Why Black–Scholes Partial Time Derivatives Hold Spot Price Fixed
Summary
The document explains why differentiating a Black–Scholes option price with respect to time does not include a derivative of the stock price. The option pricing formula treats time and spot price as separate arguments: a partial derivative with respect to time holds the current spot value fixed. This formal derivative is used in the Black–Scholes framework, where the option value is a function of state variables and time.
The answer distinguishes this partial derivative from the full change in option value as the underlying evolves. Since the stock price is a stochastic process with paths that are generally not differentiable in the ordinary sense, its joint evolution with the option price is described using Itô’s lemma. A simple example shows that partial derivatives depend on the chosen representation of a function and on which variables are held fixed. The explanation is conceptual; it does not derive the Black–Scholes equation or discuss boundary conditions and other model assumptions.
Key ideas
- A partial time derivative treats the spot price as a separate state variable held fixed.
- The stock price evolves randomly, so its path does not have an ordinary time derivative.
- Itô’s lemma captures the full change when both time and the underlying process evolve.
- The intended use of a derivative determines which variables are held fixed.
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Full text
# The partial derivative of a call option with respect to $t$
# The partial derivative of a call option with respect to $t$
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?
## Answer by Jesper Tidblom (score 1)
https://quant.stackexchange.com/a/78499
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 $t$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.