Long-Maturity European Put Bounds and Option Price Behavior

Summary

The document considers whether a European put’s no-arbitrage price bounds imply that its value must fall toward zero as time to maturity becomes very large. The stated lower bound is the discounted strike minus the spot price, while the upper bound is the discounted strike. With a positive interest rate, the discounted strike tends to zero as maturity grows. Combining the upper bound with the fact that a put price cannot be negative implies that the put price tends toward zero under those assumptions.

The response says this is consistent with the Black–Scholes pricing function and observes that put value need not keep increasing with maturity: after becoming sufficiently long, maturity can cease to raise the value. The document refers to graphs comparing this behavior with the bounds, but the graph details and parameter values are not present in the supplied text. The limiting argument relies on a positive rate and the stated bounds; it does not establish monotonicity for all maturities or cover cases such as nonpositive rates.

Key ideas

• For a positive interest rate, the discounted strike approaches zero as maturity tends to infinity.
• The upper no-arbitrage bound and nonnegative put values then imply a zero limiting price.
• A European put price need not increase monotonically with time to maturity.
• The response says the limiting behavior is consistent with Black–Scholes, but the referenced graph details are absent.

Tags

# is relating bounds to relation between time to maturity and european put option price correct?


# is relating bounds to relation between time to maturity and european put option price correct?












J.C. Hull derives the following relation $$Ke^{-rT} - S \le p \le Ke^{-rT}$$

where $p$ is european put option price, $K$ is strike price, $S$ is stock spot price,$r$ rate of interest and $T$ time to maturity. The above relation holds for no arbitrage.

The book states european put option price does not necessarily increase with increase in time to maturity. But just using the above relation as $T$ increases doesn't it mean $p$ will always go to 0 for large $T$ ?

## Answer by Sanjay (score 3, accepted)

https://quant.stackexchange.com/a/44440

There is no contradiction at all here. $Ke^{-rT}$ goes to zero for $T$ going to $\infty$ so the relation you mention suggests that $-S\leq p \leq 0$ as $T$ gets bigger. If $r>0$. Put prices can (in theory) not be negative so $p$ goes also to zero according to the relation you mention.

Is this consistent with the pricing function of Black-Scholes? Yes. Consider this graph:

When Time to maturity $T$ gets large then enough then the Put value is not longer increasing in $T$

The next graph shows that the inequalities you mention hold for the parameters ($r$, $S$, $K$, $\sigma$) being the same as in the above graph:

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.