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I'm trying to compute the implied volatility of a binary option but I cannot get some of the strikes to reach a convergent solution using either a Monte Carlo pricing model or an analytical Black Scholes model, minimizing using Newton's method. Since binaries are essentially the derivative of a vanilla call wrt the strike, is it even possible to always compute an implied volatility?

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So, I've gotten 3 answers - an emphatic yes, a strong no, and a maybe. Can anyone break the tie? – Mike Mar 31 '14 at 23:54
Mike, I think @brian-b and pc-iit would both agree that while numerical solution for a valid (0<BC<{e}^{-r\tau}) price can always be calculated, sometimes option ask is quoted above the upper value. The real answer to your question is that sometimes you cannot calculate IV. This is true for vanilla options as well (e.g. itm bid below intrinsic value) – Mar 4 at 20:27

4 Answers 4

No, there is an upper limit to a binary option's value, based on the interest rate and how much of the distribution can be packed under the payoff region. Essentially

$$C = e^{-rT} \int_K^\infty \psi(S_T) dS_T$$

for calls and

$$ P = e^{-rT} \int_0^K \psi(S_T) dS_T$$

for puts. Neither of the integrals can ever exceed 1.0 and often they take on a smaller maximum in $\sigma$.

If you are working with market ask prices, it is entirely possible they are above the Black-Scholes maximum, especially for binary puts.

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Ofcourse, It is always possible to find the implied volatility. The value of binary call is $$ {e}^{-r(T-t)}N(d_2) $$ where $$ d_2=\frac{ln(\frac{S}{E})+(r-D-\frac{\sigma^2}{2})\tau}{\sigma\sqrt\tau} $$

Now, there is nothing that can ever ever stop the newton raphson method to find a $\sigma$ for which the value of binary call is given and is

  • positive
  • if that given value is BC then essentially $$ 0<BC<{e}^{-r\tau} $$

Just, change your code to incrementally change the error value as well as starting point on the newton raphson until it reaches the correct solution. Note: Use the CDF function approximation, as it is better on the extremities than the excel built in NormsDist $$ d = \frac{1}{1 + 0.2316419 * |x|} $$ $$ a_1 = 0.31938153 $$ $$ a_2 = -0.356563782 $$ $$ a_3 = 1.781477937 $$ $$ a_4 = -1.821255978 $$ $$ a_5 = 1.330274429 $$

$$ y = d * (a_1+d*(a_2+d*(a_3+d*(a_4+d*a_5)))) $$ $$ cdf = 1 - \frac{1}{(2\pi)^2} * e^{(-0.5x^2)} * y $$

Except when If x < 0, Then $$ cdf = 1 - cdf $$

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Implied volatility is used to explain the market price, generally of vanilla options. Binary Call values goes to zero when $\sigma \rightarrow \infty $. Increasing volatility does not increase price as it does for vanilla options. You can another question that shed more light. The quoted SPZ binaries on CBOE on SPX are completely out of the range of maximum binary values. The bid and ask are both higher.

Binaries as you know, as you stated, they are derivative of vanilla options, so better to limit implied vol to vanilla options then explain the difference in binaries in terms of market factors such as skew, supply/demand, liquidity. In that sense, the implied volatility is not calculatable in the traditional sense or does not exist. I hope this explains it all.

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An out-of-the money binary call option will have two implied volatilities. After the first implied volatility keep looking at increasingly higher implied volatilities. After a while the binary call option price doesn't rise further, i.e. the binary call vega falls to zero, and then the binary call option price starts falling as implied volatility continues to rise, i.e. the vega turns negative.

Why this condition exists is down to the value of an out-of-the-money call being capped (at roughly 0.5) but as implied volatility keeps rising there is an increased likelihood that the option will be worthless.

There's no point in me proving this here so try it yourselves using the above equations.

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