Put-Call Parity as a Synthetic Forward Position
Summary
The document derives European put-call parity by comparing a long call and short put at the same strike and maturity with a long forward contract. Their terminal payoffs match in every case: above the strike, the call pays the difference; below it, the short put creates the corresponding loss. Because the positions have identical payoffs, absence of arbitrage implies that their values today must also match.
It expresses the forward’s current value as the discounted difference between the fair forward price and the delivery price. When the strike equals the fair forward price, that value is zero, so the call and put premiums are equal. The result is framed for European options and assumes consistent underlying, maturity, and pricing inputs. The stated forward price formula applies in the no-dividend case; dividends or other carrying costs change the fair forward price, while transaction costs, funding frictions, and market conventions can affect practical comparisons.
Key ideas
- A long call combined with a short put at the same strike has the same maturity payoff as a long forward.
- Absence of arbitrage equates the present values of positions with identical future payoffs.
- The forward value depends on the difference between the fair forward price and the contract delivery price.
- When the strike equals the fair forward price, the forward value is zero and European call and put premiums match.
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Full text
# Put call parity: when are the premiums the same?
# Put call parity: when are the premiums the same?
Please explain why put call parity could be compared to the payoff of a long forward contract. ie. $C_E-P_E=V_X(0)$ where $C_E,P_E$ are the call/put premiums and $V_X(0)$ is the value of a long forward contract.
Also please explain why if strike price $X$is equal to the theoretical forward price $S(0)e^{rT}$ of the asset, then the value of the forward contract is $0$, and so $C_E=P_E$
## Answer by Quantuple (score 2, accepted)
https://quant.stackexchange.com/a/25099
The intuitive explanation is given in @Alex C's comment. You should stick to that if you understand it.
Yet, if you are more comfortable with a mathematical approach:
- Payoff of being long a forward contrat with maturity $T$: $(S_T - X)$. Interpretation: at time $T$, you pay a certain price $X$ in exchange for which you receive the underlying $S_T$
- Payoff of being long a call option struck at $X$ and with maturity $T$: $(S_T - X)^+$. Interpretation: at time $T$, you exercise if the underlying's value is greater than the strike price, in which case you earn the difference between the 2, otherwise you do nothing.
- Payoff of being long a put option struck at $X$ and with maturity $T$: $(X - S_T)^+$. Interpretation: at time $T$, you exercise if the underlying's value is smaller than the strike price, in which case you earn the (absolute) difference between the 2, otherwise you do nothing.
- Payoff of being long (+) call option struck at $X$ and short (-) a put struck at $X$ at maturity $T$, is the difference of the 2 previous ones: \begin{align} (S_T - X)^+ - (X - S_T)^+ &= (S_T - X)1\{S_T \geq X\} - (X - S_T)1\{S_T \leq X\} \\ &= (S_T - X)1\{S_T \geq X\} + (S_T - X)1\{S_T \leq X\} \\ &= (S_T - X) \end{align}
As you can see, being long a forward with 'strike' $X$ and being long a call/short a put with strikes $X$ give the same payoff at $T$: $(S_T-X)$.
By absence of arbitrage opportunity, these two strategies should therefore have exactly the same value today, that is: $$ C_E(S_0;T,X) - P_E(S_0;T,X) = V_X(0) $$ which is the famous put-call parity relationship, where $V_X(0)$ represents the value of being long a forward contract struck at $X$ and with maturity $T$ as seen of today, in other words $$ V_X(0) = P(0,T)(F(0,T)-X)$$ with $P(0,T)$ representing the discount factor applying to cash-flows paid at $T$ and $F(0,T)$ the fair forward value.
By setting $X$ (strike price) equal to $F(0,T)$ (forward value $S_0e^{rT}$ in the absence of dividends), then you see that $V_X(0) = 0$ and put-call parity re-writes as $$ C_E(S_0;T,X) - P_E(S_0;T,X) = V_X(0) = 0 $$ meaning $$ C_E = P_E $$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.