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? enter image description here

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    $\begingroup$ It seems you are looking for a model-free rate of convergence. While it’s intriguing, a more fundamental question could be - does it even have to exist? (a rate of convergence to 0, independent from the valuation model) $\endgroup$ – Gabriele Pompa Jan 9 at 8:48
  • $\begingroup$ @Gabriele You’re right. I hope for a model independent answer. Mostly because the answer for $S\to\infty$ is model independent. I’m happy to restrict myself to non jump models though (in case this simplifies something). If it turns out that the answer does depend on the model, then it’d be super interesting to see how and why the orders for the Black-Scholes model and the Heston model differ. $\endgroup$ – Alex Jan 9 at 9:16

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 $n$.

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    $\begingroup$ Great answer. Your comment provoked a question in my mind: "what is the difference between 'Quant' and 'Math' question" ? In my mind, Quantitative finance is just mathematics applied to finance :). So any "math" question applicable to finance classifies as "quant" to me :). $\endgroup$ – Jan Stuller Jan 9 at 19:51

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).

enter image description here

  • $\begingroup$ Thanks for your answer @Jan. I don’t think that derivative and order are the same though? Suppose $f(x)=x+2$ with $f’(x)=1$. However, $f=\mathcal{O}(x)$ as $x\to\infty$ (unrelated to the derivative of $f$). So, the idea of $f=\mathcal{O}(g)$ is that $f$ and $g$ have the slope but just knowing $f’$ would not answer this? Big-o is more about bounding the function? $\endgroup$ – Alex Jan 8 at 22:23
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    $\begingroup$ Hey @Jan, thanks for taking the time to think a bit more about my question :) I think I never really emphasized the Black Scholes model and included all models with continuous trajectories (I feel jump models sometimes complicate the underlying intuition). And I asked for the order/rate from the beginning (stating the result for $S\to\infty$ and wondering whether it’s of polynomial or exponential order). Though I appreciate I could have made my question clearer. Please let me know if I can add some details/make my question more precise. $\endgroup$ – Alex Jan 9 at 9:22

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.

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.

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    $\begingroup$ I think a model-independent order exists for deep ITM call option. Basic no-arbitrage tells us $C(S)\leq S$ (no dividends), right? Thus, $\lim\limits_{S\to\infty}\frac{C(S)}{S}\leq1$ or put differently, $C(S)=O(S)$. This is model-independent and, by the way, equally applies to American and European calls? $\endgroup$ – Kevin Jan 9 at 17:15
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    $\begingroup$ I think you can satisfy the no arbitrage condition $C(S)/S \leq 1 $ even with a completely different ITM order, like $C(S)=O(\ln S)$ for big S, so I disagree with your conclusion that the no arbitrage condition $C(S) \leq S$ implies straightly a model-independent ITM behavior. Similarly, I wouldn’t conclude from the other arbitrage-free relation $C(S) \geq 0$ a model-free behavior for the OTM case. $\endgroup$ – Gabriele Pompa Jan 9 at 17:48
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    $\begingroup$ It's pretty easy to show some models for which the proposed ITM and OTM limits do not hold: for example if $p(S_T|S_0)$ does not depend on $S_0$, although this is rather absurd. For a better example, consider a stock with constant dollar volatility, such that $p(S_T|S_0)$ is the normal density. Then $C(S_0)$ is not $0$ when $S_0=0$. Now you may complain that this allows the stock to be negative. We may however introduce a reflecting barrier at zero. That gives a model with continuous paths, with positive definite stock prices, for which the call option is not zero when the stock is zero $\endgroup$ – dm63 Jan 10 at 16:21
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    $\begingroup$ The assertion appears to hold for log-type models (such as B-S), where you can write $p(S_T|S_0)dS_T = f(S_T/S_0)dS_T/S_T$ for some function $f$. For then $$C(S_0) = \int_K^\infty(S_T-K) f(S_T/S_0)dS_T/S_T$$. Using the substitution $u=S_T/S_0$, one gets $$C(S_0) =S_0 \int_{K/S_0}^\infty (u-K/S_0)f(u)du/u$$. This obeys both the ITM and OTM limiting cases. Intuitively, the if the model scales up and down with the initial stock price, the assertions hold $\endgroup$ – dm63 Jan 10 at 16:33
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    $\begingroup$ @GabrielePompa All I was suggesting is that $C(S)\leq S$ is a model-free order in the sense that $O$ is literally defined as an upper bound, $f=O(g)$ if $|f(x)|\leq Mg(x)$ for some $M>0$. I didn't mean to say that it is the best one can achieve: suppose $C(S)=O(\ln S)$, then it's also true that $C(S)=O(S)$, for $S\to\infty$. But yeah, as you say, $C(S)>0$ doesn't translate into big $O$ notation. $\endgroup$ – Kevin Jan 10 at 23:45

Put-Call parity:

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

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


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


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

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.

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    $\begingroup$ Thanks for your answer @ilovevolatility. Your answer convincingly shows that $\lim_{S\to0}f(S)=0$. However, I already knew this. I was wondering at what rate/speed/order does $f$ approach zero as $S\to0$? Does $f$ converge to zero quicker than $S^2$? So, what’s the order of the option price close to zero? $\endgroup$ – Alex Jan 8 at 22:23
  • $\begingroup$ @Alex Right, so since $C(S) < S$ for all $S$ clearly it converges as least as fast as $S$, that much is clear. You can try to find the exact order, and dm63 gives a good answer, but as I am struggling to see what is actually the practical reason for wanting to know this? Trying to decide whether you should use delta = 1e^-9 or delta 1.05e^-9 as $S$ goes to zero? :-) $\endgroup$ – Frido Rolloos Jan 9 at 14:41
  • $\begingroup$ It wasn't motivated by any numerical work originally. It’s just a question I had out of curiosity after discussing with someone in my class what happens if $S\to\infty$. I just wondered whether $S\to0$ is equally simple. But yeah, it may not be the most useful question for practical applications :) $\endgroup$ – Alex Jan 9 at 17:04

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