# ultra-long tenor European call option valued using Black-Scholes

For tenors > 100 years (i.e. 150 years), my Black-Scholes model is telling me the price of a call option will be equal to its current stock price. Can anyone intuitively explain this?

Thank you

• In a model without dividends, this is actually only true when $r > -\sigma^2 / 2$. When $r = -\sigma^2 / 2$, the limit is $S_0 / 2$ and otherwise it is zero. – LocalVolatility Apr 12 '17 at 10:48

You can think in terms of the Black-Scholes stock price dynamics:

Assuming a positive drift such as the conditions mentionned by LocalVolatility in its comment are satisfied, you can see that the underlying spot price will tend to infinity in an infinite amount of time.

Therefore, the spot price will be sufficiently large ($S>>K$) to assume that your payoff $(S-K)^+$ is roughly equal to the underlying spot.

You then see that the only way to replicate this call is to buy the underlying, which has the same payoff as your call when the time to maturity goes to infinity.

By no arbitrage, they must therefore have the same price: hence your call worthing the spot price $S$.

• Thank you. When you say positive drift, do you mean positive risk-free interest rate? When I set the risk-free rate to 0%, the same thing happens (value of the call is equal to the underlying spot), I think this is because the N(d2) goes to 0. Can you intuitively explain this as well? – user3117403 Apr 12 '17 at 11:31
• Yes, I do. Regarding the second part of your comment, this is more difficult to get an intuitive grasp. But following my replication idea, the delta of the call $\Delta = N(d_1)$ is still 1 when $r = 0$, so you still need to own the spot to replicate the call. – JejeBelfort Apr 12 '17 at 12:46
• Thanks for the reply. But I'm still having trouble grasping this intuitively. If the risk-free rate is 0%, then my payoff (S-K) wouldn't be sufficiently large, so it is hard to understand why I would still need to own the underlying to replicate the call, please let me know if there is a better way to understand this. Thanks – user3117403 Apr 12 '17 at 15:16

A European call price respectively admits $\max(0, P(0,T)(F(0,T)-K))$ and $P(0,T) F(0,T)$ for lower and upper bounds, where $P(0,T) = e^{-rT}$ is the price of the zero coupon bond expiring at $T$ and $F(0,T)$ the equity forward price.

Only if $\lim_{T \to \infty} F(0,T) \gg K$ will you have the result you mention since in that case both bounds coincide to $P(0,T) F(0,T)$. If the stock does not pay dividends $P(0,T)F(0,T) = S_0$.