Synthetic Forwards And Borrowing Positions From Options

Why Build a Position From Options?

A forward position is often described by its payoff at a future date, while an option portfolio is assembled from contracts with premiums paid today. Put-call parity connects those two viewpoints: a call and a put with the same strike and expiration can create a forward-like payoff. Adding or removing a risk-free bond then changes that payoff into a stock position or a fixed borrowing obligation.

The important exam habit is to check both dates. First calculate the portfolio’s signed value at inception. Then write each asset’s payoff or cash flow at expiration and verify that the components combine as claimed. A matching expiration payoff alone does not establish that the initial cost is correct.

Definition: A synthetic position is a portfolio of traded assets or contracts whose cash flows match those of a target position at the dates being considered. A synthetic long forward with delivery price $K$ has expiration payoff $S_T-K$. A risk-free bond that pays $K$ at time $T$ has present value $Kv^T$, where $v=1/(1+i)$ for an annual effective interest rate $i$ and a term of $T$ years.

Put-Call Parity as the Building Block

Start with European call and put options on the same underlying asset, with the same strike $K$ and expiration time $T$. Assume no transaction costs, taxes, or restrictions on borrowing and lending, and assume that the options can be exercised only at expiration. If the asset pays no distributions before expiration, put-call parity is

$$ C_0-P_0=S_0-Kv^T. $$

Here $C_0$ and $P_0$ are the call and put values today, and $S_0$ is the current asset price. The right-hand side is the value of holding the asset and borrowing the present value of $K$. If the asset instead pays known cash distributions before expiration, replace $S_0$ by its prepaid-forward value: subtract the present value of those distributions. With $I$ denoting their total present value,

$$ C_0-P_0=S_0-I-Kv^T. $$

The no-distribution case makes the terminal cash flows especially direct. When known distributions occur, an actual asset holding also receives those distributions before expiration; they must be included when checking full cash-flow equivalence.

Why the Call-Put Difference Is a Forward

At expiration, the call payoff is $\max(S_T-K,0)$, and the put payoff is $\max(K-S_T,0)$. Their difference is

\max(S_T-K,0)-\max(K-S_T,0)=S_T-K.

If $S_T\geq K$, the call pays $S_T-K$ and the put pays zero. If $S_T<K$, the call pays zero and subtracting the put payoff gives $-(K-S_T)=S_T-K$. Thus, long one call and short one put have exactly the expiration payoff of a long forward with delivery price $K$.

That payoff identity also explains the initial value. A forward contract with delivery price $K$ has value $S_0-Kv^T$ under these assumptions: the asset costs $S_0$ today, while the delivery payment has present value $Kv^T$. Put-call parity says that this is precisely $C_0-P_0$. If $K$ equals the fair forward price, the value is zero; for another strike, the synthetic forward can have a nonzero initial value.

Worked Example: A Synthetic Long Forward

A non-dividend-paying asset is priced at $100$. The annual effective risk-free rate is $25\%$, and European options expire in one year. A call and put with strike $125$ each cost $12$. Construct the synthetic forward and check its cash flows if the expiration asset price is $90$, $125$, or $160$.

Method. Use one long call and one short put. The discount factor is $v=1/1.25=0.8$, so the present value of the strike payment is $125(0.8)=100$. The option portfolio’s initial value is $12-12=0$, consistent with parity:

$$ S_0-Kv=100-125(0.8)=0=C_0-P_0. $$

At $S_T=90$, the call pays $0$, the short put pays $-(125-90)=-35$, and the total is $-35=90-125$. At $S_T=125$, both options pay zero, giving $0=125-125$. At $S_T=160$, the call pays $160-125=35$, the short put pays zero, and the total is $35=160-125$.

The synthetic position has zero initial value and the same payoff as a long forward with delivery price $125$ in each case. Common mistake avoided: the long call/short put payoff is $S_T-K$, not $K-S_T$. Reversing the option positions reverses the forward exposure.

Adding a Bond to Synthesize the Asset

The call-minus-put portfolio has payoff $S_T-K$, which is short of the asset payoff $S_T$ by $K$. Add a risk-free investment that pays $K$ at expiration. The combined expiration payoff becomes $S_T-K+K=S_T$. Its initial cost is $C_0-P_0+Kv^T$, which equals $S_0$ by put-call parity.

\underbrace{\text{long call}-\text{long put}+\text{bond paying }K}_{\text{synthetic asset}} \quad\longleftrightarrow\quad C_0-P_0+Kv^T=S_0.

The bond’s face amount is the strike $K$, not its present value. Its initial price is $Kv^T$; its expiration payment is $K$. Confusing these amounts is a frequent source of errors when constructing a synthetic asset.

Worked Example: Replicating a Stock Position

An asset is priced at $90$. The annual effective interest rate is $20\%$, and European options expire in one year. A call and put with strike $108$ each cost $10$. Show that a long call, a short put, and a risk-free investment replicate one unit of the asset. Check the expiration cash flows at asset prices $72$ and $150$.

Method. The discount factor is $v=1/1.20=5/6$. Investing $Kv=108(5/6)=90$ today buys a bond that pays $108$ at expiration. The initial portfolio cost is

$$ C_0-P_0+Kv=10-10+90=90=S_0. $$

At $S_T=72$, the call pays $0$, the short put pays $-(108-72)=-36$, and the bond pays $108$. The total is $0-36+108=72$, equal to the asset price at expiration. At $S_T=150$, the call pays $150-108=42$, the short put pays $0$, and the bond pays $108$. The total is $42+0+108=150$.

