Synthetic Stock, Bond, Call, and Put Positions

Why Construct Synthetic Positions?

A synthetic position uses a combination of traded instruments to reproduce another instrument’s payoff. For an investor or analyst, this is useful because it makes the economic exposure of a portfolio easier to see: a call combined with other positions may behave exactly like stock, a put, or a risk-free investment. The key test is not whether the positions look similar today, but whether they produce the same cash flows in every possible market outcome at expiration.

The standard put–call parity relationship provides the building block for these constructions. The positions considered here use European options with the same underlying stock, exercise price, and expiration date, and a risk-free investment that pays the exercise price at expiration.

Definition: A synthetic position is a portfolio of instruments whose cash flows match those of a specified target position over the relevant dates and states of the world. A risk-free bond in this tutorial means an investment that pays a known amount at the options’ expiration. A long position has positive exposure to the instrument; a short position has negative exposure.

Starting Point: Put–Call Parity

Let $S_0$ be the current stock price, $C$ the current price of a European call, $P$ the current price of a European put, $K$ the common exercise price, and $T$ the time to expiration in years. Let $i$ be the effective annual risk-free rate, assumed constant over the term. The discount factor for $T$ years is $v^T=(1+i)^{-T}$. Assume the stock pays no dividends before expiration, and that the options and stock have the same terms and underlying.

Theorem: Under these assumptions, put–call parity is $$ C+Kv^T=P+S_0. $$ The term $Kv^T$ is the amount invested today in a risk-free position that pays $K$ at expiration. It is the present value of the exercise price, not the exercise price itself.

Rearranging parity gives the initial costs of several equivalent positions:

\begin{aligned} \text{Synthetic stock: } & C-P+Kv^T=S_0,\\ \text{Synthetic call: } & S_0+P-Kv^T=C,\\ \text{Synthetic put: } & C+Kv^T-S_0=P,\\ \text{Synthetic risk-free bond: } & S_0+P-C=Kv^T. \end{aligned}

These equations also specify the positions. A minus sign means a short position; for example, $C-P$ means long one call and short one put. In a synthetic stock or call, the term $-Kv^T$ means borrowing the present value of $K$, which requires repayment of $K$ at expiration. A long risk-free bond costs $Kv^T$ today and pays $K$ at expiration. A synthetic borrowing position is the opposite of the synthetic bond: short the stock, short the put, and long the call.

Why the Payoffs Match

At expiration, a European call pays $\max(S_T-K,0)$, and a European put pays $\max(K-S_T,0)$, where $S_T$ is the stock price at expiration. Subtracting the put payoff from the call payoff leaves $S_T-K$ in either possible region: above or below $K$. This identity explains why a call and a short put together resemble stock financed by borrowing.

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

Adding a risk-free bond that pays $K$ cancels the $-K$ in that result, leaving $S_T$, the stock payoff. This is the payoff logic behind the synthetic stock. The same approach verifies the other positions: compare their cash flows state by state, then check that their costs today agree under parity.

Write the target payoff.
For stock, the target payoff is $S_T$. For a call or put, it is the corresponding maximum payoff.

Choose the parity rearrangement.
Put the target instrument alone on one side of the parity equation; read the remaining terms as the replicating positions.

Check expiration payoffs and initial cost.
Test representative values of $S_T$, and verify that the combined cost equals the target instrument’s price under the stated assumptions.

Constructing Each Position

Target position	Positions that replicate it	Expiration payoff
Long stock	Long call, short put, long bond paying $K$	$S_T$
Long call	Long stock, long put, borrow $Kv^T$	$\max(S_T-K,0)$
Long put	Long call, long bond paying $K$, short stock	$\max(K-S_T,0)$
Long risk-free bond	Long stock, long put, short call	$K$
Risk-free borrowing	Short stock, short put, long call	$-K$ at expiration; receive $Kv^T$ initially

The table describes positions with one option contract’s underlying share and payoff units, abstracting from contract multipliers. When implementing a position, contract size and any difference in deliverable must be accounted for. The parity construction also depends on matching the exercise price and expiration; options with different terms do not produce these same payoffs.

Worked Example: Creating Synthetic Stock

A non-dividend-paying stock is at $S_0=\$100.00$. A European call and put both have $K=\$104.00$ and expire in one year. The effective annual risk-free rate is $i=4\%$, and the call price is $C=\$10.00$. Use parity to find the put price and construct synthetic stock. Verify the payoff when $S_T=\$80.00,\ \$104.00,$ and $\$130.00$.

