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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Ian Apr 19 '18 at 11:43
  • $\begingroup$ See also these two related questions: math.stackexchange.com/q/1392586/36150 and math.stackexchange.com/q/81865/36150 $\endgroup$ – saz Apr 19 '18 at 12:09

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