# Simple Relation between Put Price and Zero Coupon Bond Price

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.

• 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. Jul 8, 2019 at 17:04
• I see, how would the proof for the non-strict inequality work? Jul 8, 2019 at 17:12

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.