# Option Pricing Range

In case of zero transaction fee, the option should have a fixed price determined from

``````c + K - S0 = 0
``````

where

• c = Call option price
• K = Strike price, and
• S0 = Current market price

If this is not satisfied i.e `|c + K - S0| > 0` then there is an arbitrage opportunity.

But from Hull's book, this doesn't seem to be true and the range should be

``````S0 - K <= c <= S0
``````

Assumptions -

1. Transaction fee is zero
2. Risk free real rate of interest is 0 (Just to simplify equation here, otherwise this can easily be factored by adding `e^-rT`)

Can anyone explain what I am missing and why would there be range for call price?

• Can you provide more context? Who is Hull, and what is the title of his book? Options are priced using a bit more complex model, it would be great to know where you encountered this. Commented Aug 27, 2021 at 11:19
• John Hull has written a number of option books. Commented Aug 27, 2021 at 15:23
• Most Notably "Options, Futures, and Other Derivatives" Commented Aug 27, 2021 at 15:37
• @JTP-ApologisetoMonica John C. Hull is the author of Options, Futures, and Other Derivatives and Fundamentals of Futures and Options Markets, which are introductory textbooks commonly used in university courses about options.
– Flux
Commented Aug 27, 2021 at 16:19

Your beginning premise (`c + K - S0 = 0`, or `c = S0 - K`) is wrong, even with no interest or transaction fees. This is the intrinsic value of a call option and is almost always the minimum price of an option. Options also contain some amount of "extrinsic" value (called "time value") to account for the downside protection afforded by the option.

So the rest of the question is meaningless until this premise is corrected.

• +1 - exactly what prompted my comment. The question itself seemed to be a quote taken completely out of context. Commented Aug 27, 2021 at 15:19
• c = Call option price
• K = Strike price, and
• S0 = Current market price

Either Hull or you wrote an incorrect answer. The correct equation for the discount arbitrage would be:

(c + K - S0) < 0

This would mean that the option trades for less than its intrinsic value which can occur with deep ITM options at/near expiration.

It's because the price of the underlying changes unpredictably over time, and the holder gets to choose whether to exercise it or not.

Suppose there's a stock with a price of \$20, and a call option on that stock with a strike price of \$20. Is the call option worthless? No, probably not, because the stock could increase in value before the option expires.