No-Arbitrage Bounds for European Put Prices

# Simple Relation between Put Price and Zero Coupon Bond Price


# 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.

## Answer by Kevin (score 2, accepted)

https://quant.stackexchange.com/a/46808

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.