Let $C(S, K, \sigma, r, T)$ be the price of a call option. How much can be said about the Greeks without picking a model? Or at least without full Black-Scholes?
Below, I write down everything I know in the hopes that
- people can refer me to a book that does similar things
- it will be useful for people reading it in the future
Delta
$\Delta = \frac{\partial C}{\partial S}$
The amount of stock you should hold for continuous hedging (no t-costs)
$\frac{\partial C}{\partial K} = \frac{1}{K} (C- S \frac{\partial C}{\partial S} ) $
Proof: assuming that $C$ is homogeneous in $S$ and $K$, apply Euler's homogeneity theorem.
The bull spread tells us that $\frac{\partial C}{\partial K}$ is positive. (You could also justify this for $\Delta$ directly with hedging). Hedging shows that $\Delta$ is positive and $\Delta|_{S = 0} = 0$, $\Delta|_{S=\infty} = 1$.
Gamma
$\Gamma = \frac{\partial^2 C}{\partial S ^2}$
This is also for hedging
$$C(S+dS) - C(S) \approx \Delta(S) dS + \frac{1}{2}\Gamma(S)dS^2$$
but since you can't just use the underlying, it's a little less clear to me. The butterfly spread shows that $\frac{\partial^2 C}{\partial K ^2} = \left( \frac{K}{S}\right)^2 \Gamma$ is positive.
Rho
$$\rho = \frac{\partial C}{\partial r}$$
$r$ can't be a parameter of a truly model independent $C$, so I guess I made the implicit assumption the bond price evolves as $d B_t = r B_t dt$ so $B_t = e^{-r (T - t)}$.
I suppose it would make more sense to have $C$ as a function of $B$ rather than $r$, but it's very similar to $\rho$:
$$\frac{\partial C}{\partial B} = \frac{\partial C}{\partial r} \frac{\partial r}{\partial B} \\ = \rho \frac{-1}{BT} $$
$\frac{\partial C}{\partial B}$ (not sure if it has a name) tells us how much of the risk-free bond to own to hedge, but I have less insight into it than $\Delta$.
Theta and Vega
$$\Theta = \frac{\partial C}{\partial T} \qquad \mathcal{V} = \frac{\partial C}{\partial \sigma}$$
Again, to have $\sigma$ as a parameter in $C$, some sort of stock price dynamics has to be understood. Both $T$ and $\sigma$ are reasoned about as uncertainty which shows that they are usually positive.
I don't have a very good model independent handle on either of these, but I'm sure it's possible to say something without going full Black-Scholes.
In particular it would be nice to get
$$\Theta = \frac{\sigma}{2T} \mathcal{V}$$
for when $r= 0$. This intuitively comes from the fact that volatility scales with square root of time, and so the option only depends on the volatility adjusted time $\sigma^2 T$.
