Building Synthetic Positions From Put-Call Parity

From Parity to Replication

Put-call parity is more than an equation for finding an unknown option value. Rearranging it shows how a portfolio of options, stock, and a risk-free bond can reproduce another asset's value and payoff. These combinations are called synthetic positions. They are useful for understanding exposure, checking calculations, and identifying which positions must be held to create a specified payoff.

The focus here is on constructing the positions and listing their legs. A synthetic position is a replication statement: under the stated assumptions, its value and payoff match those of the target. It does not, by itself, assert that market prices violate parity.

Assumptions and notation: Consider European calls and puts on a non-dividend-paying stock, with the same strike $K$ and expiration at time $T$. Assume frictionless trading, no taxes or transaction costs, and a risk-free effective annual interest rate $i$ over the term. Let $S$, $C$, and $P$ be the current stock, call, and put prices. The discount factor for one year is $v=1/(1+i)$; for a term of $n$ years, the present value of $K$ is $Kv^n$. A risk-free zero-coupon bond that pays $K$ at expiration costs $Kv^n$ today.

For a one-year term, the European put-call parity equation is $C+Kv=S+P$. For a term of $n$ years at the same effective annual rate, it is $C+Kv^n=S+P$. The following rearrangements use the one-year version to keep the examples compact; the same position logic applies with $v^n$ for other terms.

$$ C+Kv=S+P $$

The Four Basic Synthetic Positions

To build a synthetic position, first rearrange parity to isolate the target value. Then translate each term into a position: a positive term means holding it long, and a negative term means holding it short. A bond term means the present value of a risk-free payment of $K$ at expiration.

Synthetic call.
Rearrange to $C=S+P-Kv$: go long the stock, go long the put, and short a zero-coupon bond that pays $K$ at expiration. Shorting that bond is equivalent to borrowing $Kv$ today and repaying $K$ at expiration.

Synthetic put.
Rearrange to $P=C+Kv-S$: go long the call, go long the bond paying $K$, and short the stock.

Synthetic stock.
Rearrange to $S=C-P+Kv$: go long the call, go short the put, and lend $Kv$ by buying a bond that pays $K$ at expiration.

Synthetic lending or borrowing.
Rearrange to $Kv=S+P-C$: go long the stock and put, and short the call to replicate lending $Kv$ today for a risk-free payment of $K$. Reverse all three positions to synthesize borrowing.

Each recipe specifies one unit of the underlying stock and one call and put with matching strike and expiration. If a contract represents more than one share, quantities must be adjusted to match the share exposure; the examples below assume one-share units.

Why the Positions Match at Expiration

Let $S_T$ denote the stock price at expiration. The European call payoff is $\max(S_T-K,0)$, and the European put payoff is $\max(K-S_T,0)$. A bond paying $K$ has expiration value $K$. Checking the portfolios on either side of the strike verifies the replication; the identity $\max(S_T-K,0)-\max(K-S_T,0)=S_T-K$ gives a compact justification.

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

For example, the synthetic call payoff is stock plus put minus the bond payment. If $S_T\leq K$, it is $S_T+(K-S_T)-K=0$. If $S_T>K$, it is $S_T+0-K=S_T-K$. These are exactly the call's two payoff cases. The other positions follow by rearranging the same payoff identity.

Theorem: Under the assumptions above, each rearrangement of put-call parity gives a portfolio with the same current value as its target and a matching expiration payoff. If the term is $n$ years, replace $Kv$ in the current-value equations with $Kv^n$; the bond still pays $K$ at expiration.

Worked Examples

Worked Example: Building and Checking a Synthetic Call

A non-dividend-paying stock is priced at $$72$. A one-year European put with strike $$80$ is priced at $$14$. The effective annual interest rate is $5\%$. Find the parity-implied call price, list the synthetic call positions, and check the expiration payoff when the stock finishes at $$60$ and $$95$. Assume frictionless trading and one-share option units.

Method: isolate the call in parity. Since $v=1/(1.05)$, the bond paying $$80$ in one year costs:

Kv=\frac{80}{1.05}=76.190476\ldots

Parity gives $C=S+P-Kv$. Substituting the data:

C=72+14-\frac{80}{1.05} =86-76.190476\ldots =9.809524\ldots

Rounded to the nearest cent, the parity-implied call price is $$9.81$. The synthetic call is long one share of stock, long one put, and short a risk-free bond paying $$80$ in one year. Its initial net cost is $72+14-76.190476\ldots=\$9.809524\ldots$, which matches the calculated call price before rounding.

At $S_T=\$60$, the stock is worth $$60$, the put pays $80-60=\$20$, and the short bond requires a payment of $$80$. The portfolio payoff is $60+20-80=\$0$, matching the call payoff $\max(60-80,0)=\$0$.

At $S_T=\$95$, the stock is worth $$95$, the put pays $$0$, and the short bond still requires $$80$. The portfolio payoff is $95+0-80=\$15$, matching the call payoff $\max(95-80,0)=\$15$.

Common mistake avoided: subtracting the strike $$80$ from current prices as if it were payable today would give $72+14-80=\$6$. The bond's current price is the discounted strike, $$76.190476\ldots$, not $$80$; its $$80$ payment occurs at expiration.

Worked Example: Replicating a Put

A non-dividend-paying stock costs $$108$. A one-year European call with strike $$105$ costs $$12$, and the effective annual interest rate is $5\%$. Find the synthetic put price and list the legs. Check the portfolio when the expiration stock price is $$80$ and $$120$. Assume frictionless trading and one-share option units.