Method and assumptions. Use standard European put–call parity, with no dividends, a one-year term, and annual effective compounding. First calculate the present value of the exercise price:

v=\frac{1}{1.04}=0.9615385,\qquad Kv^T=\frac{\$104.00}{1.04}=\$100.00.

Rearrange $C+Kv^T=P+S_0$ to solve for the put:

P=C+Kv^T-S_0=\$10.00+\$100.00-\$100.00=\$10.00.

The synthetic stock position is long the call, short the put, and long a bond that pays $\$104.00$ in one year. Its initial cost is:

C-P+Kv^T=\$10.00-\$10.00+\$100.00=\$100.00=S_0.

At $S_T=\$80.00$, the call pays $\$0.00$, the short put pays $-(\$104.00-\$80.00)=-\$24.00$, and the bond pays $\$104.00$. Total: $\$0.00-\$24.00+\$104.00=\$80.00$.

At $S_T=\$104.00$, both options pay $\$0.00$, and the bond pays $\$104.00$. Total: $\$104.00$.

At $S_T=\$130.00$, the call pays $\$130.00-\$104.00=\$26.00$, the put pays $\$0.00$, and the bond pays $\$104.00$. Total: $\$130.00$.

The three checks agree with the stock payoff $S_T$; the payoff identity establishes that the match holds for every $S_T$. The common mistake this method avoids is omitting the bond: the call minus the put pays $S_T-K$, not $S_T$.

Worked Example: Replicating a European Call

A non-dividend-paying stock is at $S_0=\$72.00$. A European put has price $P=\$8.00$, and the matching call and put have $K=\$75.00$ and $T=2$ years. The effective annual risk-free rate is $i=5\%$. Find the call price and verify its synthetic payoff at stock prices of $\$60.00,\ \$90.00,$ and $\$110.00$ at expiration.

Method and assumptions. Rearrange parity as $C=S_0+P-Kv^T$. Discount the exercise price for two years:

v^T=(1.05)^{-2}=\frac{1}{1.1025}=0.9070295,\qquad Kv^T=\frac{\$75.00}{1.1025}=\$68.0272.

The call price implied by parity is:

C=\$72.00+\$8.00-\$68.0272=\$11.9728\approx \$11.97.

The synthetic call is long one share, long one put, and short a bond that would pay $\$75.00$ at expiration. The borrowing receives $\$68.0272$ initially and requires repayment of $\$75.00$ at expiration. At $S_T=\$60.00$, the total payoff is $\$60.00+(\$75.00-\$60.00)-\$75.00=\$0.00$. At $S_T=\$90.00$, it is $\$90.00+\$0.00-\$75.00=\$15.00$. At $S_T=\$110.00$, it is $\$110.00+\$0.00-\$75.00=\$35.00$.

Those are exactly the call payoffs $\max(S_T-\$75.00,0)$. The initial cost, using unrounded values, is $\$72.00+\$8.00-\$68.0272=\$11.9728$, or $\$11.97$ to the nearest cent. The method avoids treating the loan as a payment of $\$68.0272$ at expiration; its expiration repayment is $\$75.00$.

Worked Example: Creating a Synthetic Put

A non-dividend-paying stock is at $S_0=\$120.00$. A European call has price $C=\$15.00$. The call and put have $K=\$125.00$ and expire in one year. The effective annual risk-free rate is $i=5\%$. Find the put price using parity and verify the synthetic put payoff at $S_T=\$100.00,\ \$125.00,$ and $\$150.00$.

Method and assumptions. Use $P=C+Kv^T-S_0$. The present value of the exercise price is:

Kv^T=\frac{\$125.00}{1.05}=\$119.0476.

Thus:

P=\$15.00+\$119.0476-\$120.00=\$14.0476\approx \$14.05.

A synthetic put is long the call, long a bond paying $\$125.00$ at expiration, and short one share. At $S_T=\$100.00$, its payoff is $\$0.00+\$125.00-\$100.00=\$25.00$. At $S_T=\$125.00$, its payoff is $\$0.00+\$125.00-\$125.00=\$0.00$. At $S_T=\$150.00$, its payoff is $(\$150.00-\$125.00)+\$125.00-\$150.00=\$0.00$.

These are the put payoffs $\max(K-S_T,0)$. The initial cost is $\$15.00+\$119.0476-\$120.00=\$14.0476$, which rounds to $\$14.05$. The sign on the stock matters: using a long share instead would not reproduce a put’s payoff.

