Hi David,

I read from the FRM handbook, page 128 that “A long position in an asset is equivalent to a long position in a European call with a short position in an otherwise identical put, combined with a risk-free position.” I understand that the combination of a long call and a short put on the same asset would give the same payoff as holding the asset, and for me that completes the picture. I could not get it when we say that the combination has to be “combined with a risk-free position”.  What does it mean?

Thanks so much.

Hi Chinquee,

But you aren’t quite the same with the long call + short put. But this is important, this is put call parity.

Say stock today is $100 and buy the call and write the put (both with strike of $100). Then over time (end of period) stock increases to $120.

In future, holding stock worth $120
In future, call + put only worth $20 (payoff on option)

You are the same if TODAY you:
* On the one hand, buy call and write put: c - p
* On the other hand, buy the stock mostly with borrowed cash. How much should you borrow? Borrow the future strike price discounted to today. So, buy the stock and borrow the future strike price: S - (K)*EXP(-rT)

Notice without borrowing cash--i.e., this is the risk-free position you refer to. It is to be short cash or short a risk free bond with face value equal to strike price--to buy the stock, your first scenario is very cheap (almost free) to employ: proceeds from writing the put are used to buy the call. Unless you borrow (the risk free position), your second scenario is expensive. You have to buy the whole stock.

Now in the future you will be in the same place. If stock grows to $120,
* c-p pays $20
* Stock + riskfee position pays $120 for the stock but minus $100 to repay the loan = $20

So, that’s put call parity:
c - p = S - K*EXP(-rT) where the last term is the borrowed strike price

IMO, little formula leads many useful places; e.g., S - K* EXP(-rT) is the lower bound on the call option value and, you’ll note, all you need to do is “wrap in” the N()s to get the Black-Scholes.

David