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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 $T$ ?

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  • $\begingroup$ Not sure what your question is, yes $\lim_{T \to \infty} Ke^{rT} = 0.$ The nominal value of a put option is bounded so the real value goes to $0$ if $r>0$ and given enough time. $\endgroup$
    – Bob Jansen
    Commented Mar 5, 2019 at 17:00
  • $\begingroup$ @BobJansen I mean to ask why is it said that price of put option is greater if everything else is the same and time to maturity is higher, reason being there is more time for option to be in the money $\endgroup$ Commented Mar 5, 2019 at 17:06

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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: enter image description here

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:

enter image description here

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  • $\begingroup$ How does this reasoning hold then: put option price increase with time to maturity, reason being there is more time for option to be in the money. Does this reasoning contradict the above relation ? Im really sorry for bothering. $\endgroup$ Commented Mar 5, 2019 at 17:08
  • $\begingroup$ The statement "put option price increase with time to maturity" isn't correct. See also this question. $\endgroup$
    – Bob Jansen
    Commented Mar 5, 2019 at 17:14
  • $\begingroup$ @sashas, what as Bob Jansen mentions your statement is not true. I have updated my answer $\endgroup$
    – Sanjay
    Commented Mar 5, 2019 at 17:38

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