Why the Mid Implied Volatility Is Usually the Bid-Ask Average
Summary
The document considers how to calculate a mid implied volatility from bid and ask implied volatilities without choosing a pricing model. It concludes that averaging the two volatilities is the only model-free approach. The explanation maps volatility to a European option price and uses a first-order Taylor approximation of the inverse mapping around the midpoint price. Locally, price changes translate into volatility changes through vega, and the symmetric bid-ask price spread leads to the arithmetic average of the two implied volatilities.
This reasoning is exact when option value is linear in volatility, but option prices generally have curvature. The answer notes that for options far from the money, a second-order approximation brings in volga and makes the midpoint relationship nonlinear. Resolving that effect requires assumptions about a pricing model. The question explicitly sets aside whether the resulting midpoint is arbitrage-free; the discussion addresses how to average quotes, not how to enforce static-arbitrage constraints or choose a model-specific fair value.
Key ideas
- The arithmetic average of bid and ask implied volatilities is the model-free midpoint convention described.
- A first-order approximation explains the average through the local relationship between price changes and vega.
- The averaging rule is exact when price is linear in volatility.
- Material price curvature introduces volga and can make the midpoint calculation model-dependent.
- The discussion does not assess whether the midpoint quote is arbitrage-free.
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Full text
# Mid vol from bid/ask vols for equity options
# Mid vol from bid/ask vols for equity options
Given an arbitrary bid IV and ask IV is it possible to compute a mid IV in a model agnostic fashion? Is there anything else aside averaging the bid and ask vols or interpolation between bid/ask ivs and prices? Having the mid IV be arbitrage-free is not a concern
## Answer by Quantuple (score 6, accepted)
https://quant.stackexchange.com/a/71928
Averaging the bid/ask volatilities would be the only "model-free" way to do it.
Indeed, recalling that: $$ f^{-1}( f(x) ) = x $$ applying the chain rule of standard calculus gives $$ (f^{-1})'(y) = 1/f'(f^{-1}(y)) $$
Defining $f$ as the function which maps a Black-Scholes volatility to the European vanilla price (all other pricing parameters held constant) $$ f: \sigma \to V(\sigma,\cdot) $$ its inverse $f^{-1}$ becomes the function which outputs the implied volatility from an input price.
From an order 1 Taylor expansion of $f^{1}$ $$ (f^{-1})'(y) = (f^{-1})'(y_0) + 1/f'(f^{-1}(y_0))(y-y_0) + o(y) $$ of the bid/ask implied volatilities ($y\in \{C_{bid},C_{ask}\}$) around the mid level ($y_0=C_{mid}$) would get you \begin{align} \sigma_{bid} &= \sigma_{mid} + 1/\nu_{mid}(C_{bid}-C_{mid}) \\ \sigma_{ask} &= \sigma_{mid} + 1/\nu_{mid}(C_{ask}-C_{mid}) \end{align} where $\nu_{mid}$ denotes the contract's Vega (using the mid implied vol in input). Letting the bid/ask spread $$ \delta := C_{ask}-C_{bid} $$ and adding the two equations together would then indeed yield $$\sigma_{mid} = 1/2(\sigma_{bid}+\sigma_{ask})$$
Basically what this says is as long as $f$ is linear in the volatility, the rule of thumb is exact (since $f^{-1}$ will be too, assuming it exists).
Now, as soon as you have material convexity (i.e. far from the money), using an order 2 expansion rather than an order 1 expansion, you'll see that the Volga at $\sigma_{mid}$ appears thereby yielding a non-linear "fixed point" problem can only be solved assuming a model.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.