I am editing this question because it was originally unclear, and I didn't get the answers I was hoping for.

In my finance book I have the following question

T-bills currently yield 5.5 percent. Stock in Maria's Manufacturing is currently selling for $70 per share.

There is no possibility that the stock will be worth less than $65/share in one year.

What is the value of a call option with a $60 exercise price? The answer book is: C0=70–[60/1.055] = 13.13. I don't understand how the option price was found like this. We don't know the volatility for the stock, so how is it possible to calculate the option price. Does it have something to do with the fact that the option will definitely be exercised?

Thanks a lot for your help.

  • $\begingroup$ Please don't make a big mess of your question. This was horrible to look at. Also please clarify what you don't understand. The key insight here seems to be that the option will always be exercised. $\endgroup$ – Bob Jansen Apr 5 '15 at 10:35
  • $\begingroup$ I don't understand ow we can price it this way. What if the price suddenly goes up to $100 in a year. Don't we have to account for this type of variability? $\endgroup$ – bugsyb Apr 6 '15 at 6:11
  • $\begingroup$ As @studentT said: please make an account and don't add comments as answers. Also, if the call option is exercised with certainty (the minimum stock price is above the strike) it should have the same value as the stock minus the strike. $\endgroup$ – Bob Jansen Apr 6 '15 at 15:40
  • $\begingroup$ sorry about adding answers, I don't have enough reputation to add comments yet. I'm still confused as to why the equation is 70–[60/1.055]. Also we learned Black Scholes, and I was wondering how this option method compares to black scholes. Is this simply an inaccurate guess? Or is there some merit to this method? Thanks, and sorry for the badly formatted questions, I'm new on here, and I'll try to improve. $\endgroup$ – bugsyb Apr 6 '15 at 16:14

If you know the stock will finish above the strike, then the call option becomes a forward contract since it will always be exercised. We therefore price it as a forward. Its value at maturity is $$S_T - 60.$$ We can synthesize $S_T$ with one unit of stock costing $70.$ We can synthesize $60$ with $60$ ZC bonds which costs $60/1.055$ since the yield is $5.5\%.$ So the value is $$ 70 - 60/1.055. $$


The intrinsic value is $70 - $60.

However, we don't know exactly what the stock price will end up in a one-year time. But we know that it it is the best estimate for the future price in one-year.

Profit in one-year = ($70 * 1/D - $60) where D is the discount factor.

This profit needs to be discounted: $70 - $60 * D.

You should be able to relate the discount factor to your example yourself.

Note: I don't think Black-Scholes is an appropriate model for this case. The stock is not a GBM.

  • $\begingroup$ The discounted stock price WILL be a martingale in the risk neutral measure; martingales can be bounded, no problem. $\endgroup$ – q.t.f. May 7 '15 at 10:56
  • $\begingroup$ I'm not very sure that the process described is a martingale... Can you elaborate why you think it'll be a martingale? $\endgroup$ – SmallChess May 9 '15 at 0:43
  • $\begingroup$ The fundamental theorem of asset pricing says (basically) "no arbitrage iff spot prices are martingales in the pricing measure." See: en.wikipedia.org/wiki/No_free_lunch_with_vanishing_risk. Really, "no arbitrage" in continuous time becomes "no free lunch with vanishing risk", and there is some trickiness about local martingale or sigma-martingale possibilities instead of true martingales. But fundamentally, it has to have a martingale property or else there is arbitrage. $\endgroup$ – q.t.f. May 11 '15 at 12:54
  • $\begingroup$ I think you're right about the martingale part. I edited my answer. $\endgroup$ – SmallChess May 18 '15 at 1:50

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