Put–Call Parity Arbitrage and Interpreting Terminal Cash Flows
Summary
The document explains a sign confusion in a put–call parity arbitrage argument. A trader considers a position involving a call, a put, and the underlying stock, and asks whether the terminal expression represents a debt or a gain when the call is exercised. The answer distinguishes the borrowed amount, accrued to maturity, from the strike payment received when the option holder buys the stock.
The argument shows why the strike is a cash inflow in the exercised-call case, even though the call’s payoff is described as a loss from another perspective. It also notes that the strategy can produce a gain in the alternative terminal-price case when the stated pricing inequality holds. This is a conceptual explanation of arbitrage logic under put–call parity; it does not discuss frictions such as trading costs, funding constraints, or whether mispricing persists in practice.
Key ideas
- In the exercised-call case, the strike is cash received from the option holder for the stock.
- The financing amount and the strike proceeds must be assigned consistent signs in the terminal cash-flow calculation.
- A parity violation can imply an arbitrage when the specified call, put, stock, and discounted-strike relation holds.
- Practical arbitrage may be affected by trading costs and funding constraints, which the explanation does not analyze.
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Full text
# put-call parity equation
# put-call parity equation
I'm reading this book and I'm looking at page 4, and we are considering the case where $C_t - P_t - S_t$ is negative, which means that selling the call did not offset the cost of the stock and the put together. So, case 1), $S_T > K$, which means that the buyer of the call will exercise their option, so we will also have to give the buyer $K$ for the stock at time of maturity.
So, $C_t - P_t - S_t$ is money that we had to borrow in order to buy the cost of the stock and the put that was not offset by the money that we received by selling the call. So the interest at maturity of that money that we owe is $(C_t - P_t - S_t)e^{r(T-t)}$. To that money that we owe, we add the money that we owe to the contract buyer, since we are in case 1) where the strike is larger than the call price. So
$$(C_t - P_t - S_t)e^{r(T-t)} + K < 0$$
Is money that we owe. But on the reference, they put the opposite sign; $$(C_t - P_t - S_t)e^{r(T-t)} + K > 0$$ like it was profit!
Is this a mistake, or am I misunderstanding something?
## Answer by dosdel (score 1)
https://quant.stackexchange.com/a/9282
I think this is where your logic goes wrong:
$(C_t − P_t − S_t)e^{r(T−t)} + K$
With reference to the above equation, you are saying that "...To that money that we owe, we add the money that we owe to the contract buyer.."
Yes, $(C_t − P_t − S_t)e^{ r(T−t)}$ is the money that we owe, but $K$ is not referring to money that we also owe the contract buyer. $K$ is the strike price, so it is the money we receive from the contract buyer at maturity when he exercises the option. Yes, the call option is in a losing position, but $K$ is not referring to the actual loss.
So, the equation is really saying that the money we borrowed to finance this strategy (plus the accrued interest) is less than the amount we ended up receiving when the contract buyer exercised his option and bought the stock from us at the strike price. A net gain will also result if $S_t < K$ as well, which demonstrates that an arbitrage opportunity exists when $C_t - P_t > S_t - Ke^{-r(T-t)}$.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.