The portfolio therefore matches both the asset’s initial price and its expiration value in these cases; the payoff identity establishes the match for every $S_T$. Common mistake avoided: paying $90$ for the bond today does not mean it pays $90$ at expiration. It pays its face amount, $108$.

Turning the Bond Into a Borrowing Position

A long bond paying $K$ at expiration is a lending position: pay $Kv^T$ now and receive $K$ later. Reverse every position to create the opposite, a borrowing position: receive $Kv^T$ now and repay $K$ at expiration. From parity, the short bond payoff $-K$ can be synthesized with a short asset, a long call, and a short put.

$$ -S_0+C_0-P_0=-Kv^T. $$

This equation gives the portfolio’s signed value: $-Kv^T$. A negative position value means it costs nothing to buy the position; instead, establishing it generates cash of $Kv^T$. At expiration, its payoff is $-K$. Keeping the distinction between position value and cash received when establishing a negative-value position prevents a sign error.

Worked Example: Borrowing Synthetically

An asset is priced at $96$. The annual effective interest rate is $25\%$, and European options expire in one year. A call and put with strike $120$ each cost $14$. Construct a synthetic borrowing position and verify that it raises the correct amount and repayment.

Method. The discount factor is $v=1/1.25=0.8$, so $Kv=120(0.8)=96$. The portfolio consisting of a short asset, a long call, and a short put has signed initial value

$$ -S_0+C_0-P_0=-96+14-14=-96. $$

Because this value is negative, establishing the position generates $96$ in cash today. Its expiration payoff is always $-120$. For $S_T=80$, the short asset pays $-80$, the call pays $0$, and the short put pays $-(120-80)=-40$, for a total of $-120$. For $S_T=150$, the short asset pays $-150$, the call pays $150-120=30$, and the short put pays $0$, again totaling $-120$.

Thus the portfolio raises $96$ and requires repayment of $120$ one year later. The implied annual effective rate is $120/96-1=25\%$, matching the assumed rate. Common mistake avoided: interpreting the signed portfolio value of $-96$ as an initial cash payment. The cash flow to the investor is $+96$.

A Cash-Flow Checklist and Common Traps

For any synthetic forward, asset, or borrowing position, write down the contracts, their signs, and their cash flows before relying on a memorized formula. The following sequence keeps the initial and expiration calculations aligned.

Match the options.
Confirm that the call and put have the same underlying, strike, and expiration, and identify whether they are European options.

Discount the strike.
Under a constant annual effective rate, calculate $Kv^T$ for the initial value of a bond that pays $K$ at time $T$.

Set the position signs.
A long forward is long call and short put; reversing both option positions creates a short forward.

Check both dates.
Add signed option premiums and bond or asset values today, then add signed payoffs and any required cash flows at expiration.

Two errors deserve special attention. First, $K$ is the bond’s payment at expiration, whereas $Kv^T$ is its value today. Second, the payoff of a short option is the negative of its long-option payoff. For example, shorting a put creates a negative payoff when $S_T<K$; it does not create a positive payoff of $K-S_T$.

When known distributions occur, use the distribution-adjusted parity relation rather than the no-distribution formula. The call-minus-put payoff at expiration is still $S_T-K$, but the value of holding the underlying before expiration also reflects the cash distributions it pays. Ignoring them can make a stock-replication cost or a forward value incorrect even when the expiration option-payoff calculation is right.

Key takeaway: With no distributions, long call minus long put replicates a long forward with payoff $S_T-K$. Adding a bond paying $K$ replicates the asset; reversing the asset and option positions creates a borrowing position that receives $Kv^T$ and repays $K$. Verify each construction at inception and expiration.

Check Your Understanding

For each problem, use signed positions and check the stated initial value or expiration cash flow.

A non-dividend-paying asset has $S_0=100$. The annual effective rate is $25\%$, $T=1$, and $K=125$. A call and put each cost $12$. What is the initial value and payoff at $S_T=150$ of a synthetic long forward? (Answer: Initial value $0$; expiration payoff $25$.)
An asset has $S_0=75$. The annual effective rate is $20\%$, $T=1$, $K=90$, and both options cost $13$. What is the initial cost of a synthetic asset made from a long call, short put, and bond paying $90$? What is its payoff if $S_T=60$? (Answer: Initial cost $75$; payoff $60$.)
An asset has $S_0=120$. The annual effective rate is $25\%$, $T=1$, $K=150$, and both options cost $16$. Using a short asset, long call, and short put, how much cash is raised today and how much is repaid at expiration? (Answer: $120$ raised today; $150$ repaid at expiration.)
A European call and put share a strike and expiration. The asset price is $100$, the present value of known distributions before expiration is $5$, and $Kv^T=99$. If the call costs $7$, what must the put cost under parity? (Answer: $11$, since $C_0-P_0=100-5-99=-4$.)
A bond pays $100$ in two years. The annual effective rate is $5\%$. What cash amount is received today by the synthetic short-bond position, and what is its expiration payoff? Round the cash amount to the nearest cent. (Answer: $100/(1.05)^2=\$90.70$ received today; payoff $-\$100$ in two years.)

What You'll Learn