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Bounding Call Price Differences with Put-Call Parity

Article Quant Q&A · Author: Summerday
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Summary

The document explains why the prices of two calls on the same underlying and maturity cannot differ by more than the discounted difference between their strikes. Its main method combines put-call parity with the no-arbitrage ordering of option prices: a call with a lower strike is worth at least as much as one with a higher strike, while the corresponding put prices have the reverse ordering. Substituting parity into the call-price difference then gives the bound.

A second answer offers a payoff-based intuition: the difference between the two calls’ expiration payoffs is bounded by the strike gap. The discussion is an educational derivation, not empirical evidence or a trading strategy. It assumes the stated parity relationship, positive interest rates, and matching underlying and maturity; market frictions and deviations from idealized pricing are not discussed.

Key ideas

  • A lower-strike call is worth at least as much as a higher-strike call under no-arbitrage.
  • A lower-strike put is worth no more than a higher-strike put.
  • Put-call parity converts the call-price difference into a put-price difference plus a discounted strike gap.
  • The resulting absolute call-price difference is bounded by the discounted difference between strikes.
  • The payoff difference at expiration also provides an intuitive route to the bound.

Tags

OptionsDerivatives pricingArbitrage
Full text
# Upper bound for difference of two call options


# Upper bound for difference of two call options












> Let $r>0$ be the interest rate and $S_t$ the price of a stock at time $t$. Let $C(t,K_1),C(t,K_2)$ be the price of call options at time $t$ with the same underlying asset $S$, the same maturity $T$ and two different strike prices $K_1\leq K_2$.I want to show that $$|C(t,K_1)-C(t,K_2)|\leq (1+r)^{-(T-t)}|K_1-K_2|$$

My idea was to use the Put-Call-parity, I know that $$C(t,K_i)=P(t,K_i)+S_t-K_i(1+r)^{-(T-t)}$$ for $i=1,2$. Therefore $$\begin{align*}|C(t,K_1)-C(t,K_2)|&=\left|P(t,K_1)+S_t-K_1(1+r)^{-(T-t)}-P(t,K_2)-S_t+K_2(1+r)^{-(T-t)}\right|\\ &=\left|P(t,K_1)-P(t,K_2)-K_1(1+r)^{-(T-t)}+K_2(1+r)^{-(T-t)}\right|\\ &\leq\left|P(t,K_1)-P(t,K_2)\right|+(1+r)^{-(T-t)}|K_1-K_2| \end{align*}$$

but this is not what I want since I still get the $|P(t,K_1)-P(t,K_2)|$ in front. Can someone help me please?

## Answer by MrLCh (score 3, accepted)

https://quant.stackexchange.com/a/79416

You almost got the solution, but try to write out the absolute value! If you think about the relation of the option prices you will notice that (by no-arbitrage):

$$C(t, K_1) \geq C(t, K_2) \, \text{ and}$$

$$P(t, K_1) \leq P(t, K_2).$$

This leads to the following (using Put-Call-parity)

\begin{align} |C(t, K_1) - C(t, K_2) | &= C(t, K_1) - C(t, K_2) \\ &= P(t,K_1)+ S_t −K_1(1+r)^{-(T-t)} - \left(P(t,K_2)+ S_t −K_2(1+r)^{-(T-t)}\right) \\ &= P(t, K_1) - P(t, K_2) + (K_2 - K_1) (1+r)^{-(T-t)} \hspace{1cm} \text{ using } P(t, K_1) \leq P(t, K_2) \\ &\leq (K_2 - K_1) (1+r)^{-(T-t)} = |K_1 - K_2| (1+r)^{-(T-t)} \end{align}

Where the last line is due to $K_2 \geq K_1$.

## Answer by Arshdeep (score 1)

https://quant.stackexchange.com/a/79427

Isn't $Payoff(call_{K2}) - payoff(call_{K1})<=K2-K1$ as RHS is the maximum possible payoff of LHS?

I think that solves it more easily

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