Why Combine Stock With Options?
Owning stock leaves an investor exposed to a fall in its price. A put option can limit that downside at expiration, while a call written against the stock can help pay for the put. These combinations are useful only when their parts are evaluated consistently: the stock and options determine the terminal value, while premiums and the time value of money determine the investment’s net result.
This tutorial considers one share of stock and European options with the same expiration date. Assume no dividends or other cash flows before expiration. Let \(S_0\) be the stock price today, \(S_T\) its price at expiration \(T\) years from today, and let \(K_p\) and \(K_c\) denote the put and call exercise prices. Option premiums are paid or received at time 0. The annual effective interest rate \(i\) is the rate at which the initial net outlay could accumulate to time \(T\).
Protective Put: A Floor on Terminal Value
The payoff of a long put at expiration is \(\max(K_p-S_T,0)\). Adding the share’s value gives the protective put’s terminal value:
When the stock finishes below \(K_p\), the put payoff offsets the part of the stock’s decline below the exercise price, so the combined terminal value is \(K_p\). When the stock finishes at or above \(K_p\), the put expires without value and the combined position is worth \(S_T\). Thus the put creates a floor on terminal value, not necessarily on investment profit.
To calculate profit on a financed-cost basis, first find the net initial outlay and accumulate it to expiration. If the put premium is \(P_p\), the initial outlay is \(S_0+P_p\), and the net result at expiration is
This comparison treats the initial cost as an amount that could have earned interest at rate \(i\). If a question instead asks for a particular cash-flow measure, follow that measure’s stated convention; do not silently mix a terminal value with an unaccumulated initial cost.
Worked Example: Protective Put Below and Above the Exercise Price
A hypothetical investor buys one share for $100 and a one-year put with exercise price $95 for a premium of $3. The effective annual interest rate is 4%. Find the terminal value and net result if \(S_T=\$80\) and if \(S_T=\$110\).
Method. Add the stock value and put payoff in each price region, then subtract the accumulated initial outlay. This keeps the option payoff separate from the investment profit.
The initial outlay is \(100+3=\$103\). Accumulated for one year, it is \(103(1.04)=\$107.12\).
If \(S_T=\$80\), the raw exercise difference is \(95-80=\$15\), so the put pays $15. The stock is worth $80, giving terminal value \(80+15=\$95\). The net result is \(95-107.12=-\$12.12\).
If \(S_T=\$110\), the raw exercise difference is \(95-110=-\$15\), so the put payoff is \(\max(-15,0)=\$0\). The terminal value is \(110+0=\$110\), and the net result is \(110-107.12=\$2.88\).
Check. The terminal value is at least the put exercise price in both cases: it is $95 below the strike and $110 above it. The net results reflect the accumulated premium and stock cost, so a $95 terminal-value floor does not imply a nonnegative net result.
Collar: A Floor and a Cap
A collar adds a short call to the stock-and-put position. The short call payoff is \(-\max(S_T-K_c,0)\). Therefore, if the put premium is \(P_p\) and the call premium received is \(P_c\), the net initial outlay is \(S_0+P_p-P_c\). The collar’s terminal value is
Below the put exercise price, the put supports the position at \(K_p\). Between the two exercise prices, both options expire without value and the stock determines the terminal value. Above the call exercise price, the written call’s obligation offsets further increases in the stock value, leaving terminal value at \(K_c\). The accumulated initial outlay is \((S_0+P_p-P_c)(1+i)^T\), so net result is terminal value minus that amount.
Worked Example: Evaluating a Collar in Three Price Regions
A hypothetical investor buys one share for $100, buys a one-year put with exercise price $90 for $4, and writes a one-year call with exercise price $110 for $2. The effective annual interest rate is 5%. Find the terminal value and net result when \(S_T\) is $80, $100, and $125.
Method. Use the collar’s three regions, accounting for the call premium as money received at time 0. The net initial outlay is \(100+4-2=\$102\), which accumulates to \(102(1.05)=\$107.10\).
At \(S_T=\$80\), the put payoff is \(\max(90-80,0)=\$10\). The call payoff to its holder is \(\max(80-110,0)=\$0\), so the short-call position contributes $0. Terminal value is \(80+10-0=\$90\). Net result is \(90-107.10=-\$17.10\).
At \(S_T=\$100\), the put’s raw exercise difference is \(90-100=-\$10\), so its payoff is $0. The call’s raw exercise difference is \(100-110=-\$10\), so its payoff is also $0. Terminal value is $100, and net result is \(100-107.10=-\$7.10\).
At \(S_T=\$125\), the put’s raw exercise difference is \(90-125=-\$35\), so its payoff is $0. The call payoff to its holder is \(125-110=\$15\), which is a $15 loss for the call writer. Terminal value is \(125-15=\$110\), and net result is \(110-107.10=\$2.90\).
Check. The three terminal values are $90, $100, and $110: the put floor, the stock value between strikes, and the call cap. The call premium reduces the initial outlay by $2; it does not remove the obligation to pay the call’s $15 payoff in the high-price case.
Deriving the Payoff Regions
The piecewise formulas follow by adding the position’s components in each relevant region. This is useful when a problem changes the number of shares, option positions, or exercise prices: the method is to identify breakpoints and evaluate every component on each interval.
If \(S_T<K_p\), then \(K_p-S_T>0\), so the put pays \(K_p-S_T\). In a protective put, \(S_T+(K_p-S_T)=K_p\). In a collar, the call expires without value because \(S_T<K_p<K_c\).
If \(K_p\le S_T\le K_c\), then \(K_p-S_T\le0\) and \(S_T-K_c\le0\). Both option payoffs are zero, leaving terminal value \(S_T\).
