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In Black-Scholes related computations, why do we not treat the stock price $S$ as a function of $t$ when taking partial derivatives with respect to $t$? For example, if $$c(t,T)=SN(d_1)-Ke^{-r(T-t)}N(d_2)$$ is the price of a call option and we want to find $\partial C/\partial t$, we never include the term $\partial S/\partial t$ and don't consider $S$ as a function of $t$, but as a separate variable. How can this be justified?

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    $\begingroup$ S is just the current observed spot price. $\endgroup$
    – user70573
    Feb 28 at 3:57
  • $\begingroup$ So S doesn't change over time? $\endgroup$
    – user81883
    Feb 28 at 4:01
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    $\begingroup$ What's the partial derivative of x*y with respect to y? By definition, a partial derivative is holding all else equal. $\endgroup$
    – user70573
    Feb 28 at 4:07
  • $\begingroup$ I get your point; however, S = S(t), so how can we ignore its partial derivative. What if, in your example, x is a function of y? x(y)*y $\endgroup$
    – user81883
    Feb 28 at 4:08
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    $\begingroup$ It's not a function x(y) though, it's just S. What about IV? Surely S is dependent on VOL? Yet, it's not in the closed form solution, which looks at current S. That's why S is a constant in computing theta or vega. You also hold S constant when you compute a bump and reprice theta. $\endgroup$
    – AKdemy
    Feb 28 at 4:15

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If you want the full dependency on the time t of a process, as the process moves forward, you have to calculate and study the full Ito derivative.

But regarding your question : It depends on what you mean by "justified" and how you are going to use the partial derivative. Yes, $S(t)$ depends on t. However, the process $S(t)$ here is random and almost nowhere differentiable so taking the derivative would not make much sense in the first place.

The reason we don't take the time derivative of $S(t)$ here though is that we consider formal partial derivatives and not full derivatives. As a simple example, let us say $S(t) = t^2$ and we consider the function $$z(t) = S(t) \cdot t = S \cdot t$$ We now want to calculate $\partial{z} / \partial{t}$. What would you say this is? It actually depends on how you represent $z(t)$. If you use the above representation you get $\partial{z} / \partial{t} = S = S(t) = t^2$, but if you represent $z(t)$ as $$ z(t) = t^3 $$ you instead get $\partial{z} / \partial{t} = 3 \cdot t^2.$ When calculating the formal partial derivatives, you see the variables as placeholders. There is no contradiction here. And as I mentioned in the beginning, the justification depends on how you are using those partial derivatives later.

So in the Black-Scholes formula above, we see the value of $S = S(t)$ as such a placeholder for the value of the process $S(t)$ at a fixed $t$.

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