variable on the left hand side, and a deterministic variable on the right hand side.

Please can you clarify whether this is right or wrong and perhaps give an explanation for this peculiar identity.


I can clarify 100% that $(dw)^2$= $dt$ and recommend you to accept it as a fact.

Like any other differential, this differential is defined in terms of its integral: $$ \int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}[W(t_{k+1})-W(t_{k})]^{2} $$ Where $t_{k}=t_{0}+k(t_{1}-t_{0})/n$. Since $$ W(t_{k+1})-W(t_{k})=\sqrt{t_{k+1}-t_{k}}\xi_{k}=\sqrt{\frac{t_{1}-t_{0}}{n}}\xi_{k} $$ We have $$ \int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}\frac{t_{1}-t_{0}}{n}\sum_{k=0}^{n-1}\xi_{k}^{2} $$ where $\xi_{0}, \xi_{1},$ . . $\xi_{n-1}$ are independent $N(0,1)$ variables. Clearly the mean ofthe sum is $$ E[\frac{t_{1}-t_{0}}{n}\sum_{k=0}^{n-1}\xi_{k}^{2}]=\frac{t_{1}-t_{0}}{n}\sum_{k=0}^{n-1}E[\xi_{k}^{2}]=t_{1}-t_{0} $$ Since the $\xi$'s are independent, the variance ofthe sum is $$ Var[\frac{t_{1}-t_{0}}{n}\sum_{k=0}^{n-1}\xi_{k}^{2}]=\frac{(t_{1}-t_{0})^{2}}{n^{2}}\sum_{k=0}^{n-1}Var[\xi_{k}^{2}]=\frac{(t_{1}-t_{0})^{2}}{n^{2}}\sum_{k=0}^{n-1}E[(\xi_{k}^{2}-1)^{2}] $$ For unit Gaussian variables, $E[(\xi_{k}^{2}-1)^{2}]=2$, so the variance ofthe sum works out to $$ Var[\frac{t_{1}-t_{0}}{n}\sum_{k=0}^{n-1}\xi_{k}^{2}]=\frac{2}{n}(t_{1}-t_{0})^{2} $$ Thus $$ \int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}S_{n} $$ where the sum $S_{n}$ has mean $t_{1}-t_{0}$ and variance $O(1/n)$ . We conclude that in the limit

$ n\rightarrow\infty$, this integral is $t_{1}-t_{0}$ with certainty. Thus $$ \int_{t_{0}}^{t_{1}}(dW)^{2}=t_{1}-t_{0} $$

For any $t_0$ and $t_1$. Since differentials are defined only in terms of their integral, we can rewrite it as

$(dw)^2 = dt$

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  • 2
    $\begingroup$ Like any other differential, this differential is defined in terms of its integral - that's a bit of an overstatement. It certainly is the case for most of stochastic differentials, but in real analysis the basic differential on a real line is often defined formally way before integrals. $\endgroup$ – Ilya Nov 9 '14 at 13:32

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