Pricing a Cash-or-Nothing Digital with a Call-Spread Derivative

Summary

The document shows how a narrow call spread approximates a cash-or-nothing digital option: subtract the call price at a slightly higher strike from the call price at the target strike, then divide by the strike interval. In the limit, this becomes the negative derivative of the call price with respect to strike. Under the Black–Scholes setting cited, the Breeden–Litzenberger relation expresses that derivative as a discounted risk-neutral tail probability, with the discount factor determined by the risk-free rate and maturity.

This result links option prices across strikes to the risk-neutral distribution of the underlying at expiry. The explanation assumes a differentiable call-price curve and a digital payoff that pays when the terminal price exceeds the strike. The source's final wording about whether the probability is above or below the strike is inconsistent with its displayed expression: if Q is the cumulative distribution function, 1−Q(K) is the probability of exceeding K, subject to boundary conventions. The initial question's zero-rate assumption removes discounting, though the answer states the more general rate form.

Key ideas

• A narrow call spread converges to the negative strike derivative of the call price.
• The derivative is related to a discounted risk-neutral tail probability by the Breeden–Litzenberger result.
• The call-spread expression corresponds to a digital payoff triggered when the terminal underlying price exceeds the strike.
• If Q is a cumulative distribution function, 1−Q(K) denotes the probability above K, not below it.
• The formula assumes a differentiable call-price curve and specifies a risk-neutral valuation framework.

Tags

# Binary option analytical formula


# Binary option analytical formula












Given $r=0$, $\sigma(K)=\text{const}$ and:

$$ \text{Binary} = \lim_{ε → 0} \frac{(C(K,\sigma (K))-C(K+ε,\sigma(K+ε)))}{ε} $$

I have to find the analytical expression for the above.

Since $σ(K)=\text{const}$, I know that we can write the above as:

$$ \text{Binary} = \lim_{ε → 0}\frac{(C(K)-C(K+ε))}{ε} $$

Do I take the derivative next or use the Taylor's theorem?

## Answer by Daneel Olivaw (score 2, accepted)

https://quant.stackexchange.com/a/37018

As you say, you simply differentiate with respect to $K$. Assuming your binary's maturity is $T$, note that in a Black-Scholes framework with constant risk-free rate $r$, by the Breeden-Litzenberger equations:

$$ \begin{align} \text{Binary}&=\lim_{\epsilon \rightarrow 0}\frac{-C(K+\epsilon)+C(K)}{\epsilon} \\[6pt] &=-\frac{\partial C}{\partial K}(K) \\[9pt] &=e^{-rT}(1-Q(K)) \end{align}$$

where $Q(\cdot)$ is the cumulative, risk-neutral distribution and $(1-Q(K))$ gives the probability that the underlying asset's price is below $K$ at time $T$, which is consistent with the payoff of a binary option.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.