零波动 ATM Call

英文原题

Assumptions: Black–Scholes world, no dividends, $S_0=100$, $r=5\%$ (cont. comp.), $T=1$, strike $K=100$, volatility $\sigma=0$.

Price of the ATM European call (when $\sigma=0$)

With $\sigma=0$, the risk–neutral stock path is deterministic:
$$S_T = S_0 e^{rT} = 100\,e^{0.05}.$$
Hence the call payoff is certain and equals
$$(S_T - K)^+ = 100\,e^{0.05} - 100.$$
Discounting at $r$ gives the time–0 price
$$C_0 = e^{-rT}\big(100\,e^{0.05} - 100\big) = 100 - 100\,e^{-0.05} = S_0 - K e^{-rT}.$$

Numeric: $e^{-0.05}\approx 0.951229 \;\Rightarrow\; C_0 \approx 100(1-0.951229)\approx 4.877.$

How to hedge if you sold the call (short call)

Since exercise is certain ($S_T>K$ deterministically), the call is exactly replicated statically by:

• Long 1 share of stock,
• Short a zero-coupon bond with present value $K e^{-rT}$ (i.e., borrow $K e^{-rT}$).

So, if you are short the call, hold the opposite portfolio to hedge:

• Short 1 share,
• Long a zero-coupon bond of present value $K e^{-rT}$ (face value $K$ at $T$).

Payoff check at $T$:
Your hedge pays $-S_T + K$, which equals the short call payoff $-(S_T - K)$ (exercise is certain).
Initial cost of the hedge is $-S_0 + K e^{-rT} = -C_0$, matching the call sale proceeds — a perfect static hedge.