Basic Trading and Hedging Strategies¶

Options trading strategies are built on Greek exposures. Each strategy expresses a specific view on direction, volatility, or time decay.

Delta-neutral hedging¶

Continuously rebalance shares to offset directional moves. This is the foundation of all other strategies.

Key properties:

Requires frequent adjustment in high-gamma positions.
Transaction costs from rebalancing scale with gamma and realized volatility.
Perfect only in the continuous-time, zero-cost limit of Black–Scholes.

Gamma-theta tradeoff¶

Long gamma (buying options)¶

Position: buy calls/puts or straddles. Greek profile: $\Gamma > 0$, $\Theta < 0$, $\nu > 0$.

Profit when the underlying makes large moves (realized vol exceeds implied vol). The cost is daily theta bleed when the market is quiet. Best used ahead of anticipated events.

Short gamma (selling options)¶

Position: sell calls/puts or straddles. Greek profile: $\Gamma < 0$, $\Theta > 0$, $\nu < 0$.

Profit when the market stays calm (realized vol below implied vol). Risk: large, sudden moves cause disproportionate losses (concave P&L). Best suited for range-bound markets with elevated implied vol.

Vega trades¶

Long vega: buy options when expecting implied volatility to increase. Common before anticipated volatility events. Theta decay works against you while waiting.

Short vega: sell options when expecting implied volatility to decrease. Common after volatility spikes expected to mean-revert. Risk: further vol increases lead to mark-to-market losses.

Risk reversal¶

Construction: long OTM call + short OTM put (same expiry).

Greek profile: positive delta (bullish directional bet), typically positive net vega, can be structured for near-zero premium.

Use case: express a directional and volatility view simultaneously. The skew (puts typically more expensive than equidistant calls) means you receive premium from the rich put to fund the cheap call.

Risk: the short put creates downside exposure if the underlying falls significantly.

Straddle¶

Long straddle¶

Construction: buy call and put at the same strike (typically ATM) and expiry.

Greek profile:

$\Delta \approx 0$ (initially neutral).
$\Gamma > 0$ (benefits from moves in either direction).
$\Theta < 0$ (pays significant time decay).
$\nu > 0$ (benefits from vol increases).

Profit when: the underlying moves substantially in either direction, or implied vol rises. The move must be large enough to offset the premium paid (which includes time value for both options).

Breakeven: $S_T > K + \text{premium}$ or $S_T < K - \text{premium}$.

Short straddle¶

Construction: sell call and put at the same strike.

Greek profile: $\Gamma < 0$, $\Theta > 0$, $\nu < 0$.

Profit when: the market stays calm and near the strike. Earn premium from time decay. Risk is unlimited in both directions.

Strangle¶

Long strangle¶

Construction: buy OTM call (strike $K_2$) and OTM put (strike $K_1 < K_2$).

Similar to long straddle but cheaper (both options are OTM), with wider breakeven points. Requires a larger move to profit, but costs less in theta.

Short strangle¶

Construction: sell OTM call and OTM put.

Earns premium if the underlying stays between the two strikes. Wider profit zone than a short straddle, but less premium collected.

Greek summary of common strategies¶

Strategy	$\Delta$	$\Gamma$	$\Theta$	$\nu$	View
Long call	+	+	−	+	Bullish + long vol
Short call	−	−	+	−	Bearish + short vol
Long straddle	≈ 0	+	−	+	Big move expected
Short straddle	≈ 0	−	+	−	Range-bound
Risk reversal	+	mixed	mixed	mixed	Directional + skew
Calendar spread	≈ 0	mixed	+	mixed	Differential decay