Worked Example: Synthetic Lending and Borrowing

A non-dividend-paying stock is at $S_0=\$48.00$. A European put has price $P=\$6.0769$, and the matching call has price $C=\$6.00$. Both options have $K=\$50.00$, expire in one year, and the effective annual risk-free rate is $i=4\%$. Verify the synthetic risk-free lending position and its reverse, borrowing.

Method and assumptions. A synthetic bond is long stock, long put, and short call. Its initial cost is:

S_0+P-C=\$48.00+\$6.0769-\$6.00=\$48.0769.

The direct present value of $\$50.00$ at $4\%$ is:

Kv^T=\frac{\$50.00}{1.04}=\$48.0769.

At $S_T=\$35.00$, the synthetic bond pays $\$35.00+(\$50.00-\$35.00)-\$0.00=\$50.00$. At $S_T=\$50.00$, it pays $\$50.00+\$0.00-\$0.00=\$50.00$. At $S_T=\$70.00$, it pays $\$70.00+\$0.00-(\$70.00-\$50.00)=\$50.00$. The payoff is therefore $\$50.00$ in all three cases, and the identity ensures the same for any expiration price.

Reversing every position—short stock, short put, and long call—produces the borrowing position. It receives $\$48.0769$ initially and has an expiration payoff of $-\$50.00$, meaning $\$50.00$ must be repaid. The calculation checks both the constant bond payoff and its discounted cost.

Common Mistakes and Limits

A reliable synthetic-position calculation checks both the payoff and the initial cost. The payoff check catches position-sign errors; the cost check catches discounting and parity errors.

Using $K$ instead of $Kv^T$ today. A bond that pays $K$ at expiration costs $Kv^T$ at the risk-free rate, not $K$, unless the term is zero or the rate is zero.
Reversing a short position. A synthetic call is long stock and long put but borrows the present value of $K$. Buying the bond instead would add $K$, rather than subtract it, at expiration.
Checking only one expiration price. A portfolio can match a target payoff in one state and fail in another. Check below, at, and above $K$, or use the payoff identity.
Ignoring the assumptions. The formulas here are for European options on a non-dividend-paying stock with matching terms. Dividends or different option terms require an adjusted relationship.

Synthetic positions are useful for understanding exposures and for translating a portfolio into familiar building blocks. Their construction follows from payoff equivalence, while parity ensures the corresponding initial costs agree under the stated assumptions. This treatment is aligned with the SOA VEE Accounting and Finance syllabus.

Check Your Understanding

For each problem, use the stated European-option assumptions and verify signs and discounting before rounding money to the nearest cent.

A non-dividend-paying stock is at $\$78.9231$. A one-year call with $K=\$80.00$ costs $\$9.00$. The effective annual risk-free rate is $4\%$. Use parity to find the put price. (Answer: $\$7.00$)
A non-dividend-paying stock is at $\$55.00$. A two-year call costs $\$8.00$; the matching options have $K=\$60.00$. The effective annual risk-free rate is $5\%$. Find the put price. (Answer: $\$7.42$)
A non-dividend-paying stock is at $\$90.00$. A one-year put with $K=\$100.00$ costs $\$13.00$. The effective annual risk-free rate is $4\%$. Find the synthetic call price. (Answer: $\$6.85$)
A one-year risk-free bond pays $\$200.00$ at expiration. A non-dividend-paying stock is at $\$195.00$, the matching put costs $\$12.00$, and the matching call costs $\$16.5238$. Verify the synthetic bond’s initial cost and identify the three positions. (Answer: Long stock, long put, and short call; initial cost $\$190.48$)
A one-year call and put have $K=\$50.00$; the effective annual risk-free rate is $4\%$. The stock is at $\$48.00$, the call costs $\$6.00$, and the put costs $\$6.0769$. State the synthetic borrowing positions, the amount received initially, and the amount repaid at expiration. (Answer: Short stock, short put, and long call; receive $\$48.08$ initially and repay $\$50.00$ at expiration)

Target position	Positions that replicate it	Expiration payoff
Long stock	Long call, short put, long bond paying \(K\)	\(S_T\)
Long call	Long stock, long put, borrow \(Kv^T\)	\(\max(S_T-K,0)\)
Long put	Long call, long bond paying \(K\), short stock	\(\max(K-S_T,0)\)
Long risk-free bond	Long stock, long put, short call	\(K\)
Risk-free borrowing	Short stock, short put, long call	\(-K\) at expiration; receive \(Kv^T\) initially

What You'll Learn