Why an Extremely Long-Dated Call Can Approach the Stock Price
Summary
The document explains why a Black–Scholes European call with an extremely long maturity may be priced close to the current stock price. Under positive growth conditions, the stock is expected to become much larger than the strike over a sufficiently long horizon, so the call payoff increasingly resembles ownership of the stock. The replication intuition is that holding the underlying can then approximate the call’s payoff, implying similar values by no-arbitrage.
A second explanation uses lower and upper bounds for a call expressed through the discounted forward price and strike. When the forward becomes much larger than the strike, the bounds converge; for a stock without dividends, the discounted forward equals spot. The result depends on assumptions about growth, dividends, and the long-horizon forward level, so it is not a universal property of every long-dated call.
Key ideas
- With positive growth, the underlying may become much larger than the strike at very long maturities.
- In that limit, the call payoff can approximate the payoff from holding the stock.
- No-arbitrage links similar payoffs to similar prices.
- Call bounds converge when the forward price is much larger than the strike.
- For a non-dividend-paying stock, the discounted forward price equals spot.
Tags
Full text
# ultra-long tenor European call option valued using Black-Scholes
# 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
## Answer by JejeBelfort (score 4)
https://quant.stackexchange.com/a/33647
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 $S$.
## Answer by Quantuple (score 1)
https://quant.stackexchange.com/a/33646
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 $P(0,T)F(0,T) = S_0$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.