# Quick way to extrapolate call price as function of strike

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

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

## 1 Answer

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).

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