Following is a part of the text from Steven Shreve Stochastic Calculus for Finance II, for pricing the European Option in Black Scholes model.

European Option Pricing in Black-Scholes model

The argument is that today I start by selling a European call option at price $c(0,S(0))$ and build a portfolio valued at $X(0) = c(0,S(0))$ by investing in stocks/bonds.

Now if at all arbitrage opportunity arises, it would only occur at time $T$ at maturity of this option, when it can be exercised. And therefore, for no-arbitrage I would require $X(T) = c(T,S(T)) = (S(T) - K)^+$

Now I do not understand the argument for $c(t,S(t)) = X(t)$ for all $0 < t < T$.

I know $X(t)$ for all times $t$ because this is the portfolio I have constructed using stocks/bonds.

But what exactly is $c(t,S(t))$ at some time $t$? At some middle time $t$, is the call option value implicit or something I need to decide? Meaning, is there an arbitrage opportunity if I had instead set $c(t,S(t)) = \frac{t}{T}X(t) + (1 - \frac{t}{T})X(0)$ or anything else like that?

So what am I missing here? How do I understand why $X(t) = c(t,S(t))$ for all $t$?

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    $\begingroup$ If the portfolios disagree at any time $t$, long the cheaper one and short the expensive one. Because their values will converge at expiry, there will be a guaranteed arbitrage whose profit is the spread between the portfolios at the earlier time $t$ $\endgroup$ Oct 7, 2021 at 16:54
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    $\begingroup$ This is essentially shifting the time scale from $0$ to $T$ to $t$ to $T$, which is equivalent to $0$ to $\tilde{T} = T-t$. $\endgroup$ Oct 7, 2021 at 16:55
  • $\begingroup$ Ohh! I was under the impression that in the model, our assumption was that you could (riskfree) invest/borrow any amount, buy/short any number of stocks BUT there was only 1 call option, which we have shorted at t=0. At least that's how it all worked out in the binomial model. Now if I can buy/sell more call options at any time, then why is that not considered as a part of my portfolio used for hedging my initial short position in the call option? $\endgroup$
    – kishlaya
    Oct 8, 2021 at 9:52
  • $\begingroup$ No, you adjust the hedge by buying/selling stock and repaying/increasing the loan, you do not buy and sell options. That's not how dynamic hedging works. $\endgroup$
    – nbbo2
    Oct 8, 2021 at 10:02
  • $\begingroup$ @noob2 Yes, I agree. But I am not able to see an apriori reason for that. Why not use options and bonds only for hedging? Or all of stocks/bonds/options perhaps? $\endgroup$
    – kishlaya
    Oct 8, 2021 at 10:44

1 Answer 1


It is fairly standard to hedge a sold option as follows:

  • at any time $t$ buy $\alpha(t)=\frac{\partial}{\partial S}c(t,S(t))$ amounts of stock $S(t)\,,$ and invest

  • $\beta(t)=\frac{c(t,S(t))-\alpha(t)S(t)}{B(t)}$ into the money market account $B(t)=e^{rt}$

By definition, the hedge portfolio $X(t)=\alpha(t)S(t)+\beta(t)B(t)$ exactly matches the option value $c(t,S(t))$ at all times.

  • $\begingroup$ Comments from @rubikscube09 helped me understand that it is possible to buy/sell options at any time $t$ at value $c(t,S(t))$. So when you build a hedging portfolio, is there any apriori reason to not consider buying/selling more options at time $t$, but only use stocks/bonds? $\endgroup$
    – kishlaya
    Oct 8, 2021 at 10:29

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