European Option Arbitrage Bounds from Payoffs and Convexity

# Arbitrage bounds for Black-Scholes


# Arbitrage bounds for Black-Scholes












In some implied volatility code I came across, there is a check to ensure there is no violation of the arbitrage bounds based on the inputs to the method.

For the call option, if

$$P < 0.99 * (S-K*e^{-t*r})$$

(where $P$ is the market price of the option and $S, K, t $ and $r$ are underlying price, strike price, time to maturity and rate, respectively) then the price input to the method violates the bound and the method returns.

What is the comparable test for a put option and how is it derived?

## Answer by Gordon (score 5)

https://quant.stackexchange.com/a/18498

For a call option, the payoff is given by $(S_T-K)^+$. Note that the function $x^+$ is convex, then, by Jensen's inequality, the price $c$ satisfies \begin{align*} c &= e^{-rT}E\big((S_T-K)^+\big) \\ & \geq e^{-rT}\big(E(S_T-K)\big)^+\\ &=\big(S_0 - K \, e^{-rT}\big)^+. \end{align*} For the upper bound, note that \begin{align*} c &= e^{-rT}E\big((S_T-K)^+\big) \\ &< e^{-rT}E\big(S_T\big) \\ &=S_0. \end{align*} That is, \begin{equation} \big(S_0 - K \, e^{-rT}\big)^+ \leq c < S_0 . \end{equation}

Similarly, for a put option, the payoff is given by $(K-S_T)^+$. The price $p$ then satisfies \begin{align*} p &= e^{-rT}E\big((K-S_T)^+\big) \\ & \geq e^{-rT}\big(E(K-S_T)\big)^+\\ &=\big(K \, e^{-rT}-S_0\big)^+. \end{align*} For the upper bound, note that \begin{align*} p &= e^{-rT}E\big((K-S_T)^+\big) \\ &< e^{-rT}E\big(K\big) \\ &=K \, e^{-rT}. \end{align*} That is, \begin{equation} \big(K \, e^{-rT}-S_0\big)^+ \leq p < K \, e^{-rT}. \end{equation}

## Answer by emcor (score 3)

https://quant.stackexchange.com/a/18492

The No-Arbitrage bounds for a European put are:

$$ (Ke^{-rT}-S)^+ \leq P \leq K e^{-rT}$$

This is because the maximum payoff at maturity is $K$ (discounted) and the minimum value is the discounted intrinsic value (since $E(e^{-rT}S_T)=S_t$ by the martingale condition and the payoff being always semi-positive).

## Answer by Mark Joshi (score 2)

https://quant.stackexchange.com/a/18500

you can do the bounds without using a model or martingales. At maturity $$ 0 \leq C \leq S_T $$ with positive probability of strict inequalities. So before maturity, $$ 0 < C < S_t. $$ Since if these are violated, you can make an arbitrage. eg if $C \geq S_t$ hold $S_t - C$ to get a profit with positive probability and no chance of loss.

Similarly, if $B_T$ is a zero coupon bond expiring at $T,$ then we have $$ S_T - KB_T \leq C $$ at time $T$ and before maturity, we have $$ S_t - KB_T(t) < C. $$ That is $$ S_t - Ke^{-r(T-t)} < C < S_t $$ and positive as well.

(see my book concepts for more discussion.)