Python: advanced P&L simulator¶

```python import numpy as np import matplotlib.pyplot as plt from scipy.stats import norm

Parameters¶

S0 = 100 K = 100 T = 1.0 r = 0.05 sigma0 = 0.2 num_options = 100 num_steps = 200

Range of final stock prices¶

S_final = np.linspace(70, 130, num_steps) dS = S_final - S0

Simulate a volatility perturbation¶

dsigma = 0.05 * np.sin(np.pi * (S_final - 70) / 60) # vol smile effect

Compute Greeks at initial point¶

d1_0 = (np.log(S0 / K) + (r + 0.5 * sigma0**2) * T) / (sigma0 * np.sqrt(T)) delta_0 = norm.cdf(d1_0) gamma_0 = norm.pdf(d1_0) / (S0 * sigma0 * np.sqrt(T)) vega_0 = S0 * np.sqrt(T) * norm.pdf(d1_0)

Option value change (Taylor expansion)¶

option_pnl = num_options * (delta_0 * dS + 0.5 * gamma_0 * dS**2 + vega_0 * dsigma)

Hedging strategies¶

1. No hedge¶

unhedged = option_pnl

2. Static delta hedge¶

static_hedge = option_pnl - num_options * delta_0 * dS

3. Dynamic delta hedge (approximate: delta varies with S)¶

d1_final = (np.log(S_final / K) + (r + 0.5 * sigma0**2) * T) / (sigma0 * np.sqrt(T)) delta_final = norm.cdf(d1_final) avg_delta = 0.5 * (delta_0 + delta_final) dynamic_hedge = option_pnl - num_options * avg_delta * dS

Plot¶

plt.figure(figsize=(12, 7)) plt.plot(S_final, unhedged, label='Unhedged', linewidth=2) plt.plot(S_final, static_hedge, label='Static Delta Hedge', linewidth=2) plt.plot(S_final, dynamic_hedge, label='Dynamic Delta Hedge (approx)', linewidth=2) plt.axhline(0, color='gray', linestyle='--', alpha=0.5) plt.xlabel('Final Stock Price') plt.ylabel('P&L ($)') plt.title('P&L Under Different Hedging Strategies (with Gamma + Vega Effects)') plt.legend() plt.grid(True, alpha=0.3) plt.tight_layout() plt.show() ```

This demonstrates: unhedged P&L is dominated by directional moves, static delta hedging removes the linear component but leaves gamma and vega effects, and dynamic hedging further reduces the residual.

What to remember¶

Every option strategy is a combination of Greek exposures expressing a specific market view.
The gamma-theta tradeoff is inescapable: long gamma costs theta, short gamma earns theta.
Straddles and strangles provide symmetric exposure to large moves (or calm markets).
Risk reversals combine directional and volatility views, often exploiting skew.
Understanding the Greek profile of each strategy is essential before entering any position.

Exercises¶

Exercise 1. A trader constructs a long straddle by buying an ATM call and an ATM put, each priced at $\$5.00$. The combined premium is $\$10.00$. Compute the breakeven stock prices at expiry. If the underlying finishes at $S_T = 115$ with $K = 100$, what is the profit or loss?

Solution to Exercise 1

The long straddle consists of a long ATM call and a long ATM put, each costing $\$5.00$, so the total premium is $\$10.00$ with $K = 100$.

Breakeven at expiry. The position profits when the intrinsic value exceeds the premium paid.

Upper breakeven: the call is in the money and the put expires worthless, so

\[ S_T - K = 10 \implies S_T = K + 10 = 110 \]

Lower breakeven: the put is in the money and the call expires worthless, so

\[ K - S_T = 10 \implies S_T = K - 10 = 90 \]

P&L at $S_T = 115$. The call pays $115 - 100 = 15$ and the put expires worthless, so

\[ \text{P\&L} = 15 - 10 = +\$5.00 \]

The position is profitable because $S_T = 115$ lies above the upper breakeven of $110$.

Exercise 2. For a short strangle with $K_1 = 90$ (put) and $K_2 = 110$ (call), both OTM, the trader collects a total premium of $\$3.50$. Determine the breakeven levels and the maximum profit. Compute the P&L at $S_T = 85$ and $S_T = 120$.

