What is the put-call parity formula?

For a European option on a non-dividend stock, c plus PV(X) equals p plus S_0. The call premium plus the present value of the strike equals the put premium plus the spot price. Any deviation creates an arbitrage opportunity.

Why does put-call parity hold?

Both sides of the equation produce the same payoff at expiration. The call plus a zero-coupon bond worth X at expiration replicates the put plus the underlying stock. Since the payoffs are identical and there is no arbitrage, the costs today must be equal.

How is put-call parity adjusted for dividends?

Replace S_0 with S_0 minus PV(dividends). Or, in continuous form, replace S_0 with S_0 times exp(-q*T) where q is the dividend yield. The adjustment captures that the dividend reduces the value of the underlying without compensating the call holder.

What is a synthetic long call?

A long stock plus a long put plus borrowing at the risk-free rate produces the same payoff as a long call. From put-call parity, c equals S_0 plus p minus PV(X). The right side is the synthetic call. Useful when the actual call is mispriced or unavailable.

How heavily is put-call parity tested on CFA Level I?

Derivatives is 5-8% of CFA Level I (CFA Institute), but put-call parity is the most consistently tested concept within derivatives. Expect 1-3 questions every sitting, often as a missing-price calculation or a synthetic position identification.

Put-Call Parity: The One Derivatives Identity You Need to Know

Derivatives is only 5-8% of CFA Level I (CFA Institute), but put-call parity is the most heavily tested concept inside that weight. Almost every sitting has at least one direct put-call parity calculation, and several more questions where parity is the underlying logic (synthetic positions, no-arbitrage bounds, conversion trades).

If you understand parity deeply, you understand most of the option pricing intuition the curriculum requires at Level I. If you only memorize the formula, you will miss the questions that hide parity under different language.

The Formula

For a European option on a non-dividend-paying stock:

\[ c + \frac{X}{(1+r)^T} = p + S_0 \]

Where:

$c$ is the European call premium today
$p$ is the European put premium today, same strike and expiration as the call
$X$ is the strike price
$r$ is the risk-free rate (per period)
$T$ is the time to expiration (in periods)
$S_0$ is the spot price of the underlying today

The present value term $X / (1+r)^T$ is sometimes written $PV(X)$ or, in continuous-compounding form, $X e^{-rT}$. Use whichever form matches the question.

Why Parity Holds

The identity is not a result of any pricing model. It is a no-arbitrage relationship that holds under any model. The proof is the cleanest in derivatives.

Consider two portfolios held to expiration:

Portfolio A: Long call (strike X) plus a zero-coupon bond paying X at expiration.

Portfolio B: Long put (strike X) plus the underlying stock.

At expiration, the payoffs are identical:

| Final stock price | Portfolio A | Portfolio B |
|---|---|---|
| $ST < X$ | 0 + X = X | (X - ST) + ST = X |
| $ST = X$ | 0 + X = X | 0 + X = X |
| $ST > X$ | (ST - X) + X = ST | 0 + ST = S_T |

The two portfolios produce the same payoff in every state. By the law of one price, they must cost the same today. Therefore:

\[ c + PV(X) = p + S_0 \]

Key Concept

Put-call parity is not a model. It is a logical identity. If it appears violated in real markets, there is either a transaction cost barrier (bid-ask, financing) or an arbitrage opportunity. Pure parity violations on liquid options do not persist.

Synthetic Positions

Rearranging put-call parity gives the synthetic forms of each instrument. These are tested directly on the exam.

Synthetic Long Call

\[ c = S_0 + p - PV(X) \]

A long call is replicated by long stock, long put, and short bond (borrow at the risk-free rate). Useful when the call market is thin or wide-spread.

Synthetic Long Put

\[ p = c - S_0 + PV(X) \]

A long put is replicated by long call, short stock, and long bond.

Synthetic Long Stock

\[ S_0 = c - p + PV(X) \]

Long stock is replicated by long call, short put, and long bond. This is the basis of the conversion arbitrage: if the synthetic long stock costs less than the actual stock, you buy the synthetic and short the stock.

Synthetic Short Stock

\[ -S_0 = -c + p - PV(X) \]

Short stock is replicated by short call, long put, and short bond. The basis of the reversal arbitrage.

Synthetic Bond

\[ PV(X) = S_0 + p - c \]

A risk-free bond is replicated by long stock, long put, short call. This is interesting because it implies an option-derived risk-free rate, which professional traders monitor for box-spread arbitrage.

Note

Memorize the master identity once. Derive synthetics by rearranging. Trying to memorize each synthetic separately is a recipe for a sign error under exam pressure.

The Most Common Exam Pattern

Three of the four prices ($c$, $p$, $X$, $S0$, $r$, $T$) are given, and the candidate must find the fourth. Usually $X$, $r$, $T$, and $S0$ are given, plus one of $c$ or $p$. Solve for the missing one.

