Option Theta as a Partial Derivative of the Pricing Function

# Answer by Kevin (score 6, accepted)


# Since $S = e^{(\mu-\frac{\sigma^2}{2})t+\sigma W_t}$, why treat it as a constant when calculating the greek Theta (dC/dt) for a European call option?












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?

## Answer by Kevin (score 6, accepted)

https://quant.stackexchange.com/a/73681

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*}