Model-Free Option Measures with Finite Strike Coverage

# How many options are necessary in computing a "model-free" measure?


# How many options are necessary in computing a "model-free" measure?












The VIX itself is computed via a "model free" measure, or rather, using a continuum of OTM option prices to come up with an "P-measure" of implied volatility.

It is perhaps obvious that the more options there are, the more accurate the approximation. Are there any ideas/papers about how one should go about computing model-free measures (implied volatility, risk-neutral skewness/kurtosis)? Or does it vary case-by-case?

Should one try to go for as many options as possible? Therefore, (always) using a type of function to fit a plot (such as the call option payoff) and then discretizing to generate this continuum of prices?

## Answer by Andrea (score 6, accepted)

https://quant.stackexchange.com/a/81171

The answer is "depends".

If one can replicate a given payoff using observable prices, then this is by definition a model free price (unless there is arbitrage of course).

The easier case is that any European payoff can be replicated using a continuum of calls and puts (OTM, ITM does not matter, as long as they cover all strikes), and we can assume their bid-ask spread to be zero.

This is often not the case in reality as there will be only a finite number of options available to build the strategy.

In this situation, one will only be able to build a over/under replication strategy, and so one can only generate bid and ask prices for the desired payoff.

The question is: what is the cheapest way to replicate a payoff from above? So that the final payoff is always above the target?

And the same for the opposite, "below the target".

Formally, if $T(s)$ is the target payoff and I can use calls for strikes $K_i$ and price $P_i$ at the same maturity, I have to solve these 2 problems

$ASK = \min_{\alpha_i} \sum_i \alpha_i P_i$ subject to $\sum_i \alpha_i C_i(s) \ge T(s) \, \forall s$

and

$BID = \max_{\alpha_i} \sum_i \alpha_i P_i$ subject to $\sum_i \alpha_i C_i(s) \le T(s) \, \forall s$

The closer the 2 prices are, the more model-free price one obtains.

The problem is simple to solve if one accepts to check the over/under replication on a finite grid (and not for all $s$). It then becomes an easy Linear Programming problem.

It goes without saying that there will be cases where this works better than others, and that same unfeasible cases are more benign than others.

EDIT: The Lagrange multiplier thing was not correct, removed it.

## Answer by Frido (score 5)

https://quant.stackexchange.com/a/81188

Andrea has largely answered your question (+1), namely that a continuum of options are needed to synthesize or price a model free quantity. For instance $$ E[F(X)] = \int_{-\infty}^{\infty} F(x) p(x) dx $$ is model-free assuming $p(x)$ can be computed from observable vanilla options prices via Breeden-Litzenberger.

The integral can be approximated from below or above as also explained by Andrea.

What I'd like to add is that model-free is actually a very strong condition / requirement. A slightly weaker notion might be called "parameter-free".

A case in point is the continuously monitored variance swap / VIX index you mentioned. Given only vanilla European options on the underlying, the variance swap / VIX is "model-free" only for the class of diffusion models. In the presence of jumps the variance swap is no longer given by the usual replication formula involving puts and calls. Hence the variance swap is, imo, more aptly called parameter-free for diffusion models as it does not depend on the parameters (and hence specific form) of the diffusion model generating the smile/skew.

And an even weaker notion, but still quite general, is parameter-free approximation. An example of this is the volatility swap, which has a parameter-free approximation for the class of stochastic volatility models.