2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Have you tried just linear interpolation? Crude I know. $\endgroup$ – Kian Apr 12 '15 at 12:41
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ well that was pretty straightforward, thanks. $\endgroup$ – Dark Apr 12 '15 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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