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Basic Option Strategies

Consider two investors who both own 100 shares of the same stock at $100. The first wants extra income and will tolerate giving up unusually large upside; she sells a call struck at $110 and pockets the premium. The second is worried about a crash before earnings; he buys a put struck at $95 and pays a small premium for that protection. Same starting position, opposite goals, and yet both arrive there by attaching exactly one option to the stock. The strategies in this section are little more than these two moves — a few elementary combinations of long, short, call, put, and stock — but they already display the central lesson: options let an investor reshape a payoff curve almost piece by piece.

An option on its own provides leveraged, directional exposure. But the real utility of options emerges when they are combined with other positions — with the underlying stock, with other options, or with both. This section catalogs the four elementary single-option positions, establishes the symmetry between long and short sides, and then introduces the two most important option-stock combinations: the covered call and the protective put. These strategies illustrate the core principle that will drive the rest of this chapter: options allow investors to reshape the payoff profile of a portfolio, and pricing them correctly is essential for any rational hedging decision.

The Four Elementary Positions

Every option strategy is built from four atomic building blocks. Let \(S_T\) denote the stock price at maturity, \(K\) the strike price, and \(c\) the call premium and \(p\) the put premium paid at inception. Recall (see § Long and Short Positions): the writer's payoff is the negative of the holder's, so the market is zero-sum at expiration. From this point on, we distinguish payoff (the value received at maturity, always non-negative for long positions) from profit (payoff minus premium, which can be negative).

Subtracting the premium from each long/short payoff in turn gives the four atomic profit functions and their breakevens, summarized below:

Position Profit at Maturity Max Profit Max Loss Breakeven
Long call \((S_T - K)^+ - c\) \(\infty\) \(c\) \(K + c\)
Short call \(c - (S_T - K)^+\) \(c\) \(\infty\) \(K + c\)
Long put \((K - S_T)^+ - p\) \(K - p\) \(p\) \(K - p\)
Short put \(p - (K - S_T)^+\) \(p\) \(K - p\) \(K - p\)

For any position, \(\Pi_{\text{short}} = -\Pi_{\text{long}}\): every dollar gained by one party is lost by the other.

Covered Call

A covered call consists of a long position in the stock combined with a short call on the same stock:

\[ \Pi_{\text{covered call}} = (S_T - S_0) + [c - (S_T - K)^+] \]

where \(S_0\) is the initial stock price. Evaluating piecewise:

\[ \Pi_{\text{covered call}} = \begin{cases} S_T - S_0 + c & \text{if } S_T \leq K \\ K - S_0 + c & \text{if } S_T > K \end{cases} \]
  • Max profit: \(K - S_0 + c\) (capped at the strike)
  • Max loss: \(S_0 - c\) (if the stock falls to zero)
  • Breakeven: \(S_T = S_0 - c\)

The covered call sacrifices upside beyond \(K\) in exchange for the premium income \(c\). It is the most widely used option strategy in practice because it converts uncertain upside into immediate income, favored when the investor holds stock and has a neutral-to-mildly-bullish outlook.

xychart-beta
    title "Covered Call Profit (S0 = 100, K = 105, c = 3)"
    x-axis "Stock Price at Maturity" [80, 85, 90, 95, 100, 105, 110, 115, 120]
    y-axis "Profit" -20 --> 10
    line [-17, -12, -7, -2, 3, 8, 8, 8, 8]

Protective Put

A protective put consists of a long position in the stock combined with a long put on the same stock:

\[ \Pi_{\text{protective put}} = (S_T - S_0) + [(K - S_T)^+ - p] \]

Evaluating piecewise:

\[ \Pi_{\text{protective put}} = \begin{cases} K - S_0 - p & \text{if } S_T \leq K \\ S_T - S_0 - p & \text{if } S_T > K \end{cases} \]
  • Max profit: unlimited (reduced by the premium \(p\))
  • Max loss: \(S_0 - K + p\) (bounded, regardless of how far the stock falls)
  • Breakeven: \(S_T = S_0 + p\)

The protective put is effectively an insurance policy: the investor pays premium \(p\) to guarantee that losses on the stock never exceed \(S_0 - K + p\), while retaining full participation in any upside.

