# How does volatility affect the price of binary options?

In theory, how should volatility affect the price of a binary option? A typical out the money option has more extrinsic value and therefore volatility plays a much more noticeable factor. Now let's say you have a binary option priced at .30 as people do not believe it will be worth 1.00 at expiration. How much does volatility affect this price?

Volatility can be high in the market, inflating the price of all options contracts, but would binary options behave differently? I haven't looked into how they are affected in practice yet, just looking to see if they would be different in theory.

Also, the CBOE's binaries are only available on volatility indexes, so it gets a bit redundant trying to determine how much the "value" of volatility affects the price of binary options on volatility.

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Hi CQM, and thanks for the question. Have you considered registering on the site? It’s easy to register and you will be able to do more on the site, such as vote. –  olaker Sep 29 '11 at 9:24

The price of a binary option, ignoring interest rates, is basically the same as the CDF $\phi(S)$ (or $1-\phi(S)$ ) of the terminal probability distribution. Generally that terminal distribution will be lognormal from the Black-Scholes model, or close to it. Option price is

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

for calls and

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

for puts.

Volatility widens the distribution and, under the Black-Scholes model, shifts its mode a bit. Generally speaking, increased volatility will

• Increase the density in the "payoff region" for out-of-the-money options, thereby increasing their theoretical value. Assuming your option was worth 0.30 due to probabilities and not high risk-free rates $r$, more volatility will increase its value.

• Increase the density in the "no-payoff region" for in-the-money options, thereby decreasing their theoretical value. An option now worth 0.70 will lose value, as the probability of ending outside the payoff region is increased.

As volatility $\sigma$ approaches $\infty$, all option prices converge toward 0 for calls and 1 for puts. In Black-Scholes land, even though the term $\frac{\log(S_0/K)}{\sigma \sqrt{T}} \to 0$ and the probability distribution is spreading out all the way to infinity on the positive as well as negative side of the exponential of its distribution, it concentrates lognormally on values less than any finite strike.

Therefore, out-of-the-money calls will take on a maximum value at some volatility that concentrates as much probability as possible below the strike before concentrating the distribution too close to zero.

Edit: A huge thank-you to @Veeken to pointing out that it is out-of-the-money calls, rather than puts, which take on a maximum theoretical value.

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Brian, isn't it the other way around? as vol goes to $\infinity$ cash binaries with put payoffs go to 1 while those with call payoffs go to 0, for any strike K. this is because of the negative skew, which makes the left side of the distribution far more 'heavy' than the right side as vol goes to $\infinity$. see here: en.wikipedia.org/wiki/File:Some_log-normal_distributions.svg –  Veeken May 7 '13 at 20:43
@Veeken: Once can construct models where put payoffs go to 1, but they are not the Black-Scholes model, which has "flat" skew in the options-trading sense. –  Brian B May 8 '13 at 19:10
i don't understand what you mean by 'flat' skew in the BS model. As soon as $\sigma>0$, there is skew in the BS model. Allow me to cast the first integral above into BS terms: BinaryCashCall = $e^{-rT}*N(d_2)$ with $d_1, d_2$ given here:en.wikipedia.org/wiki/…. as $\sigma \to \infty$, $d_1 \to \infty$ while $d_2 \to -\infty$. This makes $N(d_2) \to 0$, and thus makes the binary call price 0. By obvious symmetry, the binary put goes to 1 in the event. All this is in the BS world. Thanks for your time... –  Veeken May 8 '13 at 20:48
@Veeken: thank you for pointing out the error. By "flat skew in the options-trading sense" I mean that an options trader would perceive option implied vols to be the same across strikes if the option prices were generated by the BS model. In the sense of distributional moments, you are quite correct that the 3rd moment (skew) is negative for this model. It is an unfortunate collision of terminology between traders and mathematicians that the same word is used both ways. –  Brian B May 10 '13 at 0:35
phew, thanks Brian. as a former eq deriv trader myself, now i see what you mean -- should have interpreted where you were aiming. –  Veeken May 10 '13 at 1:02

all of the volatility effects on a binary option struck at 105 with a one dollar payoff are approximately the same as the volatility effects on the following portfolio of options:

short 100 of the 104.99 calls / long 200 of the 105 calls / short 100 of the 105.01 calls

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I have a mathematical proof with no graphs or pictures. Suppose $r=0$, what we want is to see what happens if volatility changes in $E^Q[1_{S_T>K}]$.

The latter quantity is $Q(S_T>K)=Q(\log S_T > \log K)$.

Under Q, we know that $S_T=S_0 \exp\left(-\frac12 \sigma^2T + \sigma W_T\right)$, so $\log S_T$ is distributed as $N(\log S_0 -\frac12\sigma^2T, \sigma^2 T)$.

So we can write $Q\left(\sigma \sqrt{T} N + \log(S_0) -\frac12 \sigma^2T > \log K\right)$ which equals $Q\left(N>\frac{\log{\frac{K}{S_0}}+\frac12 \sigma^2T}{\sigma \sqrt T}\right).$

Since $f(y)=Q(N>y)$ decreases in $y$, it is enough to study $y=y(\sigma)=\frac{\log{\frac{K}{S_0}}+\frac12 \sigma^2T}{\sigma \sqrt T}$.

If $K>S_0$ (out of the money option), then if $\sigma \to 0$, $y(\sigma)\to +\infty$ and the same happens if $\sigma \to +\infty$. Hence there is a minimum for $\sigma=\sqrt{\log{\frac{K}{S_0}}}$. We deduce (by continuity) that $f(y(0))=0$, $f(y(+\infty))=0$, and we have a maximum for $\sigma=\sqrt{\log{\frac{K}{S_0}}}$.

If instead $K<S_0$ (in the money option), $\sigma \to 0$ gives $-\infty$, $\sigma\to \infty$ still gives $\infty$ and the function $y(\sigma)$ is strictly increasing. So $f(y(0))=1$, $f(y(+\infty))=0$ and $f$ is strictly decreasing.

Finally, for an at the money option $S_0=K$, we have $f(y)=Q\left(N > \frac12 \sigma \sqrt T\right)$, so $f(0)=\frac 12$, and $f$ strictly decreases to the value $0$.

Hope this helps.

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Looks interesting way to do it. how do you go from $E^Q[1_{ST>K}] "to" Q(ST>K)=Q(logS_T>logK)? – user12348 Jun 10 '14 at 16:56 An expectation of an indicator function of a set is the probability of that set, by definition. Then you have$S_T>K$if and only if$\log S_T > \log K\$, so the probabilities are the same. –  AmericanCaller Jun 10 '14 at 21:15