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?
Thanks in advance
Best Answer
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 $r>0\,.$