Solution to Exercise 2

The short strangle collects a total premium of $\$3.50$ by selling an OTM put at $K_1 = 90$ and an OTM call at $K_2 = 110$.

Breakeven levels. The seller loses money when the payoff on either short option exceeds the premium collected.

Upper breakeven: the short call is exercised, so

\[ S_T - K_2 = 3.50 \implies S_T = 110 + 3.50 = 113.50 \]

Lower breakeven: the short put is exercised, so

\[ K_1 - S_T = 3.50 \implies S_T = 90 - 3.50 = 86.50 \]

Maximum profit. This occurs when both options expire worthless, i.e., $K_1 \leq S_T \leq K_2$. The maximum profit is the full premium: $\$3.50$.

P&L at $S_T = 85$. The put is in the money: payoff $= 90 - 85 = 5$, and the call expires worthless.

\[ \text{P\&L} = 3.50 - 5.00 = -\$1.50 \]

P&L at $S_T = 120$. The call is in the money: payoff $= 120 - 110 = 10$, and the put expires worthless.

\[ \text{P\&L} = 3.50 - 10.00 = -\$6.50 \]

Exercise 3. Using the Greek summary table, explain why a long straddle has $\Delta \approx 0$ initially but $\Gamma > 0$. As the underlying moves from $S = 100$ to $S = 108$, how does the portfolio delta change? What must the trader do to maintain delta neutrality?

Solution to Exercise 3

Why $\Delta \approx 0$ initially. A long straddle at the money consists of a long call with $\Delta_{\text{call}} \approx +0.5$ and a long put with $\Delta_{\text{put}} \approx -0.5$. The portfolio delta is

\[ \Delta_{\text{straddle}} = \Delta_{\text{call}} + \Delta_{\text{put}} \approx 0.5 + (-0.5) = 0 \]

Why $\Gamma > 0$. Both long options have positive gamma. Since gamma is always non-negative for long vanilla options:

\[ \Gamma_{\text{straddle}} = \Gamma_{\text{call}} + \Gamma_{\text{put}} > 0 \]

Delta change as $S$ moves from 100 to 108. As the underlying rises, both the call delta and the put delta increase (both move toward $+1$ and $0$, respectively). The rate of change is governed by gamma. With $\delta S = 8$:

\[ \Delta_{\text{new}} \approx \Delta_{\text{old}} + \Gamma_{\text{straddle}} \cdot \delta S = 0 + \Gamma_{\text{straddle}} \cdot 8 > 0 \]

The portfolio becomes net long delta. The call delta increases toward $1$ while the put delta moves toward $0$, so the positive gamma ensures the straddle picks up positive delta on an up-move.

Maintaining delta neutrality. The trader must sell $\Gamma_{\text{straddle}} \times 8$ shares of the underlying to offset the newly acquired positive delta and restore $\Delta_{\text{portfolio}} = 0$.

Exercise 4. A risk reversal is constructed by buying a call at $K = 110$ and selling a put at $K = 90$ with $S_0 = 100$. If the call costs $\$3.00$ and the put premium received is $\$3.50$, what is the net premium? Compute the P&L at $S_T = 80$, $100$, and $120$. Explain why this strategy expresses both a directional and a volatility view.

Solution to Exercise 4

Net premium. The trader buys the call for $\$3.00$ and receives $\$3.50$ from selling the put:

\[ \text{Net premium} = -3.00 + 3.50 = +\$0.50 \text{ (credit)} \]

P&L at $S_T = 80$. The call expires worthless. The short put is exercised: the trader must buy at $K = 90$ while the stock is worth $80$, costing $90 - 80 = 10$.

\[ \text{P\&L} = 0.50 - 10.00 = -\$9.50 \]

P&L at $S_T = 100$. Both options expire worthless (call strike 110, put strike 90).

\[ \text{P\&L} = +\$0.50 \]

P&L at $S_T = 120$. The call pays $120 - 110 = 10$. The put expires worthless.

\[ \text{P\&L} = 0.50 + 10.00 = +\$10.50 \]

