Interpreting Implied Volatility and Straddle Break-Even Moves
Summary
The document clarifies that Black–Scholes volatility is a parameter governing proportional price changes, rather than the standard deviation of the price level itself. In the geometric Brownian motion model, the instantaneous volatility scales the underlying price in the random component. Market shorthand that an implied volatility predicts a percentage move over the option horizon is therefore an approximate interpretation, not a literal statement about the distribution of price levels.
It uses the Brenner–Subrahmanyam approximation for an at-the-money call and doubles that estimate to approximate the price of an at-the-money straddle. Equating this estimated premium with a breakeven move gives a rough move of volatility times the square root of time as a percentage of spot. The argument is an approximation based on Black–Scholes assumptions and near-the-money options; it does not guarantee realized movement or account for skew, jumps, transaction costs, or the exact option payoff.
Key ideas
- Black–Scholes volatility scales proportional price changes in the underlying’s stochastic process.
- The horizon volatility is commonly expressed as instantaneous volatility multiplied by the square root of time.
- An at-the-money straddle’s breakeven move can be approximated from its estimated premium.
- The implied-volatility move interpretation is approximate and does not predict what the asset will realize.
Tags
Full text
# Answer by Magic is in the chain (score 3, accepted)
# Since implied volatility is the standard deviation of returns, why do people treat it as the standard deviation of the price process?
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?
## Answer by Magic is in the chain (score 3, accepted)
https://quant.stackexchange.com/a/50296
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.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.