American Put Calendar Spreads and Early Exercise Risk

Summary

The document considers whether an American put calendar spread can have a negative price and examines an attempted no-arbitrage argument based on selling the shorter-maturity put and buying the longer-maturity put at the same strike. The key issue is assignment: the holder of the short-dated put decides whether to exercise, so the spread seller cannot assume that exercise occurs only in a particular spot-price scenario.

The response argues that if the short put is exercised, the short position must pay its intrinsic value, which can be funded by exercising the longer-dated put when appropriate. If the short put is exercised out of the money, the longer-dated holder can wait rather than exercise. This addresses the assignment concern raised in the question, but the short reply does not provide a full proof of calendar-spread price bounds or discuss other contract details, costs, or market frictions.

Key ideas

• The holder of the short-maturity American put controls whether it is exercised.
• A spread analysis must account for assignment rather than assume exercise occurs only at expiration.
• The response says the longer-dated put can fund the intrinsic-value payment when the short put is exercised in the money.
• When assignment occurs out of the money, the longer-dated option can be held for a later decision.
• The brief argument does not establish a complete pricing result under all market conditions.

Tags

# Can an American option calendar spread have a negative price?


# Can an American option calendar spread have a negative price?












Consider an American put calendar spread on strike K with maturities T1 < T2. Is the longer-dated put always more expensive than the shorter-dated one?

How valid is the following no-arbitrage argument?

Assume P(K,T1) > P(K,T2). Then one could sell the put expiring on T1 and buy the put expiring on T2. Then, on T1, if spot S > K, then the T1 put expires worthless and you’re left with long the T2 put. If S <= K, you exercise the T2 put early and get assigned on the T1 put and are left with no position. In all cases, you have made at least the difference in price.

Are there some considerations on assignment of the T1 put that make this argument invalid?

## Answer by Andrea (score 1)

https://quant.stackexchange.com/a/80553

I think you need to consider the (optimal) decision of the T1 buyer.

You are short T1 and long T2.

As soon as the T1 buyer exercises (you can't control that), you need to pay $(K - S)^+$, which you can get exercising your option at T2.

Obviously, if she exercises when her option is out of the money, you don't (and wait to make more money).

This way, you always have enough money to pay the T1.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.