Consider your standard European Put Option, with strike price $K$ and maturity $T$, and denote by $P_t(K,T)$ the price of this option at time $t$. Moreover, consider a standard Zero-Coupon bond with maturity at time $T$ and face value $1$, and price $Z_t(T)$ at time $t$. Why is it necessarily the case that: $$ P_t(K,T) \ < KZ_t(T) $$ for all times $t \neq T$? I know equality occurs at the terminal time $T$ if $S_T = 0$, however I am unsure how one obtains the above relation. Moreover, I believe this is a model-independent result, however I am not entirely sure.
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$\begingroup$ For this to be true you have to assume that $S_t$ cannot go to zero before $T$ and then stay there. Without this assumption you can prove only $P_t\le K Z_t$. In this sense it is somewhat model dependent. $\endgroup$– Alex CJul 8, 2019 at 17:04
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$\begingroup$ I see, how would the proof for the non-strict inequality work? $\endgroup$– rubikscube09Jul 8, 2019 at 17:12
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1 Answer
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Your question is answered by the no-arbitrage principle. The payoff of your put option is $\max\{K-S_T,0\}\leq K$. Thus, their time $t$ prices need to have the same relationship (everthing else creates arbitrage opportunities), i.e. $$ P_t(K,T)\leq KZ_t(T).$$
This statement is kind of related to the law of one price which is implied by assuming an arbitrage-free market. As Alex said, if you know that $S_T>0$ almost surely, then $\max\{K-S_T,0\}< K$ and you get$$ P_t(K,T)< KZ_t(T).$$
The question is whether $S_T$ may be zero or not. This is indeed model dependent, for instance it is impossible in the Black-Scholes and Heston model. To sum up, the $\leq$ case is model-independent, the slightly stronger version with $<$ is model-dependent.
Please note that these inequalities derive from a lot of financial intuition: a put option gives you the right to purchase the underlying asset for $K$, so this claim can hardly be worth more than $K$ discounted since $K$ is the most you can get out of it.