Protective Puts in Practice: Portfolio Insurance

Institutional investors frequently use SPX put options to protect equity portfolios. With the S&P 500 near 6,600, an ATM put costs roughly 160 index points, or $16,000 per contract. Since each contract controls a notional exposure of \(6{,}600 \times 100 = \$660{,}000\), the cost of one month of downside protection is approximately \(16{,}000 / 660{,}000 \approx 2.4\%\) of the notional value.

This "cost of insurance" is one reason OTM puts are more popular in practice: a put struck 200 points below the index might cost only half as much, providing protection against a decline beyond roughly 3% at a lower premium. The trade-off between the level of protection and its cost is a central consideration in portfolio risk management.

xychart-beta
    title "Protective Put Profit (S0 = 100, K = 95, p = 4)"
    x-axis "Stock Price at Maturity" [75, 80, 85, 90, 95, 100, 105, 110, 115]
    y-axis "Profit" -10 --> 15
    line [-9, -9, -9, -9, -9, -4, 1, 6, 11]

From Strategies to Pricing Theory

Both strategies reshape the portfolio's payoff distribution — but the cost of that reshaping is the premium, and a mispriced premium combined with the stock creates arbitrage. Recall (see § Why Pricing Matters): the no-arbitrage premium is the cost of a replicating portfolio of stock and bond.

Exercises

Exercise 1. A stock trades at \(S_0 = \$50\). A European call with strike \(K = 55\) and maturity \(T = 3\) months costs \(c = \$2.50\). Compute the profit \(\Pi\) for the call buyer when (a) \(S_T = 60\), (b) \(S_T = 55\), and (c) \(S_T = 48\). State the breakeven stock price.

Solution to Exercise 1

The profit for a long call is \(\Pi = (S_T - K)^+ - c\).

(a) \(S_T = 60\):

\[ \Pi = (60 - 55)^+ - 2.50 = 5 - 2.50 = 2.50 \]

(b) \(S_T = 55\):

\[ \Pi = (55 - 55)^+ - 2.50 = 0 - 2.50 = -2.50 \]

(c) \(S_T = 48\):

\[ \Pi = (48 - 55)^+ - 2.50 = 0 - 2.50 = -2.50 \]

The breakeven stock price is \(K + c = 55 + 2.50 = 57.50\). Below this level, the call buyer incurs a net loss; above it, the buyer profits.

Exercise 2. Verify the zero-sum property: if a call buyer's profit at maturity is \(\Pi_{\text{long}}\), show algebraically that the call writer's profit satisfies \(\Pi_{\text{short}} = -\Pi_{\text{long}}\) in both cases \(S_T > K\) and \(S_T \leq K\).

Solution to Exercise 2

Case 1: \(S_T > K\). The call is exercised.

\[ \Pi_{\text{long}} = (S_T - K) - c \]
\[ \Pi_{\text{short}} = c - (S_T - K) \]

Adding: \(\Pi_{\text{long}} + \Pi_{\text{short}} = (S_T - K) - c + c - (S_T - K) = 0\), so \(\Pi_{\text{short}} = -\Pi_{\text{long}}\).

Case 2: \(S_T \leq K\). The call expires worthless.

\[ \Pi_{\text{long}} = 0 - c = -c \]
\[ \Pi_{\text{short}} = c - 0 = c \]

Again \(\Pi_{\text{long}} + \Pi_{\text{short}} = -c + c = 0\). In both cases the short profit is the exact negative of the long profit, confirming the zero-sum property. \(\square\)

Exercise 3. An investor holds 100 shares of a stock currently priced at \(S_0 = \$80\) and writes a covered call with strike \(K = 90\) for a premium of \(c = \$3\) per share. (a) Compute the profit on the combined position if \(S_T = 95\). (b) Compute the profit if \(S_T = 70\). (c) At what stock price does the covered call position break even?

Solution to Exercise 3

Per share, the covered call profit is

\[ \Pi = (S_T - S_0) + c - (S_T - K)^+ \]

(a) \(S_T = 95 > K = 90\):

\[ \Pi = (95 - 80) + 3 - (95 - 90) = 15 + 3 - 5 = 13 \]

Total for 100 shares: \(100 \times 13 = \$1{,}300\).

