In a nutshell, if S is dependent on 't', why treat it as a constant when calculating the partial derivative $\frac{dC}{dt}$?

The equation for $\frac{dC}{dt}$ in a European call option is: $\frac{SN'(d_1)\sigma}{2\sqrt{T-t}}-rKe^{-r(T-t)}N(d_2)$. This is calculated by taking the derivative of a European call option with respect to 't'. However, S is treated as a constant here! Isn't S a function of t as well? $S = e^{(\mu-\frac{\sigma^2}{2})t+\sigma W_t}$ I understand that there is a white noise component in the equation, but $\frac{dS}{dt} = (\mu-\frac{\sigma^2}{2})e^{(\mu-\frac{\sigma^2}{2})t+\sigma W_t} = (\mu-\frac{\sigma^2}{2})S $, so I don't see why that would matter.

Could someone please enlighten me on this?

  • 1
    $\begingroup$ Hi, Theta is defined as a partial derivative, i.e. $\frac{\partial C}{\partial t}$. To this end, we do not consider higher order effects of $t$, e.g. thru $S_t$. The full PDE is of $C$, $dC$, is the usual Black-Scholes PDE that includes $dt$ and $dS$ alike. $\endgroup$ Nov 3, 2022 at 7:43
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    $\begingroup$ Hi Benedict (: I doubt the question will be reopened, but I grant to you that it is a common question that many students ask! The confusion stems from a constant abuse of notation, either by people who know what they do (which is then okay but still confuses students) or by people who don't really know what they're doing. The value of a call is $C=V(t,S_t)$, where $V(t,x)$ is a standard function of two real inputs. You can calculate $V_x(t,x)=\frac{\partial V}{\partial x}$ as in any calculus class. The option delta is then $\Delta=V_x(t,S_t)$. Similar confusion applies to proving Itô's Lemma. $\endgroup$
    – Kevin
    Nov 3, 2022 at 16:29
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    $\begingroup$ I think the comments here are to valuable to remain comments and it would be better if they are answers so I reopened $\endgroup$
    – Bob Jansen
    Nov 4, 2022 at 6:28

1 Answer 1


Let me heed @Bob's suggestion and turn my comment into a full answer:

Like other disciplines, finance uses lots of shortcuts to achieve brevity and convenience. That can be awfully confusing for students. For example, we write $\text{d}S_t$ instead of integrals, just because it's shorter. Formally, $\text{d}S_t$ has no meaning. It's just a symbol that people know how to interpret.

The worst form of this notational shortcutting is the expression $\frac{\partial C}{\partial S_t}$. This symbol again has no meaning whatsoever. What on earth should be a partial derivative wrt to a stochastic process? It's not defined. People only use this symbol because it's easy to write down and everyone knows what is meant by it (hopefully).

What we actually mean

Let $V:\mathbb{R}_+\times\mathbb{R}_+\to\mathbb{R}$ be a sufficiently smooth function with inputs $t$ and $x$ which solves the following PDE \begin{align*} \frac{\partial V}{\partial t} + (r-q)x\frac{\partial V}{\partial x} + \frac{1}{2}\sigma^2x^2\frac{\partial^2V}{\partial x^2}-rV=0, \end{align*} alongside some boundary conditions. This is just a normal PDE and $V$ is a standard function mapping parts of $\mathbb{R}^2$ to $\mathbb{R}$. There's no finance, randomness or stochastic calculus here. This is pure analysis and you can solve the PDE using usual techniques from analysis.

As it turns out, the value of a call option written on a stock with value $S_t$ is given by $C=V(t,S_t)$. So you take your normal function $V$ and then you substitute the stock price for the spatial variable of $V$. This gives you the value of the option.

How do you calculate delta? You take your function $V(t,x)$. You partially differentiate wrt $x$ as you learnt in calculus, $V_x(t,x)=\frac{\partial V(t,x)}{\partial x}$ and then you replace $x$ by $S_t$ to get the option's delta, $\Delta = V_x(t,S_t)$.

It might look like a technical point - because it is. Intuitively and for brevity, we often write $\Delta=\frac{\partial C}{\partial S_t}$ because we know what we actually mean. But this often causes questions about what $\frac{\partial C}{\partial S_t}$ actually means and whether $\partial S_t$ is something like $S_{t+\text{d}t}-S_t$. The answer is no and that's all non-sense because the symbol $\frac{\partial C}{\partial S_t}$ is just a shortcut with no meaning.

Another example

Think of Itô's Lemma. You start with a sufficiently smooth function $V(t,x)$. Then you write down its Taylor expansion \begin{align*} \text{d}V(t,x) = \frac{\partial V(t,x)}{\partial t} \text{d}t + \frac{\partial V(t,x)}{\partial x} \text{d}x + \frac{1}{2}\frac{\partial^2 V(t,x)}{\partial t^2} (\text{d}t)^2 + \frac{\partial^2 V(t,x)}{\partial t\partial x} \text{d}t\text{d}x + \frac{1}{2}\frac{\partial^2 V(t,x)}{\partial x^2} (\text{d}x)^2. \end{align*} It's only now that stochastic calculus comes in and we substitute $S_t$ for $x$. Remember that $x$ is just a placeholder: a symbol for which we can substitute other things. [It's like writing down a polynomial $p(X)$ in linear algebra and then replacing $X$ by a matrix or endomorphism. No problem, $X$ is merely a placeholder.] \begin{align*} \text{d}V(t,S_t) = \frac{\partial V(t,S_t)}{\partial t} \text{d}t + \frac{\partial V(t,S_t)}{\partial x} \text{d}S_t + \frac{1}{2}\frac{\partial^2 V(t,S_t)}{\partial t^2} (\text{d}t)^2 + \frac{\partial^2 V(t,S_t)}{\partial t\partial x} \text{d}t\text{d}S_t + \frac{1}{2}\frac{\partial^2 V(t,S_t)}{\partial x^2} (\text{d}S_t)^2. \end{align*} Importantly,$\frac{\partial V(t,S_t)}{\partial x}$ means the following: take the function $V(t,x)$, differentiate with respect to $x$ as you learnt in calculus and then replace $x$ by $S_t$. Note how the meaning changes: $\text{d}V(t,x) $ is an object from real analysis that is easy to deal with. The expression $\text{d}V(t,S_t)$ is now a random variable because we inserted $S_t$ for $x$. [Of course, to be fully precise, we should write down Itô's Lemma in integral form only but we’re all too lazy to do so.] Using the properties of a geometric Brownian motion, the above equation turns into \begin{align*} \text{d}V(t,S_t) = \left(\frac{\partial V(t,S_t)}{\partial t}+\mu S_t\frac{\partial V(t,S_t)}{\partial x}+\frac{1}{2}\sigma^2 S_t^2\frac{\partial^2 V(t,S_t)}{\partial x^2} \right) \text{d}t + \sigma S_t \frac{\partial V(t,S_t)}{\partial x} \text{d}W_t. \end{align*}

  • $\begingroup$ It's still not clear why we ignore $\partial S/ \partial t.$ $\endgroup$
    – user81883
    Feb 28 at 4:10

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