# Stochastic Differential

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
$(dw)^2 = dt$