If \(S_T>K_c\), the put payoff is zero and the written call costs \(S_T-K_c\). The collar is worth \(S_T-(S_T-K_c)=K_c\).
At each boundary the adjacent formulas agree. At \(S_T=K_p\), both the put payoff and the difference between \(K_p\) and \(S_T\) are zero. At \(S_T=K_c\), the call payoff is zero and the middle-region value equals \(K_c\). The choice of strict or non-strict inequality at those boundaries therefore does not change the terminal value.
Worked Example: Finding a Collar’s Break-Even Stock Price
A hypothetical investor buys one share for $120, buys a one-year put with exercise price $105 for $6, and writes a one-year call with exercise price $135 for $4. The effective annual interest rate is 3%. Find the net result for \(S_T=\$95\), \(S_T=\$125\), and \(S_T=\$150\), and find the break-even stock price if it lies in the uncapped region.
Method. Calculate the accumulated net outlay once, then apply the collar’s floor, middle region, and cap. In the middle region, terminal value equals \(S_T\), so break-even occurs when \(S_T\) equals the accumulated net outlay, provided that value lies between the strikes.
The net initial outlay is \(120+6-4=\$122\). Its value after one year is \(122(1.03)=\$125.66\).
At \(S_T=\$95\), the put payoff is \(105-95=\$10\), and the call payoff is $0. Terminal value is \(95+10=\$105\); net result is \(105-125.66=-\$20.66\).
At \(S_T=\$125\), both options expire without value: the put’s raw exercise difference is \(105-125=-\$20\), and the call’s is \(125-135=-\$10\). Terminal value is $125; net result is \(125-125.66=-\$0.66\).
At \(S_T=\$150\), the put payoff is $0 and the call payoff is \(150-135=\$15\). Terminal value is \(150-15=\$135\); net result is \(135-125.66=\$9.34\).
In the middle region, break-even requires \(S_T-125.66=0\), so \(S_T=\$125.66\). This is between $105 and $135, as required. Check. The three computed terminal values respect the floor of $105 and cap of $135, and the middle-region result changes sign at the calculated break-even price.
Comparing the Two Hedges and Avoiding Traps
A protective put retains the stock’s upside above \(K_p\), but its premium increases the initial outlay. A collar generally reduces the net premium outlay because the call premium is received, but caps the terminal value at \(K_c\). Its put still establishes a downside floor at expiration. The appropriate comparison depends on the investor’s objective and on the stated premiums and rate; a collar is not automatically less costly or more suitable in every case.
| Position | Terminal value below put strike | Terminal value between strikes | Terminal value above call strike |
|---|---|---|---|
| Protective put | \(K_p\) | \(S_T\) | \(S_T\), with no call cap |
| Collar | \(K_p\) | \(S_T\) | \(K_c\) |
A frequent exam trap is to call the terminal-value floor a profit guarantee. For example, in the first worked example the protective put’s terminal value at \(S_T=\$80\) is $95, but the accumulated initial outlay is $107.12. The net result is therefore \(95-107.12=-\$12.12\), not a gain of $15. The $15 is the put payoff, not the profit on the full position.
Another trap is to leave a negative exercise difference as an option payoff. For a put with \(K_p=\$95\) and \(S_T=\$125\), the raw difference is \(95-125=-\$30\), but the payoff is \(\max(-30,0)=\$0\). Similarly, a call with \(K_c=\$110\) and \(S_T=\$100\) has raw difference \(100-110=-\$10\), but its payoff is zero. The negative difference does not represent a payment by the option holder.
- Use the payoff of a long put as \(\max(K_p-S_T,0)\), and subtract the call holder’s payoff for a short call.
- Include the call premium received with a minus sign in the net initial outlay.
- Accumulate the net initial outlay to the same focal date as the terminal value when calculating the financed net result.
- State whether a requested quantity is an option payoff, a combined terminal value, or a net result; these are different quantities.
These distinctions make the position’s risk and return transparent. The formulas in this tutorial apply to the stated one-share, common-expiration setup without intervening distributions. If a problem specifies dividends, additional cash flows, or a different valuation convention, those must be included explicitly rather than inferred from these formulas. This treatment is aligned with the SOA Exam FM (Financial Mathematics) syllabus.
Check Your Understanding
For each problem, distinguish the combined terminal value from the net result after accumulating the initial outlay.
- A hypothetical investor pays $80 for a share and $2 for a one-year put with exercise price $75. At an effective annual interest rate of 5%, the share finishes at $60. Find terminal value and net result. (Answer: terminal value $75; net result \(-\$11.10\).)
- A hypothetical investor pays $90 for a share and $3 for a one-year put with exercise price $85. The effective annual interest rate is 4%, and the share finishes at $92. Find terminal value and net result. (Answer: terminal value $92; net result \(-\$4.72\).)
- A hypothetical collar consists of a share bought for $100, a put with exercise price $92 purchased for $4, and a call with exercise price $108 written for $2. If the effective annual interest rate is 5% and the share finishes at $115 after one year, find terminal value and net result. (Answer: terminal value $108; net result $0.90.)
- A hypothetical collar consists of a share bought for $140, a put with exercise price $125 purchased for $8, and a call with exercise price $155 written for $5. The effective annual interest rate is 2%, and the share finishes at $130 after one year. Determine which options pay off, then find terminal value and net result. (Answer: neither option pays off; terminal value $130; net result \(-\$15.86\).)
- A hypothetical investor buys a share for $75, buys a one-year put with exercise price $80 for $9, and writes a one-year call with exercise price $95 for $4. The effective annual interest rate is 3%, and the share finishes at $100. Find terminal value and net result. (Answer: terminal value $95; net result $12.60.)