4
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ I suspect that the condition $P<0.99(S−Ke^{−tr})$ should be $P<0.99(S−Ke^{−tr})^+$ instead $\endgroup$
    – Gordon
    Commented Jun 23, 2015 at 19:26

3 Answers 3

Reset to default
5
$\begingroup$

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}

Improve this answer
$\endgroup$
2
  • $\begingroup$ What is the upper bound..? $\endgroup$
    – emcor
    Commented Jun 24, 2015 at 7:51
  • $\begingroup$ @emcor: The upper bound is now added. $\endgroup$
    – Gordon
    Commented Jun 24, 2015 at 12:44
3
$\begingroup$

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).

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ I think you mean convex not semi-positive. The conclusion would not hold for a put spread for example. $\endgroup$
    – AFK
    Commented Jun 23, 2015 at 22:14
  • $\begingroup$ @AFK Yes, the argument for the lower bound is then analogous to (at)Gordon,MarkJoshi. $\endgroup$
    – emcor
    Commented Jun 24, 2015 at 7:18
2
$\begingroup$

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.)

Improve this answer
$\endgroup$
11
  • 1
    $\begingroup$ 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? $\endgroup$
    – emcor
    Commented Jun 24, 2015 at 7:06
  • $\begingroup$ The answer is not correct. Let $S<K$. Then your lower bound would be negative. $\endgroup$
    – emcor
    Commented Jun 24, 2015 at 7:41
  • $\begingroup$ a negative certainly is a lower bound. I also say that it is positive, $\endgroup$
    – Mark Joshi
    Commented Jun 24, 2015 at 10:31
  • 2
    $\begingroup$ at maturity the pay-off satisfies $C_T \geq S_T - K = S_T - KB_T.$ $\endgroup$
    – Mark Joshi
    Commented Jun 24, 2015 at 10:32
  • $\begingroup$ The proof by @Gordon shows that the lower arbitrage bound can be $0$? $\endgroup$
    – emcor
    Commented Jun 24, 2015 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.