You buy both a European put option which costs pt and a European call option which costs ct at t with strike K on the same underlying asset with the same expiration at T. Do you make an overall profit or loss if STK+(pt+ct)er(T−t)?

Question

You buy both a European put option which costs pt and a European call option which costs ct at t with strike K on the same underlying asset with the same expiration at T. Do you make an overall profit or loss if ST>K+(pt+ct)er(T−t)?

BlackTom AI · Accepted Answer

First, restating the problem helps ensure we’re analyzing the exact setup: you purchase a European put with price pt and a European call with price ct, both on the same underlying asset, same strike K, and same expiration T. You want to know whether you make an overall profit or loss given the price condition ST > K + (pt+ct) e^{r(T−t)} at time t. Since you hold both a put and a call with the same strike, you effectively hold a long straddle. The payoff at expiration is: payoff = max(K − ST, 0) + max(ST − K, 0) = |ST − K|. However, you paid pt for the put and ct for the call at time t, so the total outlay at time t is Pt = pt + ct. To assess profit, we should compare the discounted payoff at T to the initial cost. The present value of the payoff is e^{−r(T−t)} |ST − K|. A profit occurs if this discounted payoff exceeds the initial cost, i.e. e^{−r(T−t)} |ST − K| > pt + ct. Rearranging the inequality gives |ST − K| > (pt + ct) e^{r(T−t)}. This is the same as two cases depending on ST relative to K: - If ST > K, then |ST − K| = ST − K, and the condition becomes ST − K > (pt + ct) e^{r(T−t)}, or equivalently ST > K + (pt + ct) e^{r(T−t)}. - If ST < K, then |ST − K| = K − ST, and the condition would be K − ST > (pt + ct) e^{r(T−t)}, which would imply ST < K − (pt + ct) e^{r(T−t)}. This is a different region, but your given condition ST > K + (pt + ct) e^{r(T−t)} specifically concerns the case ST above the strike by a margin that exceeds the discounted total premium. Thus, under the scenario where ST indeed ends up greater than K by more than (pt + ct) e^{r(T−t)}, the inequality is satisfied and the present value of the payoff exceeds the total premium paid, yielding a net profit. If we consider the other potential outcomes for ST (e.g., ST ≤ K or ST moderately above K but not enough to satisfy the inequality), the inequality would not hold, and the strategy would not yield a profit; the payoff could be zero (if ST = K) or less than the total initial outlay when discounted, leading to a loss or a break-even scenario depending on the exact numbers. Key subtle point: the comparison uses the discounted payoff e^{−r(T−t)}|ST − K| against the undiscounted total premium pt + ct. That’s why the threshold involves e^{r(T−t)} on the right-hand side when you isolate the condition in terms of ST. In summary, the given condition ST > K + (pt + ct) e^{r(T−t)} corresponds to the long straddle producing a profitable outcome after accounting for the time value of money, for the branch ST > K. If the stock ends above K by more than that discounted premium amount, you have a positive profit; otherwise, you do not. Note that alternative ST regions (ST < K) would require a different inequality to determine profitability.

You buy both a European put option which costs pt and a European call option which costs ct at t with strike K on the same underlying asset with the same expiration at T. Do you make an overall profit or loss if ST>K+(pt+ct)er(T−t)?单项选择题

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