Directional and volatility view. The risk reversal expresses a bullish directional view because it profits when the underlying rises and loses when it falls. It also exploits the volatility skew: OTM puts are typically more expensive than equidistant OTM calls due to the skew, so the trader receives a net credit by selling the "rich" put and buying the "cheap" call. The strategy implicitly takes a view that the skew overstates downside risk relative to upside potential.

Exercise 5. Explain the gamma-theta tradeoff for a short gamma position. If a trader sells an ATM straddle with $\Gamma = -0.08$ (portfolio) and $\Theta = +0.15$ per day, compute the P&L after one day if the underlying: (a) does not move; (b) moves by $2\%$; (c) moves by $5\%$. At what daily move does the gamma loss exceed the theta gain?

Solution to Exercise 5

The portfolio is short gamma with $\Gamma = -0.08$ and earns $\Theta = +0.15$ per day. Let $S_0 = 100$ for concreteness.

(a) Underlying does not move ($\delta S = 0$).

\[ \text{P\&L} = \Theta \cdot 1 + \frac{1}{2}\Gamma(\delta S)^2 = 0.15 + 0 = +\$0.15 \]

Pure theta gain.

(b) Underlying moves by 2% ($\delta S = 2$).

\[ \text{P\&L} = 0.15 + \frac{1}{2}(-0.08)(2)^2 = 0.15 - 0.16 = -\$0.01 \]

The gamma loss nearly offsets the theta gain.

(c) Underlying moves by 5% ($\delta S = 5$).

\[ \text{P\&L} = 0.15 + \frac{1}{2}(-0.08)(5)^2 = 0.15 - 1.00 = -\$0.85 \]

A large move produces a substantial loss.

Breakeven move. Set the gamma loss equal to the theta gain:

\[ \frac{1}{2}|\Gamma|(\delta S^*)^2 = \Theta \implies \frac{1}{2}(0.08)(\delta S^*)^2 = 0.15 \]

\[ (\delta S^*)^2 = \frac{0.30}{0.08} = 3.75 \implies \delta S^* = \sqrt{3.75} \approx 1.94 \]

Any daily move exceeding approximately $\$1.94$ (or $1.94\%$ of $S_0 = 100$) causes the gamma loss to exceed the theta gain.

Exercise 6. Extend the P&L simulator to include a long strangle strategy. Compute and plot the P&L as a function of the final stock price for $K_1 = 90$, $K_2 = 110$, with both options having $\tau = 0.5$, $\sigma = 0.25$, $r = 0.03$. How does the profile compare to a long straddle at $K = 100$?

Solution to Exercise 6

Using Black--Scholes pricing with $S_0 = 100$, $r = 0.03$, $\sigma = 0.25$, and $\tau = 0.5$:

Long strangle: Buy an OTM put at $K_1 = 90$ and an OTM call at $K_2 = 110$. The total cost is $P(90) + C(110)$.

Long straddle at $K = 100$: Buy a call and a put at $K = 100$. The total cost is $C(100) + P(100)$.

The P&L at expiry for each strategy is:

Long strangle: $\max(K_1 - S_T, 0) + \max(S_T - K_2, 0) - \text{premium}_{\text{strangle}}$
Long straddle: $|S_T - K| - \text{premium}_{\text{straddle}}$

Comparison. The strangle is cheaper because both options are OTM, but it requires a larger move to reach profitability. The straddle has narrower breakeven points but costs more. Specifically:

The straddle has breakevens at $K \pm \text{premium}_{\text{straddle}}$, while the strangle has breakevens at $K_1 - \text{premium}_{\text{strangle}}$ and $K_2 + \text{premium}_{\text{strangle}}$.
For moves within the strangle's strikes ($90 < S_T < 110$), the straddle still generates intrinsic payoff while the strangle generates none.
For very large moves, both strategies converge to the same slope ($\pm 1$ per dollar of underlying move), but the straddle's P&L is shifted up by the difference in strikes minus the difference in premiums.
The strangle is preferred when the trader expects a very large move but wants to minimize the upfront cost and theta bleed.