How is implied vol calculated if the quoted prices are out of the range for any possible volatility? E.g. Current quote on CBOE for options expiring on Aug 16, 2014

Calls   Last Sale   Net     Bid     Ask     Vol     Open Int    
BSZ1416H1900-E  0.0     0.0     0.66    0.81    0   0   14 Aug 1900.00      BSZ1416T1900-E  0.0     0.0     0.20    0.32    0   0
BSZ1416H1950-E  0.0     0.0     *0.45   0.60*   0   0   14 Aug 1950.00       BSZ1416T1950-E     0.0     0.0     0.40    0.55    0   0
BSZ1416H2000-E  0.0     0.0     0.20    0.32    0   0   14 Aug 2000.00      BSZ1416T2000-E  0.0     0.0     0.67    0.82    0   0

For BSZ1416H1950-E both the bid and ask are outside of the range. (cannot upload the picture I created, it showed that max value attainable is .4 around 50% volatility.

Normally supply/demand, liquidity are all figured into IV. How do you find IV? What is the mechanism to handle such things for building vol surface? How are such differences used, in a control variate like manner, perhaps, to estimate the price at some time in the future? Thanks.

  • $\begingroup$ A similar question was already asked and should appear on the right on your screen: quant.stackexchange.com/questions/10749/…? $\endgroup$
    – AFK
    Jun 11, 2014 at 20:25
  • $\begingroup$ This question is about how to use the ask price for forecasting IV or future prices, when it exceed the BS maximum possible price $\endgroup$
    – user12348
    Jun 11, 2014 at 22:45

2 Answers 2


Are you sure you are using the correct pricing formula.

For a binary (digital) call that pays $1$, the simple Black-Scholes price at time $t=0$ is

$$ C_d = e^{-rT}N(d_2)$$ $$d_2 = \frac{\text{ln}(F/K) - \frac1{2}\sigma^2T}{\sigma \sqrt{T}}$$ where $N$ is the standard normal distribution function, $F=Se^{(r-q)T}$ is the forward index price, $S$ is the spot index price, $K$ is the strike price, $T$ is the time to expiration and $\sigma$ is the implied volatility.

Here are some current values

$$S = 1942, r = 0.25\%, q = 1.97\%.$$

For the 14 Aug 1950 call

$$K = 1950, T = 64 \ \text{days}=0.1752 \ \text{years} ,$$

and assuming an implied volatility of $\sigma = 12\%$, the binary call price is $0.43$.

So the quotes you are showing look reasonable under the current implied volatility conditions.

In terms of your general question about finding implied volatility, there are two issues. (1) How to build a no-arbitrage pricing model that will correctly match market prices of vanilla calls and puts, and (2) how to price more exotic options (such as binary options) in the new framework.

In general, observed market prices of SPX index options are not consistent with simple Black-Scholes assumptions -- an underlying that follows geometric Brownian motion with constant volatility. The actual prices look like expecations under a probability distribution that is not lognormal -- perhaps more skewed. Implied volatility -- that value which makes the Black-Scholes formula match the market price -- varies both with strike price and time to expiration. In theory, if we knew the market price of a call option $C(S,t;K,T)$ for every conceivable strike price $K$ when the index price is $S$ at time $t$, then for a given time-to-expiration $T$ we could find the implied probability density function as

$$f(S) = e^{r(T-t)}\frac{\partial^2}{\partial K^2}C(S,t;K,T).$$

In practice, there are not enough market price observations to use this formula directly in a meaningful way -- but it suggests there are broader stochastic models (with more degrees of freedom) that can be used to generate no-arbitrage option prices that match market prices. One of the more popular approaches is the local volatility model that assumes the underlying index price follows a stochastic process of the form

$$dS_t=\mu S_t dt + \sigma(S_t)S_tdW_t$$

where $W_t$ is a Brownian motion and the volatility $\sigma(\cdot)$ is not a constant but a deterministic function of the underlying price. There is an extensive literature on the local volatility model indicating how to calibrate the function $\sigma(\cdot)$ to match market prices.

