# Expected value of delta-hedged portfolio

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

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*}