Quadratic Variation in Delta-Hedging Simulations
Summary
The document addresses an apparent contradiction in a Black–Scholes delta-hedging calculation when one simulated stock-price increment is zero. A literal reading of the differential notation suggests that the gamma term vanishes at that step, while the time decay term remains negative; with a zero risk-free rate, this seems inconsistent with the stated hedge accounting.
The resolution is that the squared differential in Itō calculus is shorthand for the increment in quadratic variation, not an ordinary algebraic square that must vanish whenever a realized price increment is zero. Under geometric Brownian motion, the quadratic variation accumulates according to instantaneous variance over time. Thus a single zero observed increment does not make the quadratic-variation contribution zero. The exchange provides a conceptual correction to interpreting stochastic differentials; it does not discuss simulation design, discretization error, or hedging performance beyond this point.
Key ideas
- In Itō calculus, the squared price differential represents quadratic variation rather than ordinary multiplication.
- A realized zero price increment does not imply zero quadratic variation over the time step.
- The volatility contribution in the Black–Scholes hedge relation therefore need not disappear when the sampled increment is zero.
- Treating differential shorthand as ordinary algebra can create apparent contradictions in a hedging derivation.
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# A bug in delta hedging, when for a certain step dS=0
# A bug in delta hedging, when for a certain step dS=0
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
$\frac{1}{2}\frac{\partial^2 V}{\partial S^2}(dS)^2=\frac{1}{2}\frac{\partial^2 V}{\partial S^2}\sigma^2S^2$
where
$dS=\mu Sdt+\sigma SdW$
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
$dS=0$
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?
## Answer by Gordon (score 2)
https://quant.stackexchange.com/a/32351
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 $d[S, S]_t=0$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.