Raising the bar The payoff for put debit spread is derived and illustrated

Apr 21

Put debit spread

Put debit spread

This post parallels the one about the call debit spread. A combination of several options in one trade is called a strategy. Here we discuss a strategy called a put debit spread. The word "debit" in this name means that a trader has to pay for it. The rule of thumb is that if it is a debit (you pay for a strategy), then it is less risky than if it is a credit (you are paid). Let $p(K)$ denote the price of the put with the strike $K,$ suppressing all other variables that influence the put price.

Assumption. The market values higher events of higher probability. This is true if investors are rational and the market correctly reconciles views of different investors.

We need the following property: if $K_{1}<K_{2}$ are two strike prices, then for the corresponding put prices (with the same expiration and underlying asset) one has $p(K_{1})<p(K_{2}).$

Proof. A put price is higher if the probability of it being in the money at expiration is higher. Let $S(T)$ be the stock price at expiration $T.$ Since $T$ is a moment in the future, $S(T)$ is a random variable. For a given strike $K,$ the put is said to be in the money at expiration if $S(T)<K.$ If $K_{1}<K_{2}$ and $S(T)<K_{1},$ then $S(T)<K_{2}.$ It follows that the set $\{ S(T)<K_{1}\}$ is a subset of the set $\{S(T)<K_{2}\} .$ Hence the probability of the event $\{S(T)<K_{2}\}$ is higher than that of the event $\{S(T)<K_{1}\}$ and $p(K_{2})>p(K_{1}).$

Put debit spread strategy. Select two strikes $K_{1}<K_{2},$ buy $p(K_{2})$ (take a long position) and sell $p(K_{1})$ (take a short position). You pay $p=p(K_{2})-p(K_{1})>0$ for this.

Our purpose is to derive the payoff for this strategy. We remember that if $S(T)\ge K,$ then the put $p(K)$ expires worthless.

Case $S(T)\ge K_{2}.$ In this case both options expire worthless and the payoff is the initial outlay: payoff $=-p.$

Case $K_{1}\leq S(T)<K_{2}.$ Exercising the put $p(K_{2})$ , in comparison with selling the stock at the market price you gain $K_{2}-S(T).$ The second option expires worthless. The payoff is: payoff $=K_{2}-S(T)-p.$

Case $S(T)<K_{1}.$ Both options are exercised. The gain from $p(K_{2})$ is, as above, $K_{2}-S(T).$ The holder of the long put $p(K_{1})$ sells you stock at price $K_{1}.$ Since your position is short, you have nothing to do but comply. The alternative would be to buy at the market price, so you lose $S(T)-K_{1}.$ The payoff is: payoff $=\left(K_{2}-S(T)\right) +\left( S(T)-K_{1}\right) -p=K_{2}-K_{1}-p.$

Summarizing, we get:

payoff $=\left\{\begin{array}{ll} -p, & K_2\le S(T) \\ K_{2}-S(T)-p, & K_{1}\leq S(T)<K_{2}\\ K_{2}-K_{1}-p, & S(T)<K_{1} \end{array}\right.$

Normally, the strikes are chosen so that $K_{2}-K_{1}>p.$ From the payoff expression we see then that the maximum profit is $K_{2}-K_{1}-p>0,$ the maximum loss is $-p$ and the breakeven stock price is $S(T)=K_{2}-p.$ This is illustrated in Figure 1, where the stock price at expiration is on the horizontal axis.

Figure 1. Payoff from put debit spread. Source: https://www.optionsbro.com/

Conclusion. For the strategy to be profitable, the price at expiration should satisfy $S(T)< K_{2}-p.$ Buying a put debit spread is appropriate when the price is expected to stay in that range.

In comparison with the long put position $p(K_{2}),$ taking at the same time the short call position $-p(K_{1})$ allows one to reduce the initial outlay. This is especially important when the stock volatility is high, resulting in a high put price. In the difference $p(K_{2})-p(K_{1})$ that volatility component partially cancels out.

Remark. There is an important issue of choosing the strikes. Let $S$ denote the stock price now. The payoff expression allows us to rank the next choices in the order of increasing risk: 1) $S<K_1<K_2$ (both options are in the money, less risk), 2) $K_1<S<K_2$ and 3) $K_1<K_2<S$ (both options are out of the money, highest risk). Also remember that a put debit spread is less expensive than buying $p(K_{2})$ and selling $p(K_{1})$ in two separate transactions.