Example

A stock trades at $50. A 1-year European call with strike $48 is priced at $5.50. The risk-free rate is 4%. What is the price of the corresponding 1-year European put?

\[ p = c - S_0 + \frac{X}{(1+r)^T} = 5.50 - 50 + \frac{48}{1.04} = 5.50 - 50 + 46.15 = 1.65 \]

The put is priced at $1.65.

If the market is showing a put price different from $1.65, an arbitrage trade is available. A trader can lock in a riskless profit by buying the cheap side and selling the expensive side, plus the bond and stock to neutralize.

Conversion and Reversal

Conversion: long stock, long put, short call, short bond. Locks in PV(X) at expiration. Used when the synthetic short stock (short call + long put) is cheaper than shorting the actual stock.

Reversal: short stock, short put, long call, long bond. The mirror image. Used when the synthetic long stock is cheaper than buying the actual stock.

In institutional options trading, these trades absorb most of the small parity discrepancies that appear during volatile sessions. They are the reason parity holds so tightly in liquid markets.

Adjusting for Dividends

If the underlying pays dividends during the option's life, the parity formula adjusts:

\[ c + PV(X) = p + S_0 - PV(\text{Dividends}) \]

Or, in continuous form with dividend yield $q$:

\[ c + Xe^{-rT} = p + S_0 e^{-qT} \]

The intuition: a dividend reduces the value of the stock without compensating the call holder, who is not entitled to the dividend during the option's life. So the synthetic long stock must reflect the dividend leakage.

Common Trap

A common exam trap describes a high-dividend stock and asks for a parity-derived put price. Candidates who use the no-dividend formula get the wrong answer. The dividend term must be subtracted from $S_0$.

Continuous-Time Form (Black-Scholes)

The Black-Scholes pricing model satisfies put-call parity by construction. Substituting the Black-Scholes call and put formulas into the parity equation reduces to an identity. This is one reason the model is used as a baseline: it is internally consistent with the no-arbitrage structure.

\[ c - p = S_0 - X e^{-rT} \]

This form is convenient because it isolates the call-minus-put difference, which equals the forward-minus-strike present value. CFA Level I tests this rearrangement directly: "Given $c - p$ and $X e^{-rT}$, find $S_0$."

American Options

Put-call parity in its strict form holds for European options only. American options can be exercised early, breaking the equal-payoff argument. The parity relation becomes an inequality:

\[ S0 - X \leq C - P \leq S0 - X e^{-rT} \]

Where $C$ and $P$ are American call and put prices. The bounds are loose enough that exam questions usually flag whether the option is American or European. If American is mentioned, do not apply strict parity.

For non-dividend stocks, an American call has the same value as a European call, because early exercise of a call is never optimal. American puts on dividend stocks can be optimally exercised early, which is why the inequality is needed.

What Level I Tests

From my own sitting and from working through hundreds of CFA Level I derivatives questions:

Direct calculation. Three of the four prices given, find the fourth.
Synthetic identification. Given a position description, identify the equivalent synthetic. Long stock plus long put plus short bond equals what?
Arbitrage detection. Two of the four prices are given alongside an actual market price for the third. The candidate must identify whether an arbitrage exists and on which side.
Dividend adjustment. A dividend-paying stock parity question. Watch for the $PV(\text{Dividends})$ term.
No-arbitrage bounds. Stating the lower bound on a European call ($c \geq \max(0, S_0 - PV(X))$) is a parity-derived result.

Common Mistakes

Forgetting to discount the strike. Parity uses $PV(X)$, not $X$. With a 5% rate over 1 year, the difference is meaningful.
Wrong sign on the put or call. Synthetic constructions are sign-sensitive. Derive from the master identity, do not memorize separately.
Using American options in strict parity. The exam usually specifies. If not specified and the topic is parity, assume European.
Skipping the dividend term. The most expensive trap on dividend-stock questions.
Compounding mismatch. If $r$ is annual and $T$ is in years, use $(1+r)^T$. If $r$ is continuous and $T$ is in years, use $e^{-rT}$. The exam is usually explicit about which form to use.

How to Practice

Write the master identity at the top of every derivatives practice question:

\[ c + PV(X) = p + S_0 \]

Derive the synthetic you need from there. Do not try to memorize four separate synthetics, you will sign-error one of them under time pressure.

Review the CFA Level I formula sheet for the parity equation in both discrete and continuous forms, with and without dividends. Then drill 30-50 derivatives questions back-to-back on FreeFellow.

The free CFA Level I question bank has a Derivatives topic filter. The solutions walk through the parity setup explicitly, so you train the habit of writing the identity first and solving second.

If you can write put-call parity in 5 seconds and derive any synthetic from it in 15 seconds, you have mastered the most-tested derivatives concept on the exam. That is 1-3 free points every sitting.