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In the Black and Scholes model, when it is needed to immunize the portfolio from variations in the stock the argument given is the following. If $\alpha_t$ is the amount of invested in the stock, $\beta_t$ the amount in the bond, we construct a portfolio whose value is

$$ V_t = -C_t+\alpha_t\,S_t+\beta_t\,B_t, $$

where $C_t=f\left(t,S_t\right)$ is the price of a Call option and where $S_t$ and $B_t$ are, respectively, the price of the stock and the price of the bond at time $t$. So we are short selling one unit of the call and buying the portfolio whose $\alpha_t\,S_t+\beta_t\,B_t$. Now one chooses $x$ such that

$$ \frac{\partial V_t}{\partial S_t} = 0\Leftrightarrow -\frac{\partial C_t}{\partial S_t}+\alpha_t=0\Leftrightarrow \alpha_t = \frac{\partial C_t}{\partial S_t}\equiv \Delta_t. $$

What puzzles me is the fact that when deriving the value of the portfolio we assume that $\alpha_t$ does not depend from $S_t$ whereas the final solution does depend. So in principle one should do this computation (assuming that $\beta_t$ does not depend on $S_t$)

$$ \frac{\partial V_t}{\partial S_t} = -\frac{\partial C_t}{\partial S_t}+\alpha_t+\frac{\partial \alpha_t}{\partial S_t}\,S_t = 0 $$

whose solution is of course different from $\alpha_t=\Delta_t$. Where am I wrong?

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$\alpha_t$ must be chosen prior to stock price movements so the expression $S_t d\alpha $ does not make sense: we can't take a position in a stock based off information that we don't know yet.

The missing step is that the replicating portfolio is required to be self financing: that is, for all $t$ the following equations hold: $$X_t=\Delta S_t+\Gamma M_t$$ $$dX=\Delta dS+\Gamma dM$$ Where $X$ is the portfolio value and $S$ and $M$ are the stock and riskless asset. The first equation states that no external asset is injected or removed at any time. The second states that we cannot take a position in an asset based off information that is not available at time $t$ (since naively applying Ito's lemma to $X_t$ would yield a $d\Delta$ and $d\Gamma$ term).

Combining the two equations yields $$dX=\Delta dS +r(X_t-\Delta S_t)dt$$

Matching this equation with $df(S, t)$ yields the correct PDE.

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