enter image description here"Final stock" means the stock price at expiration, and "strike" means strike price. If a call option had to be purchased for more than the max(final stock - strike,0)then you would never make money off buying the call option right? I found the no-arbitrage assumption in Merton's Theory of Rational Option Pricing.

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    $\begingroup$ I do not think that Merton wrote nonsense like that. Since we do not know today what "Final stock" $S_T$ at expiration will be the question how the call price relates to this is pointless. A more reasonable question is how the call price relates to the discounted expectation of that payoff $\max(S_T-K,0)$ under the risk-neutral measure. $\endgroup$
    – Kurt G.
    Aug 25 at 20:11
  • $\begingroup$ Well I've attached the screenshots from what I read and I've found the same assumption in other sources so I think it's valid. $\endgroup$ Aug 25 at 21:15
  • $\begingroup$ first screenshot tells you a warrant is just a call option, second shows this is from Merton,and shows the assumption I mentioned in my question $\endgroup$ Aug 25 at 21:19
  • $\begingroup$ This screen shot is from a 1973 paper doi.org/10.2307/3003143 . $\endgroup$ Aug 25 at 21:22

1 Answer 1


Merton writes

Further it follows from conditions of arbitrage that $$\tag{3}F(S,\tau;E)\ge{\rm Max}(0,S-E)\,.$$ In general , a relation like (3) need not hold for a European warrant.

This is the well-known relationship between continuation value $F(S,\tau;E)$ of the American option and the payoff you get from it when you exercise at time $t=T-\tau\,.$ In (3), $S$ is not the final stock but the price of the stock at $t\,.$

Now we have a decent QSE question we can ask: Why does (3) hold?

Hint: Will you as the holder of the American option be rational when you exercise at $t$ in the case that the payoff you get is less than the value of the option you could keep by not exercising?

As a further exercise highly recommend to try to find the simple proof that when the stock does not pay dividends and interest rates and volatility are strictly greater than zero then (3) always holds strictly. In other words: We never exercise that American call before maturity.


First, the American call is worth more than the European call $G(S,\tau;E)$ and that is strictly worth more than its intrinsic value ${\rm Max}(0,S-\color{red}{e^{-r\tau}}E)\,.$ That intrinsic value is strictly worth more than ${\rm Max}(0,S-E)\,.$

Finally, it is instructive to think about the question why this does not hold for the American put.

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    $\begingroup$ The OP should further ponder how everything flips when the interest rates turn negative. $\endgroup$ Aug 26 at 11:42
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    $\begingroup$ @DimitriVulis +1 which wasn't an issue in the days when Merton wrote that paper. Things change sometimes. $\endgroup$
    – Kurt G.
    Aug 26 at 11:55
  • $\begingroup$ inflation-adjusted ("real") interest rates were often very negative in the 1970s :) chicagofed.org/publications/economic-perspectives/2017/… $\endgroup$ Aug 26 at 19:49
  • $\begingroup$ This could be a further exercise question: Is the riskless rate in option pricing a real rate or a nominal rate? $\endgroup$
    – Kurt G.
    Aug 26 at 19:54
  • $\begingroup$ I guess, the real interest rate if the strike and all prices were inflation-adjusted. $\endgroup$ Aug 26 at 20:00

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