Method: isolate the put in parity. The present value of the bond payment is:

Kv=\frac{105}{1.05}=100

Using $P=C+Kv-S$, the put price is:

$$ P=12+100-108=4 $$

The synthetic put costs $$4$ initially. Its legs are long one call, long a bond paying $$105$ in one year, and short one share of stock. The initial net value is $12+100-108=\$4$, consistent with the parity-implied put price.

At $S_T=\$80$, the call pays $$0$, the bond pays $$105$, and the short stock position is worth $-\$80$. The portfolio payoff is $0+105-80=\$25$. The put also pays $\max(105-80,0)=\$25$.

At $S_T=\$120$, the call pays $120-105=\$15$, the bond pays $$105$, and the short stock position is worth $-\$120$. The portfolio payoff is $15+105-120=\$0$, equal to the put payoff $\max(105-120,0)=\$0$.

Common mistake avoided: the put requires a short stock position, not a long one. The minus sign in $P=C+Kv-S$ is a portfolio instruction as well as an algebraic sign.

Worked Example: Synthesizing Stock and Risk-Free Lending

A non-dividend-paying stock costs $$101$. A one-year European call with strike $$104$ costs $$8$, and the matching put costs $$7$. The effective annual interest rate is $4\%$. Construct a synthetic stock position and a synthetic one-year lending position, then verify both at expiration prices of $$70$ and $$125$. Assume frictionless trading and one-share option units.

Method: rearrange parity for each target. The bond paying $$104$ costs $104/1.04=\$100$. For stock, use $S=C-P+Kv$: go long the call, short the put, and buy the bond. The initial value is $8-7+100=\$101$, matching the stock price.

For lending, use $Kv=S+P-C$: go long the stock and put, and short the call. The initial value is $101+7-8=\$100$, which is the amount invested in the risk-free bond. The synthetic lending position is therefore intended to pay $$104$ in one year.

At $S_T=\$70$, the call pays $$0$, the put pays $104-70=\$34$, and the bond pays $$104$. The synthetic stock pays $0-34+104=\$70$, matching the stock. The lending portfolio pays $70+34-0=\$104$, matching its fixed bond payment.

At $S_T=\$125$, the call pays $125-104=\$21$, and the put pays $$0$. The synthetic stock pays $21-0+104=\$125$. The lending portfolio pays $125+0-21=\$104$. Both replications match their targets at this second stock price as well.

Common mistake avoided: the synthetic lending portfolio is long stock and put but short call. Reversing the call and put signs would make the payoff vary with $S_T$ instead of remaining fixed at $$104$.

Reading the Signs and Avoiding Traps

A reliable way to convert an equation into positions is to write the target alone on the left, then interpret the signs on the right. A plus sign means long that asset; a minus sign means short it. For bond terms, keep separate the bond's current cost and its payment at expiration: $Kv$ is the amount invested or borrowed today, while $K$ is the maturity cash flow.

Use matching contracts. The call and put in a parity construction must have the same underlying stock, strike, and expiration, and the standard equality here assumes European exercise.
Distinguish price from payoff. $Kv$ belongs in the current-value equation; $K$ belongs in the expiration payoff calculation.
Check both sides of the strike. A single expiration price may not reveal a misplaced option sign. Testing one value below and one above $K$ checks both payoff regions.
Keep the intended target clear. A synthetic call and a synthetic put use different stock and option signs, even though both are formed from the same parity identity.

These constructions matter because they translate an algebraic identity into explicit holdings and cash flows. A candidate can use them to interpret how an option position changes exposure, or to check whether a proposed portfolio really matches the intended asset. The initial values and expiration payoffs should both agree; a match at expiration alone is not a substitute for using the correct discounted bond value today.

Key takeaway: Put-call parity can be rearranged to synthesize a call, a put, stock, or a risk-free bond. List every long and short leg, use the discounted strike for the bond's current value, and verify the target payoff at expiration.

Check Your Understanding

For each problem, assume frictionless trading, European options on a non-dividend-paying stock, matching strikes and expirations, and one-share option units unless stated otherwise.

A one-year put costs $$8$, the stock costs $$97$, the strike is $$105$, and the effective annual rate is $5\%$. Find the parity-implied call price and state its synthetic legs. (Answer: Call price $$5$; long stock, long put, short a bond paying $$105$.)
A one-year call costs $$18$, the stock costs $$120$, the strike is $$126$, and the effective annual rate is $5\%$. Find the put price and list its synthetic legs. (Answer: Put price $$18$; long call, long a bond paying $$126$, short stock.)
A one-year call costs $$11$, a matching put costs $$8$, the stock costs $$103$, and the strike is $$104$. The effective annual rate is $4\%$. Give the synthetic stock portfolio and verify its initial value. (Answer: Long call, short put, and long a bond paying $$104$; initial value $11-8+104/1.04=\$103$.)
A one-year stock costs $$90$, a matching put costs $$16$, the strike is $$105$, and the effective annual rate is $5\%$. Find the call price and the synthetic lending portfolio. (Answer: Call price $$6$; long stock, long put, short call, with initial value $$100$ and a $$105$ payment at expiration.)
A stock costs $$107$, a one-year call costs $$12$, and a matching put costs $$5$. The strike is $$104$, and the effective annual rate is $4\%$. State the synthetic call and put prices and positions. (Answer: Call price $$12$, replicated by long stock, long put, and short the bond paying $$104$; put price $$5$, replicated by long call, long that bond, and short stock.)

What You'll Learn