# 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?

• I suspect that the condition $P<0.99(S−Ke^{−tr})$ should be $P<0.99(S−Ke^{−tr})^+$ instead – Gordon Jun 23 '15 at 19:26

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, $$\big(S_0 - K \, e^{-rT}\big)^+ \leq c < S_0 .$$

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, $$\big(K \, e^{-rT}-S_0\big)^+ \leq p < K \, e^{-rT}.$$

• What is the upper bound..? – emcor Jun 24 '15 at 7:51
• @emcor: The upper bound is now added. – Gordon Jun 24 '15 at 12:44

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).

• I think you mean convex not semi-positive. The conclusion would not hold for a put spread for example. – AFK Jun 23 '15 at 22:14
• @AFK Yes, the argument for the lower bound is then analogous to (at)Gordon,MarkJoshi. – emcor Jun 24 '15 at 7:18

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.)

• In the first case you say $C\leq S_T$ and then $C\geq S_T-KB_T$. I dont see why this follows directly? – emcor Jun 24 '15 at 7:06
• The answer is not correct. Let $S<K$. Then your lower bound would be negative. – emcor Jun 24 '15 at 7:41
• a negative certainly is a lower bound. I also say that it is positive, – Mark Joshi Jun 24 '15 at 10:31
• at maturity the pay-off satisfies $C_T \geq S_T - K = S_T - KB_T.$ – Mark Joshi Jun 24 '15 at 10:32
• The proof by @Gordon shows that the lower arbitrage bound can be $0$? – emcor Jun 24 '15 at 11:34