Delta-Hedging a Long Call After a Sudden Jump

Worked Solution

How to Think About It: A long call has positive delta, so its value rises and falls with the stock. Hedging means adding a position whose delta is the negative of the call's, which for one call is $-\Delta$ shares, and parking the cash difference in the risk-free asset so the whole package has zero initial cost. Since delta itself moves with $S$, the hedge must be rebalanced, and the direction of rebalancing follows from the sign of gamma.

Quick Estimate: Take $K = 100$, $r = 5\%$, $\sigma = 20\%$, $y = 0$, six months to expiry. At $S = 100$: $d_1 = 0.247$, $\Delta = 0.598$, $c = 6.89$. Hedge: short $0.598$ shares (receive $59.77$), and lend $59.77 - 6.89 = 52.88 = Ke^{-r\tau}N(d_2)$. If GM jumps to $110$: $d_1 = 0.921$, $\Delta = 0.822$. You now need to be short $0.822$ shares, so you sell a further $0.224$ shares at $110$ and lend the extra proceeds.

Formal Solution:

Part (a): Constructing the hedge

*Step 1 -- Delta of the call.* With dividend yield $y$, the call is worth $c = Se^{-y\tau}N(d_1) - Ke^{-r\tau}N(d_2)$ with $d_1 = [\ln(S/K) + (r - y + \sigma^2/2)\tau]/(\sigma\sqrt{\tau})$, and $$\Delta = \frac{\partial c}{\partial S} = e^{-y\tau}N(d_1).$$

*Step 2 -- Replicating portfolio and its mirror.* The formula says a call equals $e^{-y\tau}N(d_1)$ shares minus $Ke^{-r\tau}N(d_2)$ of risk-free borrowing. To hedge a long call, take the opposite position per call: short $\Delta = e^{-y\tau}N(d_1)$ shares of GM and lend $Ke^{-r\tau}N(d_2)$ at the risk-free rate. The cash flows balance: shorting brings in $\Delta S = c + Ke^{-r\tau}N(d_2)$, of which $c$ paid for the call and the remainder is lent. The net position (call, short stock, loan) has zero delta and zero value, and it is instantaneously immune to small moves in $S$.

*Step 3 -- Dynamic.* $\Delta$ changes with $S$ and with time, so the share position must be adjusted continuously (in practice, at discrete intervals or when delta drifts past a tolerance).

Part (b): Rebalancing after an up-jump

*Step 4 -- Sign of gamma.* $\Gamma = \partial\Delta/\partial S = e^{-y\tau}N'(d_1)/(S\sigma\sqrt{\tau}) > 0$, so the call's delta is increasing in $S$. After GM jumps up, $\Delta$ is larger; the existing short of $\Delta_{\text{old}}$ shares is no longer enough.

*Step 5 -- Action.* Short an additional $\Delta_{\text{new}} - \Delta_{\text{old}} > 0$ shares and lend the proceeds (the required loan $Ke^{-r\tau}N(d_2)$ also rises because $d_2$ rose). Symmetrically, after a drop you would buy back shares. Note that you sell after the price rises and buy after it falls, which is where a long-gamma hedger's profits from realized volatility come from, paid for by the option's time decay.

Answer:

(a) Short $\Delta = e^{-y\tau}N(d_1)$ shares of GM per call and lend $Ke^{-r\tau}N(d_2)$ at the risk-free rate; rebalance as delta changes.

(b) Delta increases when $S$ jumps up (positive gamma), so short additional shares to bring the short position up to the new $e^{-y\tau}N(d_1)$ and lend the extra proceeds.