Why European Put Prices Increase with Strike
Summary
The document proves that a European put’s value is nondecreasing as its strike increases, assuming standard risk-neutral pricing and no-arbitrage conditions. For two ordered strikes, the put payoff at the higher strike is at least as large in every terminal underlying-price state. Monotonicity of conditional expectation therefore carries that ordering through to the discounted option prices.
It also gives a portfolio argument: buy the lower-strike put and sell the higher-strike put. The portfolio’s terminal payoff is never positive, so its current value cannot be positive without creating an arbitrage; this implies the higher-strike put costs at least as much. These are general pricing arguments rather than a numerical example, and the result concerns a fixed maturity and otherwise comparable European puts.
Key ideas
- For any terminal underlying price, a put with a higher strike has a payoff at least as large as a lower-strike put.
- Risk-neutral valuation and monotonicity of conditional expectations imply the same ordering for option prices.
- A long lower-strike and short higher-strike put spread has a nonpositive terminal payoff.
- The no-arbitrage argument implies that put value is nondecreasing in strike for a fixed maturity.
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Full text
# Is the european put option an increasing function?
# Is the european put option an increasing function?
My question is to show that the function $K \rightarrow p(T,K)$ is increasing. T being maturity time,K being any strike and $p(T,K)$ is a european put option. My only approach to this question has been the case $K_1 \leq K_2$. It would be great if someone could help me through this proof.
## Answer by Kevin (score 2, accepted)
https://quant.stackexchange.com/a/48902
I firstly fill in the gaps from Slade whose comments outline the answer and then I provide an alternative approach.
Let $K_1\leq K_2$. You want to prove $P(S_t,K_1,T)\leq P(S_t,K_2,T)$.
Recall firstly that $P(S_t,K,T)=e^{-r(T-t)} \mathbb{E}^\mathbb{Q}[\max\{K-S_T,0\}\mid\mathcal{F}_t] $ which is the result from risk-neutral pricing. Probably you know that if $X\leq Y$, then $\mathbb{E}[X]\leq\mathbb{E}[Y]$ which also holds for conditional expectations, i.e.$\mathbb{E}[X\mid\mathcal{F}_t]\leq\mathbb{E}[Y\mid\mathcal{F}_t]$. This is called monotonicity. Then, you are already done since
\begin{align*} P(S_t,K_1,T) &= e^{-r(T-t)} \mathbb{E}^\mathbb{Q}[\max\{K_1-S_T,0\}\mid\mathcal{F}_t] \\ &\leq e^{-r(T-t)} \mathbb{E}^\mathbb{Q}[\max\{K_2-S_T,0\}\mid\mathcal{F}_t]\\ &= P(S_t,K_2,T). \end{align*}
As an alternative to this answer, you may want to consider a classical no arbitrage argument and look at a portfolio which owns one put option with strike price $K_1$ and is short one put option with strike price $K_2$, with $K_1\leq K_2$. Then, the portfolio value is given by $\pi(t,S_t)=P(S_t,K_1,T)-P(S_t,K_2,T)$ and has the payoff $\pi(T,S_T)=\max\{K_1-S_T,0\}-\max\{K_2-S_T,0\}\leq 0$. By no arbitrage, $\pi(t,S_t)\leq 0$ for all $t\leq T$ and thus, $P(S_t,K_1,T)\leq P(S_t,K_2,T)$.
It always boils down to the intuitive idea that a put option with larger strike price has a higher payoff and thus needs to cost more than a put option with a lower strike price.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.