Consider portfolio in black-scholes world

$\Pi = \Delta S - V$, where $S$ is the stock price and V is the price of the option.

I have read that if we set $\Delta = \frac{\partial V}{\partial S} $ then we obtain $d\Pi = (...)dt + 0 * dW$, where $W$ is brownian motion. And by no-arbitrage we have $d\Pi = r \Pi dt$, where is risk-free interest rate, so that $\Pi_T = (\Delta_0S_0 - V)\exp(rT)$.

I came across some lecture notes, that claim that if $\Pi = \Delta S - V$ is $\Delta$-hedged then value of such portfolio is $0$ at time of expiration of the option $T$.

But I would be expecting such a portfolio to have a value of $\Pi_T = (\Delta_0S_0 - V)\exp(rT)$, could someone help to figure out what is going on?

Thank you


2 Answers 2


It should indeed grow at the risk free rate, as explained in the Black Scholes paper (excerpt below):

enter image description here

Would be worth knowing the context around the presentation in the lecture notes.

Note though the delta is assumed to be continuously rebalanced (dynamic hedging), so think of it as very local approximation, and there would be many rebalancing between 0 and T.

PS: Black Scholes hedging portfolio is different than how it is normally presented in the textbooks as the delta is the other way around. Also note some controversy around Black Scholes arguments (see for example: the hypothesis underlying the pricing of options by Bartlets, and FAQs in Option pricing theory by Peter Carr).


In the Black-Scholes' setting, as we discussed in this question, the portfolio $\Pi = \Delta S -V$, where $V= \frac{\partial V}{\partial S}= N(d_1)$, is not self-financing. Moreover, \begin{align*} \Pi = \Delta S -V=Ke^{-r(T-t)}N(d_2) \end{align*} does not satisfy the equation \begin{align*} d\Pi = r\Pi dt. \end{align*} In fact, let \begin{align*} \Delta_t^1 = \frac{\frac{\partial V}{\partial S} e^{rt}}{V_t - \frac{\partial V} {\partial S}S},\quad \Delta_t^2 =\frac{-e^{rt}}{V_t - \frac{\partial V}{\partial S}S}. \end{align*} Then, it can be checked that the portfolio \begin{align*} \Pi_t = \Delta_t^1 S + \Delta_t^2 V = e^{rt} \end{align*} is self-financing, and \begin{align*} d\Pi = r\Pi dt. \end{align*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.