No-Arbitrage Bounds for Extrapolating Call Prices Across Strikes

Summary

The document derives bounds for a call price at a lower strike when prices at two higher strikes are known. Its example gives call values at strikes 100 and 90, then asks what can be inferred at strike 80. Convexity supplies a lower bound, while a payoff comparison gives an upper bound: lowering the strike from 90 to 80 can add at most 10 to the call payoff in every outcome. Thus, under the stated zero-rate simplification, the strike-80 call cannot exceed the strike-90 call plus 10.

The argument uses the risk-neutral expected payoff representation and assumes no arbitrage. With a positive risk-free rate, the increment is discounted, so the upper bound is lower. These bounds constrain a plausible price but do not determine the actual price or provide a full interpolation method; market information about the underlying distribution and volatility is still needed.

Key ideas

• Call prices are convex in strike, which can provide a lower bound for an unobserved call price.
• The payoff from lowering a call strike by a fixed amount can increase by no more than that amount.
• Under zero rates and no arbitrage, the lower-strike call is bounded above by the higher-strike call plus the strike difference.
• A positive risk-free rate discounts the strike-difference increment in the upper bound.
• The bounds restrict possible prices but do not identify a unique price.

Tags

# Quick way to extrapolate call price as function of strike


# Quick way to extrapolate call price as function of strike












Let's say I know the price of a call for two different values of strike. Is there a quick way to guess the price for another value of strike ?

Actually, I know that C(100)=15 and C(90)=20 and I have to guess the value of C(80).

I know that C(K) is a convex function of K. Hence we deduce that C(80) $\geq$ 25. Is it possible to find an upper bound for C(80) ?

Thanks

## Answer by ocstl (score 2, accepted)

https://quant.stackexchange.com/a/17357

The upper bound for the 80 call is C(90) + 10, or 30. At least assuming no arbitrage.

Let's start by assuming the risk-free rate is 0 (this isn't a problem, but the math is clearer without it), so we don't have to discount the price. Then, the call price is given by $C(K) = E_t[(S_T - K)^+]$, which gives:

\begin{array} $C(K - 10) &= E_t[max(S_T - (K - 10), 0)] \\ &= E_t[max(S_T - K + 10, 0)] \\ &\leq E_t[max(S_T - K, 0) + 10] = E_t[max(S_T - K, 0)] + 10 \\ \end{array}

Replacing K with 90, we get: \begin{array} $C(90 - 10) &\leq E_t[max(S_T - 90, 0)] + 10 \\ C(80) &\leq C(90) + 10 = 30 \\ \end{array}

Obviously, given a positive risk-free rate, the upper bound would be smaller, by discounting the 10\$.

Another way to see this is that the most one can earn over and above the 90\$ call with an 80\$ call is 10\$, with probability at most 1 (only the case if the 90\$ call has probability 1 of finishing ITM).

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.