How Volatility Skew Affects a Binary Call Option

# Flaw in the following argument with Binary Options and Skew


# Flaw in the following argument with Binary Options and Skew












A Binary option is ATM and expires tomorrow. If the skew of the vanilla options steepens (left side up, right side down) what happens to the price of the Binary Option.

I know that using a replication argument we are long call at $K_1$, short a call at $K_2$, where $K_1 < K_2$. Therefore when the skew steepens the $K_1$ call becomes more expensive and the call at $K_2$ becomes less expensive, so the overall price is that the Binary option will increase in price.

However, I have an argument that as the skew steepens the market is saying that it expects the volatility to be higher on the downside (more chance of expiring OTM) and lower on the upside (more change of expiring ITM). This would mean the binary option would become less expensive. What is the flaw in this argument, as it all seems logical to me?

## Answer by RRL (score 5, accepted)

https://quant.stackexchange.com/a/38579

Let $f_0(S_T) =f(S_T|S_0)$ be the risk-neutral PDF for the underlying asset price at time $T$ (conditional on the price $S_0$ at present time $t=0$). The probability that the price is above a strike price $K$ at time $T$ is

$$P(S_T \geqslant K) = \int_K^\infty f_0(x) \, dx.$$

This is just definitional regardless of the shape of the distribution (eg. symmetric, skewed, etc). It could be the distribution implied in the known option prices at $t= 0$.

The price of a vanilla call option at $t=0$, expiring at time $T$ and with strike price $K$, is the discounted risk-neutral expected value

$$C(K) = e^{-rT} \int_0^\infty\max(x-K,0) \, f_0(x) \, dx = e^{-rT} \int_K^\infty(x-K) \, f_0(x) \, dx. $$

Here we have ignored dividends and suppressed the dependence of the option price on other parameters in writing $C(K)$.

We can apply the Leibniz rule and differentiate the integral once with respect to $K$ to obtain

$$\frac{\partial C}{\partial K} = -e^{-rT}\int_K^\infty f_0(x) \, dx \\ \implies P(S_T \geqslant K) = - e^{-rT}\frac{\partial C}{\partial K}$$

In the presence of an implied-volatility skew, the underlying distribution is not lognormal. However, we can represent the option price as a composition of the Black-Scholes formula with a (smooth) implied volatility as a function of strike:

$$C(K) = C_{BS}(K,\sigma(K)).$$

Hence,

$$P(S_T \geqslant K) = -e^{-rT}\frac{\partial C_{BS}}{\partial K} - e^{-rT}\frac{\partial C_{BS}}{\partial \sigma}\sigma'(K) \tag{*} $$

Typically for an equity index, the skew exhibits a negative slope, $\sigma'(K) < 0$, and vega, the partial derivative of $C_{BS}$ with respect to $\sigma$, is positive.

> All else the same, $P(S_T \geqslant K)$ increases as $-\sigma'(K)$ increases -- ie., the skew steepens.

As you observed, the binary or digital call option $C_D$ can be replicated approximately with a call spread according to

$$C_{D }(K) \approx \frac{C(K-\delta) - C(K+\delta)}{2\delta}.$$

As the strike spread $2\delta$ tends to $0$ and the notional $1/(2\delta) $ tends to infinity, the replication is more accurate (although impractical) and

$$C_D(K) = \lim_{\delta \to 0} \frac{C(K-\delta) - C(K+\delta)}{2\delta} = - \frac{\partial C}{\partial K}.$$

This, of course, shows that the digital option price itself is directly related to the probability of the underlying ending above the strike at expiry.

## Answer by Ivan (score 3)

https://quant.stackexchange.com/a/38577

Here is an intuitive explanation: you conclude that there is more chance of expiring otm than itm if there is skew but that isn’t correct. The atm volatility is unchanged vs the flat vol case, and atm is where you (you’re the stock) start from.

Once you have moved from there (ATM), either you’re on the downside where vol is (only now) higher and the payoff is 0 (OTM), or you’re on the upside where vol is (only now) lower and the payoff is 1 (ITM).

In the former case (OTM) higher vol is good because you don't want to be stuck here and want a chance to move ITM. In the latter case (ITM) lower vol is good because you do want to stay here and not move OTM. The net result is a more favourable situation than if you kept moving at the ATM vol in all situations.

Skew makes it more likely than flat vol to move far to the downside from atm and less likely to move far to the upside, but it makes it more likely to move up to and stay in the immediate upside vicinity of atm.