Asymptotics of Call Option as $S\to0$

Question

Let $C(S)$ denote the (initial) value of a call option with underlying spot price $S$. I assume that the underlying has continuous sample paths (not necessarily a geometric Brownian motion though).

As $S\to\infty$, we know that $C=\mathcal{O}(S)$ (using big O notation) because the option is essentially linear in the underlying ($\Delta=1$ and $\Gamma=0$ for deep ITM options).

Do we know at which rate the option price converges to zero as $S\to0$? I'd guess it's more polynomial rather than exponential decay? I'm asking what is the best function $g(S)$ in $C=\mathcal{O}(g(S))$ as $S\to0$?

Here is an example of initial call option prices with $T=1$ and $K=8$. The option value is (or will be) linear for large stock prices but what's the order for small stock prices?

Answer 1

This is more of a math question than a quant question. Under Black Scholes dynamics (assuming $r=0$ for simplicity), as everyone knows we have $$C=SN(d_1)-KN(d_2)$$. In this case, we are interested in large negative $d$, since $lnS$ is large and negative. There is an asymptotic series for $N(x)$ whose first term for large negative x is $$N(x)=-\phi(x)/x$$, where $\phi$ is the normal distribution. Plugging this in, we get $$C=S(-\phi(d_1)/d_1) - K(-\phi(d_2)/d_2)$$, and using the relation $d_2=d_1-\sigma\sqrt(T)$ one can derive that to first order$$C = S\sigma T^{1/2}\phi(d)/d^2$$ where $d=\ln(S)/\sigma T^{1/2}$. This is essentially your $g(S)$. You can show it goes to zero faster than $S^n$ for any fixed $n$.

Answer 2

Edit: the original question didn't specify "model independence" and so the below focuses on the BS framework. Also, i focused on speed of convergence rather than order of convergence: I will try to update my answer with some thoughts on order of convergence later on.

Not sure this answers your question, but the speed at which the option prices changes with respect to the underlying is the delta, which is just equal to $N(d_1)$, with $d_1$:

$$d_1=\frac{ln\left(\frac{S}{K}\right)+rt+0.5\sigma^2t}{\sigma \sqrt{t}}$$

As $S \to 0$, $ln\left(\frac{S}{K}\right) \to (-\infty)$, and therefore $d_1 \to (-\infty)$ and therefore $N(d_1) \to 0$ (so the speed at which the option price goes to zero as $S$ goes to zero: goes itself to zero asymptotically)

Delta attains its maximum value of 1 for deep ITM options, and then the delta gradually declines to zero as the option becomes OTM. Gamma (which can be thought of as the speed at which delta itself changes with respect to the underlying) is highest for ATM options.

So the way I think about it is:

• Option price declines fastest (with respect to the underlying $S$) when delta is 1 (obviously)

• As the underlying gradually declines, Delta starts declining from 1 and the speed at which it declines gets gradually faster as the underlying approaches the strike (i.e. as the option approaches ATM from above): so that actually means that the rate at which the option price decreases, gets smaller at a faster pace as the underlying decreases

• Below the strike, as the option turns OTM, the gamma starts to decrease again, so the rate at which delta decreases starts to decrease (nonetheless, the delta still does decrease as the underlying gets lower, so again: the rate at which the option value decreases, itself decreases as the underlying value decreases: although at an increasingly slower pace).

Can we mathematically quantify the rate at which the option value decreases? Yes, this value is the delta (i.e. $N(d1)$). Can we quantify the rate at which delta itself decreases? Yes, the value is the Gamma.

Aside from the graph below, I guess we could try to quantify the rate of convergence to zero further in terms of what type of function the "option price w.r.t. strike" is dominated by - is that what you had in mind? Graphically, we can see its slower than linear towards the end (somewhat obviously, because it becomes asymptotic).

Answer 3

I'm not 100% sure, please double-check. I think that both in the ITM and the OTM case (requested), a model-free answer cannot exist. In particular, the rate at which:

• ITM: $C(S_0) \rightarrow S_0$ as $S_0 \rightarrow \infty$ and
• OTM: $C(S_0) \rightarrow 0$ as $S_0 \rightarrow 0$

depends on the model-specific risk-neutral transition density $p^Q(S_T, T | S_0, 0)$ from $S_0$ at time $0 $ to a value $S_T$ at time T.

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My idea is the following. Let $C_{K,T}(S_0)$ be the initial (that is, at time $t=0$) price of a call option of strike $K$ and maturity $T$. This is, by risk-neutral evaluation (wlog, let's assume constant short-rate $r$ for simplicity):

\begin{align} C_{K,T}(S_0) &= e^{-rT}\mathbb{E}^{Q}[(S_T-K)^+|\mathbb{F}_0] \\ &= \int_0^{\infty} (S_T-K)^+ p^Q(S_T, T | S_0, 0) dS_T \end{align}

where (informally) the information content of the filtration $\mathbb{F}_0$ is "$S(t=0)=S_0$".

The risk-neutral transition density $p^Q(S_T, T | S_0, 0)$ is the solution of the Kolmogorov-forward equation (aka, Fokker-Plank) equation. Since this density is model-dependent (it will be a lognormal in the case of the Black-Scholes model, gaussian in the case of a gaussian diffusion and even different in the case of SV models like Heston), the price $C_{K,T}(S_0)$ is dependent on the model.

Therefore, there is no reason for the ITM and OTM rates of convergence to be model-free.

Answer 4

Put-Call parity:

$C = P + (S-K),$

taking $r=0$ for simplicity but without loss of generality.

Now,

$\lim_{S\rightarrow 0} \,P = K$,

hence

$\lim_{S\rightarrow 0}\, C = 0$

Another way to see this is to note that both $N(d_1)$ and $N(d_2)$ tend to zero as $S$ tends to 0.