Why a Black-Scholes Hedge Portfolio Tracks a Call’s Value
Summary
The note clarifies the relation between a European call’s value and a replicating portfolio in the Black-Scholes framework. A trader who sells the call can hedge it by holding a dynamically adjusted quantity of stock, with the remainder invested in the money market account. The stock position is set by the call’s sensitivity to the underlying price, while the cash position makes the portfolio’s current value equal to the call value.
This equality through time follows from the self-financing replication strategy under the model assumptions, rather than from choosing arbitrary intermediate portfolio values. At maturity, the hedge matches the call payoff; before maturity, rebalancing maintains replication. The excerpt gives the hedge construction but does not explain the full no-arbitrage proof or discuss frictions, model error, or constraints on continuous trading, all of which matter when applying the result outside the idealized setting.
Key ideas
- A Black-Scholes call hedge combines stock and a money market account.
- The stock holding is determined by the call’s sensitivity to the underlying price.
- The cash holding is chosen so the hedge portfolio’s value matches the call at each time.
- The replication argument assumes idealized conditions and does not address trading frictions or model error.
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Full text
# Why is call option value same as portfolio value at all times in Black Scholes model?
# Why is call option value same as portfolio value at all times in Black Scholes model?
Following is a part of the text from Steven Shreve Stochastic Calculus for Finance II, for pricing the European Option 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 $t$?
## Answer by Kurt G. (score 1)
https://quant.stackexchange.com/a/68285
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.