Exercise: Where does $\mu$ go?

The Black-Scholes PDE

V_t + \tfrac12\sigma^2 S^2 V_{SS} + rSV_S - rV = 0

contains no

\mu

. Yet Itô's expansion of

dV

does contain a

\mu

term. The hedging construction is supposed to eliminate risk; how does it also happen to eliminate

\mu

Tasks

Write out $dV$ and $d(V - \Delta S)$ step by step, identifying all $\mu$ terms. Verify that choosing $\Delta = V_S$ cancels not only the $dW$ term but also the $\mu$ in the $dt$ term.
Explain in one sentence why it's not coincidence that the same hedge ratio cancels both terms simultaneously.
Consider an option whose payoff depends on two correlated assets: $dS_1 = \mu_1 S_1 dt + \sigma_1 S_1 dW_1$ and $dS_2 = \mu_2 S_2 dt + \sigma_2 S_2 dW_2$ with $dW_1\,dW_2 = \rho\,dt$ . Derive the hedging argument and the resulting 2D PDE. Does $\mu_1$ or $\mu_2$ appear?
Suppose instead the asset jumps: $dS = \mu S\,dt + \sigma S\,dW + S\,dJ$ where $dJ$ is a compound Poisson jump. Show that $\Delta = V_S$ no longer creates a riskless portfolio — the jump risk is not hedgeable by trading only $S$ .
What does the non-uniqueness of jump-option pricing tell you about the role of the hedging assumption in Black-Scholes?

For part 3, hedge with both $S_1$ and $S_2$ : $\Pi = V - \Delta_1 S_1 - \Delta_2 S_2$ , choose $\Delta_i = V_{S_i}$ .