2
$\begingroup$

In the Black Scholes framework, the parameter sigma (volatility) is the standard deviation of the underlying's returns NOT the standard deviation of the underlying price process. But I often see when people talk about implied volatility that they say e.g: implied vol on this ATM straddle is 10% so the stock would have to move by 10% up or down to break even. Would this be wrong since the 10% refers to the volatility of the returns and not the price process?

$\endgroup$
1
  • 1
    $\begingroup$ This is wrong. To know the breakeven of any vanilla you need to know the price, at certain maturities there will be approximations, but it's not possible (at least not in the form you mention here) to give a general price simply the implied vol L. $\endgroup$
    – will
    Dec 14 '19 at 3:04
3
$\begingroup$

I think there are two questions here.

First, this abuse of terminology regrading a) the volatility term in the equation describing the dynamics of the process, $dS=rSdt+\sigma S dW$, which is sometimes referred to as the instantaneous volatility, and b) the volatility of the price itself is quite 'standard'! Probably because in most cases the meaning is clear from the context, though that's not a great excuse. Also you will hear of the total instantaneous volatility, which is $\sigma \sqrt{T}$.

A related confusion arises in the terminology around the diffusion coefficient - finance people would identify the diffusion coefficient as the coefficient of the brownian in the SDE, say $\sigma$, whereas physicists would would identify it as $\frac{1}{2}\sigma^2$ (think the diffusion equation).

Second, as @will commented, the precise statement regrading the breakeven is in the approximation sense. As a simple explanation, recall the Brenner Subrahmanyam's approximation of Black Scholes:

$C_0 \approx 0.4 S_0 \sigma \sqrt{T}$

We can approximate the price of the straddle by just doubling the call price:

$\mathrm{Straddle} \approx 0.8 S_0 \sigma \sqrt{T}$

So to recover the cost, the price has to move by roughly $\sigma$ percent.

$\endgroup$

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.