# On the time value asymptotic behaviour of call option in the generalized Black-Scholes model

In the context of generalised Black-Scholes models, $$\frac{\partial V}{\partial t}+\frac{1}{2} \sigma^2(S, t) S^2 \frac{\partial^2 V}{\partial S^2}+(r(t)-D(t)) S \frac{\partial V}{\partial S}-r(t) V=0,$$or

$$\frac{\partial V}{\partial t}+\frac{1}{2} \sigma^2(S, t) S^2 \frac{\partial^2 V}{\partial S^2}+(r(S,t)-D(S,t)) S \frac{\partial V}{\partial S}-r(S,t) V=0,$$

the price-asymptotic behaviour of call options can be mathematically described by, $$V(S,t) \approx S\,e^{-\int_0^t D(s)\,ds}-Ee^{-\int_0^t r(s)\,ds} ~~\text{as}~S \to \infty,$$where, $$V(S,t)$$ is the value of call option which depends on the asset price $$S$$ and time $$t$$ to the expiry $$T$$, $$D(t)$$ is the dividend yield, $$E$$ is the strike price, and $$r(t)$$ is the interest rate. This makes sense as in the long price run, there is a high likelihood that the underlying asset's price will eventually equal or exceed the strike price, and thereby, the option's intrinsic value approaches the difference between the underlying asset's price and the strike price.

However, while thinking about the case $$T \to \infty,$$ Of course, the distribution of option value tends to become more tightly centred around the intrinsic value $$\max\{S-E,0\},$$ and it is observable that option premium is higher than the terminal condition $$\max\{S-E,0\},$$ in some neighbourhood of expiry and it diffuses to $$\max\{S-E,0\},$$ near the strike. But I feel that the diffusion process is not much explicit (as in the case of heat distribution on a finite rod) in the distribution of option value although the underlying dynamics agrees with diffusion-dominated Brownian motion. Why it is so?

• It is an elementary calculus exercise to write down the BS-Formula and take the limit $T\to\infty\,.$ I don't want to do that. If I am not mistaken the limit is zero when the dividend yield and the risk-free rate are both positive. Can you proceed? Feb 8 at 8:34
• Ya...I got your point, thnks...Actually I intended to deal with generalized Black-Scholes equation (with time-price depending volatility) having no exact solution. The behavior is same, right?
– Riaz
Feb 9 at 4:19
• What makes you think that? Feb 9 at 5:44
• How can we justify the asymptotic phenomenon in strong assumptions ( generalized models)?
– Riaz
Feb 9 at 6:27

It is well-known and easy to see that in any model where the stock price solves an SDE of the form $$\tag{1} \frac{dS_t}{S_t}=(r-d)\,dt+\sigma_t\,dW_t$$ the option price can be written as $$\tag{2} C=e^{-dT}S_0 P_1-e^{-rT}KP_2$$ where $$P_2$$ is the probability that the option expires in the money: $$\tag{3} P_2=\mathbb Q(S_T>K)\,,$$ and $$P_1$$ is a similar probability but under a measure that has Radon-Nikodym density $$\tag{4} \exp\Big(\int_0^T\sigma_t\,dW_t-\frac{1}{2}\int_0^T\sigma^2_t\,dt\Big)$$ w.r.t. the risk-neutral measure $$\mathbb Q\,.$$

My assumptions in (1) are

• risk-free rate $$r$$ and dividend yield $$d$$ are constants (making them time-dependent isn't really an interesting generalization)

• the volatility $$\sigma_t$$ is an arbitrary stochastic process. In particular it can be $$\sigma(t,S_t)$$ (local volatility) or stochastic volatility.

Looking at (2) and taking into account that the probabilities $$P_1$$ and $$P_2$$ are bounded by $$1$$ it is clear that $$\lim_{T\to\infty}C=0$$ when $$d>0$$ and $$r>0\,.$$

• Nice...In what sense you have mentioned ''making them time-dependent isn't really an interesting generalization'' ?
– Riaz
Feb 9 at 11:44
• If you have time-dependent but deterministic $r(t)$ and $d(t)$ you have their integrals in the exponentials. It is then easy to formulate conditions under which those exponentials converge to zero. Feb 9 at 12:46
• Thanks...Once again, Can we monitor the phenomenon for the most generalized case $\frac{\partial V}{\partial t}+\frac{1}{2} \sigma^2(S, t) S^2 \frac{\partial^2 V}{\partial S^2}+(r(S,t)-D(S,t)) S \frac{\partial V}{\partial S}-r(S,t) V=0.$
– Riaz
Feb 10 at 4:04
• A whole industry was developed to solve for option prices in such general cases, albeit, a practitioner would probably rightfully ask why the risk-free rate should depend on the stock price. Nonetheless, solve the PDE numerically or by simulation and look at large $T$. The cases I answered are in my opinion as good as it gets without solving numerical problems. Feb 10 at 6:13