Let's say you are long a call and want to replicate that call buy being short underlying and long bonds.

If the underlying moves up in the next period but not enough to cover theta, the option loses money. Is this loss offset by your long bond position?

Now let's say the underlying does not move in the next period. The option loses money. Does the BSM model stop working in this situation?


  • $\begingroup$ The model does not stop working because the stock doesn’t move. You simply lose money because the option was mispriced in the first place. $\endgroup$
    – Ivan
    Jul 11, 2019 at 20:05
  • $\begingroup$ How does that fit with the no-arbitrage derivation? If we are long call, short underlying, long bonds, the portfolio should grow at risk free rate. I'm probably confusing two concepts. Thanks $\endgroup$
    – confused
    Jul 11, 2019 at 20:25
  • 1
    $\begingroup$ The BSM model by definition assumes that the volatility is known and constant, i.e. that the underlying path will move exactly according to the specified variation. If it doesn't, in a sense the BSM model 'breaks', although 'break' is too strong a word. $\endgroup$
    – user34971
    Jul 12, 2019 at 6:04
  • $\begingroup$ @ilovevolatility thanks. the future volatility is "known" is the assumption I forgot about. $\endgroup$
    – confused
    Jul 12, 2019 at 16:02
  • $\begingroup$ You;re welcome. And if the volatility (and hence the model) is mis-specified you will have gamma/theta, vega, vanna, volga p/l, but that's a slightly different topic. $\endgroup$
    – user34971
    Jul 13, 2019 at 15:24

1 Answer 1


You're replicating the value of the long call. As with your small move example, other things equal, if the underlying doesn't move, your position loses money (as an actual call would) due to theta. It works exactly as it should, the position itself just loses money as a result.


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