For a binary option, it is not entirely clear what simple pricing appoach should be used when vanilla calls and puts exhibit an implied volatility skew. One possibility is to find the price in terms of a replicating portfolio of vanilla options. If a binary option pays $1$ when the index is above a strike price $K$ then it can be replicated, in theory,approximately using a call spread. We would buy a number $1/\delta$ of ordinary calls with strike price $K$ and sell the same number of calls with strike price $K+\delta.$ In this way

$$C_d(S,t;K,T) \approx \frac1{\delta} \big[C(S,t;K,T)-C(S,t;K+\delta , T)\big]$$

Ideally we would make $\delta$ as small as possible, but there are practical limitations in terms of available strikes and the eventual leverage that would be applied. Nevertheless, this replication model suggests how the binary option might be priced in the presence of a volatility skew. Taking the limit as $\delta \rightarrow 0$ we get

$$C_d(S,t;K,T) \approx \frac{\partial}{\partial K} C(S,t;K,T),$$

and this relationship indicates how to extract the price of the binary option that is consistent with the prices of vanilla options in a framework (eg. local volatility) where implied volatility depends on strike.

  • $\begingroup$ The largest value using your inputs is .471 at .15% IV. The ask side of .6 is clearly out of the possibility of occurring at any volatility. The point is even with the skew, .6 will be unattainable. So, how do you use the current pricing structure to forecast or modify BS prices? $\endgroup$
    – user12348
    Jun 11, 2014 at 22:42
  • 1
    $\begingroup$ Regardless of model the binary call price is the discounted risk-neutral probability that the index is above the strike at expiration, so we always have a value of 1 as an upper bound. As you are finding, there is an even lower limit that depends on the current forward price (currently below spot) relative to the strike no matter how high the implied vol. So it means an ask of 0.6 is high relative to any no-arbitrage model. It could be because of low liquidity, supply/demand or just a bad print. At times OTC dealers have wide bid/ask for deep OTM options because they don't want the risk. $\endgroup$
    – RRL
    Jun 12, 2014 at 2:28
  • $\begingroup$ Well expressed, led me to dig into several interesting things. $\endgroup$
    – user12348
    Jun 16, 2014 at 14:28

The skew plays an important part for pricing binaries. In S&P the VIX increases on declines and decreases on rises. We can explain a part of the premium by assuming the Black-Schole Call option captures the underlying volatility. Let us represent call option as $C(K,\sigma)$ and binary $V_{Binary}(K,\sigma)$ Then we can write $$ V_{Binary}(K,\sigma) = -\frac {\partial C}{\partial K}(K,\sigma)$$ Apply chain rule, $$= -\left[\frac{\partial C(K,\sigma)}{\partial K} \frac{\partial K}{\partial K} + \frac{\partial C(K,\sigma)}{\partial \sigma} \frac{\partial \sigma}{\partial K}\right] $$ The term $-\frac{\partial C(K,\sigma)}{\partial K} \texttt{ is quoted} \space V_{Binary} $ and $ \frac{\partial C(K,\sigma)}{\partial \sigma}$ is vanilla call option vega and $\frac{\partial \sigma}{\partial K} $ is the skew of the vanilla call option. Therefore, $$ $V_{Binary}(K,\sigma) = V_{Binary}(S,t) - C_{Vega}(S,t) * C_{Skew}(K,\sigma) $$ $$ =e^{-rT} N(d_2) - S \sqrt {T} N'(d_1) * Skew $$ Skew can be estimated from the SPX options close to the strike K. The above equation has all the values that are readily available. Of course, on top of that there is premium for supply/demand, liquidity etc.

As a note, the SPX contracts are very liquid but the liquidity in Binaries is another add on. Binaries are embedded in many structured products and there are many way to use them for creating new products or risk management. BSZ maybe are being used as lotto play. Hopefully more players will recognize its importance and increase demand.


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