Model-Free Implied Volatility

Ask one weather forecaster for tomorrow's temperature and you get one model's opinion. Instead, poll a thousand people who've each staked money on a specific temperature bracket, and add up how they've bet — you get a market-implied forecast that doesn't depend on any single forecaster's model. That's what "model-free" implied vol does with options: instead of asking "what vol makes Black-Scholes match one option's price," it reads expected variance directly off the entire strip of traded prices across all strikes.

The idea, defined

The key result: a portfolio of out-of-the-money options across every strike, weighted by $1/K^2$, replicates a claim that pays off based on realized variance — no assumption about how the underlying moves is needed, only that a full range of strikes trades.

$$\sigma_{\text{MF}}^2 = \frac{2}{T} \int_0^\infty \frac{P(K)}{K^2}\, dK$$

In words: take every out-of-the-money option price $P(K)$ across the full range of strikes $K$ (puts below the forward, calls above it, roughly), divide each by the square of its strike, add them all up (the integral), and scale by $2/T$ where $T$ is time to expiry. The result, $\sigma_{\text{MF}}^2$, is the market's risk-neutral expected variance over that expiry — recoverable from prices alone, without ever fitting a model or solving for any single option's Black-Scholes vol. This is, in essence, the fair strike of a variance swap, and it's the actual computation behind the CBOE's VIX index.

Worked example 1 — a discretized toy version

In practice the integral becomes a discrete sum over the actual traded strikes, spaced $\Delta K$ apart: $\sigma_{\text{MF}}^2 \approx \frac{2}{T}\sum_i \frac{\Delta K}{K_i^2} P(K_i)$. Suppose strikes are spaced $5 apart and, for a 1-month expiry ($T = 1/12$), three relevant OTM option prices are: $90 put at $0.40, $95 put at $1.00, $105 call at $1.10. Approximating with just these three terms: $\frac{5}{90^2}(0.40) + \frac{5}{95^2}(1.00) + \frac{5}{105^2}(1.10) = 0.000247 + 0.000554 + 0.000499 = 0.00130$. Multiply by $2/T = 2/(1/12) = 24$: $\sigma_{\text{MF}}^2 \approx 24 \times 0.00130 = 0.0312$, so $\sigma_{\text{MF}} \approx \sqrt{0.0312} \approx$ 17.7%. A real calculation sums dozens of strikes; this toy version shows the mechanics with three.

Worked example 2 — contrast with a single-strike Black-Scholes vol

Suppose the $100 ATM call alone, priced with Black-Scholes, backs out an implied vol of 18.5% — the "one number, one option" approach. The model-free number from example 1, 17.7%, is close but not identical, and that's expected: the ATM Black-Scholes vol reflects only the level at one strike, while $\sigma_{\text{MF}}$ blends information from the entire smile, including the wings, weighted by how much variance each strike contributes. If the smile has a pronounced skew (OTM puts trading rich relative to the ATM level, as equity index smiles typically do), $\sigma_{\text{MF}}$ will sit above the ATM Black-Scholes vol, because the rich put wing pulls the strike-weighted average up — a genuinely different, and arguably more complete, measure of expected variance.

What this means in practice

Model-free implied vol underlies the VIX and equivalent indices in other markets, and it's the theoretically correct fair strike for a variance swap — a contract that pays the difference between realized and this pre-agreed implied variance. Traders use it as the cleanest available forward-looking measure of the market's variance expectation, precisely because it doesn't inherit the biases of a single strike or a single pricing model.

Practice

1. Using the discretized formula, if a single relevant strike at $100 with 5-wide spacing has an OTM option price of $1.20 and $T = 1/12$, what is that strike's contribution to $\sigma_{\text{MF}}^2$ before summing with the rest of the strip?
2. Why would a steep downside skew (rich OTM puts) push the model-free implied vol above the plain ATM Black-Scholes vol?