# Why does $\left( dW_t \right)^3 = 0$, where $W_t$ denotes a standard Brownian motion process?

I am currently trying to prove that, for a standard Brownian motion $W_t$ and a twice-differentiable and bounded function $f(x)$, $$dg(W_t) = g'(W_t) dW_t + \frac{1}{2} g''(W_t) dt$$

To do this I have expressed $dg(W_t)$, using the Taylor series expansion, as $$dg(W_t) = g'(W_t) dW_t + \frac{1}{2} g''(W_t) (dW_t)^2 + \frac{1}{3!} g'''(W_t) (dW_t)^3 + \dots$$

From here, I have shown that $(dW_t)^2 = dt$ and I am now trying to complete this proof by showing that $$(dW_t)^k = 0$$ for all integers $k \geq 3$.

Can anyone help me to do this?

• Strictly speaking none of this makes sense; you need to work with the actual sums. But informally speaking, $\Delta W_t$ scales like $\Delta t^{1/2}$ and a Riemann sum has a number of terms scaling like $\Delta t^{-1}$ which gives the result. – Ian Apr 19 '18 at 11:43
• See also these two related questions: math.stackexchange.com/q/1392586/36150 and math.stackexchange.com/q/81865/36150 – saz Apr 19 '18 at 12:09