Proving European Put-Call Parity with Replication and No-Arbitrage

# Proving the put call parity


# Proving the put call parity












In my course notes on the put-call parity, the proof is presented by going over two inequalities, namely $\text{RHS} > \text{LHS}$ implies arbtirage and $\text{RHS} < \text{LHS}$ implies arbitrage. Therefore, they conclude, $\text{RHS} = \text{LHS}$.

This strategy is legit, but I have the feeling the following proof is more straightforward.

$\textbf{Lemma 1 (law of one price):}$ If two portfolios have the same profit at maturity time $T$, then for all prior times $t<T$ the price of the portfolio's must be equal.

$\textbf{Proof:}$ The proof can easily be done by deriving arbitrage by contradiction.

$\textbf{Theorem (put-call parity)}:$ Let $P_0$ be the price of a European put with strike $K$ and maturation date $T$. Let $C_0$ be the price of a European call with same parameters as the put, and $r$ be a risk-free rate. Let $S_0$ be the price of a stock at $t=0$. Then $$S_0 + P_0 = D(r)K + C_0,$$ where $D(r)$ is the discount of the risk-free bank account.

$\textbf{Proof:}$ Work out that the portfolios $\{\text{own a put, stock}\}$ and $\{D(r)K\text{ in risk-free bank, own a call}\}$ make the same profit at time $T$. Then by lemma 1 at all times $t<T$ they must be worth the same, so too for $t=0$.

Is there something wrong with this proof?

## Answer by Stijn D'hondt (score 4, accepted)

https://quant.stackexchange.com/a/50224

There are usually two ways to write proofs of equalities (like put-call parity) in quantitative finance.

- By replication,

- by constructing arbitrage.

Both of these are actually the same, since the first one is done by making, say, two portfolios, $A$ and $B$, and showing that they have the same outcome at time $t=T$. Then, by argument of LOOP (law of one price), one can argue that the two portfolios must be priced the same for all $t\leq T$.

But note that the LOOP is actually just a corollary of the no-arbitrage assumption. So the two methods of proof are just arguments by no-arbitrage assumption.

So, the proof posted in the question is correct, yet you might come across a proof in a textbook that argues directly from the no-arbitrage assumption in the following way:

- suppose $S_0 + P_0 > D(r)K + C_0$, and derive an arbitrage position and,

- suppose $S_0 + P_0 < D(r)K + C_0$, and derive an arbitrage position.

Conclude $S_0 + P_0 = D(r)K + C_0$.

Although the argument by LOOP is often shorter, sometimes it is preferable to argue directly from a no-arbitrage assumption.