Note that the upside is capped: even though the stock rose $15, the profit is only $13 because the call was exercised against the investor.

(b) \(S_T = 70 < K\):

\[ \Pi = (70 - 80) + 3 - 0 = -10 + 3 = -7 \]

Total for 100 shares: \(100 \times (-7) = -\$700\).

The premium partially offsets the stock decline.

(c) Breakeven requires \(\Pi = 0\). For \(S_T \leq K\), the profit is \(S_T - S_0 + c = S_T - 80 + 3 = S_T - 77\). Setting this to zero gives \(S_T = 77\). (For \(S_T > K\), the profit is \(K - S_0 + c = 13 > 0\), so breakeven occurs only on the downside.) The breakeven price is \(\$77\).

Exercise 4. An investor buys a stock at \(S_0 = \$100\) and simultaneously buys a protective put with strike \(K = 95\) for a premium of \(p = \$4\). (a) What is the maximum loss on this position? (b) Derive the breakeven stock price. (c) Compare the protective put to simply holding the stock without a put: for what range of \(S_T\) does the unprotected stock position outperform?

Solution to Exercise 4

The protective put profit per share is

\[ \Pi = (S_T - S_0) + (K - S_T)^+ - p \]

(a) The worst case occurs when \(S_T \leq K = 95\), giving

\[ \Pi = (S_T - 100) + (95 - S_T) - 4 = -9 \]

The maximum loss is $9 per share, regardless of how far the stock falls. Without the put, a drop to \(S_T = 0\) would lose $100.

(b) For \(S_T > K\), the put expires worthless and

\[ \Pi = S_T - 100 - 4 = S_T - 104 \]

Setting \(\Pi = 0\) gives \(S_T = 104\). The breakeven price is \(\$104\).

(c) The unprotected stock profit is \(S_T - 100\). The protective put profit is

\[ \Pi_{\text{pp}} = \begin{cases} -9 & \text{if } S_T \leq 95 \\ S_T - 104 & \text{if } S_T > 95 \end{cases} \]

For \(S_T > 95\), the unprotected stock outperforms by exactly the put premium: \((S_T - 100) - (S_T - 104) = 4\). For \(S_T \leq 95\), the unprotected stock outperforms only when \(S_T - 100 > -9\), i.e., \(S_T > 91\). Therefore the unprotected stock outperforms for \(S_T > 91\), and the protective put outperforms for \(S_T < 91\). The insurance is valuable precisely in severe downturns.

Exercise 5. Using the payoff formulas from this section, show that the following identity holds for all \(S_T \geq 0\):

\[ \underbrace{(S_T - S_0) + [(K - S_T)^+ - p]}_{\text{protective put}} = \underbrace{[(S_T - K)^+ - c] + [Ke^{-rT} \cdot e^{rT} - S_0] + (c - p)}_{\text{long call + bond + premium difference}} \]

under the assumption that put-call parity holds (i.e., \(c - p = S_0 - Ke^{-rT}\)). Explain why this means a protective put can be replicated by a long call plus a risk-free bond.

Solution to Exercise 5

Start with the protective put profit:

\[ \Pi_{\text{pp}} = (S_T - S_0) + (K - S_T)^+ - p \]

Consider two cases.

Case 1: \(S_T > K\). The put expires worthless:

\[ \Pi_{\text{pp}} = S_T - S_0 - p \]

A long call plus bond gives: \((S_T - K) - c + (K - S_0) + (c - p) = S_T - S_0 - p\). These are equal.

Case 2: \(S_T \leq K\). The put is exercised:

\[ \Pi_{\text{pp}} = S_T - S_0 + K - S_T - p = K - S_0 - p \]

A long call (expires worthless) plus bond gives: \(0 - c + (K - S_0) + (c - p) = K - S_0 - p\). These are equal.

The identity holds in all states because put-call parity ensures \(c - p = S_0 - Ke^{-rT}\), which means the net initial cost of the call-plus-bond portfolio equals the cost of the protective put. Since both portfolios produce identical payoffs for every \(S_T\) and cost the same, they must be equivalent by the law of one price. This shows that a protective put is synthetically equivalent to holding a long call and investing the present value of \(K\) in a risk-free bond — a result that follows directly from put-call parity. \(\square\)