EDIT: As pointed out by Gordon in the comments, the portfolio I considered in my original post is neither self-financing nor (locally) risk-free. Though the central question is still open. Suppose that we have a portfolio $P_t$ satisfying $$dP_t=a(W_t,t)dt+b(W_t,t)dW_t.$$ Then, apparently, the portfolio is called (locally) risk-free iff $b(W_t,t)$ vanishes. My question is why this definition makes sense. After all, the coefficient $a(W_t,t)$ might depend on the Wiener process, and thus on the path we are given.

This was the original post: I'm trying to understand the derivation of the Black-Scholes equation for an option by arbitrage considerations, and feel puzzled by the notion of a risk-free portfolio.

As usual, let the price of the underlying stock be given by the Ito-process $$dS_t=\mu S_t dt+\sigma S_t dW_t,$$ and let $V$ denote the price of the option. We then consider the portfolio $$P_t=V_t+\Delta S_t.$$ If we assume that this portfolio is self-financing it satisfies $$dP_t=\left(\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^2 S_t^2 \frac{\partial^2 V}{\partial S^2}\right)dt+\left(\frac{\partial V}{\partial S}+\Delta\right)dS.$$ Then it is claimed that the choice $\Delta=-\frac{\partial V}{\partial S}$ makes the portfolio risk-free because the $dS$ term vanishes. Hence it must grow at the risk-free rate $$dP_t=rP_tdt.~(1)$$

I have some problem understanding why such a portfolio is considered to be risk-free. After all in the equivalent integral equation $$P_t-P_0=\int_0^t a(W_s,s)ds,$$ where $a$ is the first bracket, we still integrate over the paths of the Wiener process (or some function thereof). So it still depends on the path we are given.

I came up with the following (heuristic) idea: since the paths of the Wiener process are continuous (almost surely), on a very small interval $[t_0-\epsilon,t_0+\epsilon]$ the term $a$ is bounded $$a_0-k\leq a(W_t(\omega),t)\leq a_0+k.$$ By monotony of the integral the growth of $P_t$ is thus approximately linear in $t$ with rate between $a_0-k$ and $a_0+k$ (where $k$ may be arbitrarily small). Hence at this small time scale the portfolio should be approximately risk-free, and taking the limit should give equation $(1)$. Such an argument does not seem to work for a potential $dS$ term because integration with respect to the Wiener process lacks monotony.

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    $\begingroup$ But your integral $dPt=rP_td_t$ is deterministic, there is no Wiener process there. That's why it's risk-free. $\endgroup$ – dbluesk Feb 21 '17 at 12:07
  • $\begingroup$ The portfolio $P_t=V_t+\Delta S_t$ is neither self-financing nor (locally) risk free; see discussion here. $\endgroup$ – Gordon Feb 21 '17 at 13:46
  • $\begingroup$ @Gordon - While I agree with $P_t$ not being self-financing, it surprises me that you say it is not locally risk-free? $\endgroup$ – LocalVolatility Feb 21 '17 at 14:10
  • $\begingroup$ @dbluesk: I agree that this equation is deterministic. But if I understand correctly it should be a consequence of the riskfreeness (and the no arbitrage assumption). $\endgroup$ – Handschuh Feb 21 '17 at 14:18
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    $\begingroup$ @LocalVolatility and Hanschuh, this is a common mis-understanding even in John Hull's book. For a vanilla European call option $V_t + \Delta S_t= Ke^{-r(T-t)}N(d_2)$, which is not risk-free.Yes, locally risk free implies that the dW terms goes to zero. $\endgroup$ – Gordon Feb 21 '17 at 14:23

Regarding the self-financing and locally risk-free issues of the portfolio $P_t= V_t +\Delta S_t$, see this question and also discussions on Page 100 of the book Mathematical Methods for Financial Markets by Jeanblanc, Yor, and Chesney.

Note that a portfolio is locally risk-free means that it earns the risk free rate $r$ over the infinitesimal interval $[t, t+dt]$ (see the bottom part on Page 99 of the above book). That is, \begin{align*} dP_t = rP_t dt. \end{align*} In other words, \begin{align*} a(W_{t}, t) dt + b(W_{t}, t)dW_t = rP_t dt. \end{align*} Consequently, $b(W_{t}, t)=0$ and \begin{align*} a(W_{t}, t) = rP_t. \end{align*}

  • $\begingroup$ This is the kind of answer I was looking for, thanks! (Again, I can not upvote your answer because I lack reputation.) $\endgroup$ – Handschuh Feb 21 '17 at 22:40

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