Suppose we are doing a delta hedging simulation according to Black Scholes, where the initial condition are [stockPrice, strike, timeToExpire ,riskFreeRate, dividend, sigma, isCall] = [100, 100, 1, 0, 0, 0.2, True]. Let's say they are denoted as [S, K, t, r, q, $\sigma$, ] in Black Scholes. Hence particularly we haver=0 Black Scholes is usually written as

$r\frac{\partial V}{\partial S}S+\frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}+ \frac{\partial V}{\partial t}-rV=0$

where V is the value of option. Rewriting it gives

$rdt(V-\frac{\partial V}{\partial S}S)=(\frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}+\frac{\partial V}{\partial t})dt$

By assumption of Geometric Brownian Motion and approximations of Wiener process we have

$\frac{1}{2}\frac{\partial^2 V}{\partial S^2}(dS)^2=\frac{1}{2}\frac{\partial^2 V}{\partial S^2}\sigma^2S^2$


$dS=\mu Sdt+\sigma SdW$

Hence equivalently,

$rdt(V-\frac{\partial V}{\partial S}S)=(\frac{1}{2}\frac{\partial^2 V}{\partial S^2}(dS)^2+\frac{\partial V}{\partial t})dt------(1)$

By no arbitrage assumption,

$rdt(V-\frac{\partial V}{\partial S}S)=dV-\frac{\partial V}{\partial S}dS ------(2)$

Now let's say on a certain step of hedging, we found that


such that

$\frac{1}{2}\frac{\partial^2 V}{\partial S^2}(dS)^2=0$

And it always holds that $\frac{\partial V}{\partial t}<0$, by (1) and (2) (or simply Itō's lemma) we have

$dV-\frac{\partial V}{\partial S}dS = (\frac{1}{2}\frac{\partial^2 V}{\partial S^2}(dS)^2+\frac{\partial V}{\partial t})dt<0$

Whereas since $r=0$, we have $rdt(V-\frac{\partial V}{\partial S}S)=0$. By (2) we have

$dV-\frac{\partial V}{\partial S}dS=rdt(V-\frac{\partial V}{\partial S}S)=0$

Looking at the above two equations we find a contradiction. So what's the problem out there?


This is a common misunderstanding. Note that $(dS_t)^2$ is just a short hand or heuristic notation for $d[S, S]_t$. Here $[S, S]_t$ is the quadratic variation. You should not literally take $(dS_t)^2$ as the algebraic product of $dS$ and $dS$. In fact, note that \begin{align*} d[S, S]_t = \sigma^2 S^2 dt. \end{align*} Then, even if $dS=0$, it does not mean $d[S, S]_t=0$.

  • $\begingroup$ Sorry for the half-year-lagged follow-up. It was really helpful. Many thanks. $\endgroup$ – wangsrii Jul 31 '17 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.