# 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.

• You can probably just use this: the put price is the conditional expectation of the payoff $(K - S_T)^+$. So using linearity of expectation you can write the difference between two puts of different strikes as the conditional expectation of the difference between two positive part functions. I think from there you can show that the conditional expectation must be nonnegative (almost surely) and so put prices are increasing in $K$. I'm not sure if this is enough as a proof though, so hopefully you can take it from there or someone else chimes in. Hope this helps – Slade Sep 25 '19 at 4:11
• Adding on to my previous comment, you should be able to prove the expectation of a nonnegative random variable (the difference of payoffs) is nonnegative using Markov's inequality or something along those lines. So from there you'd have your proof. I'm still not sure if this is what is expected but it seems to work. – Slade Sep 25 '19 at 4:19

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.

• in the alternative approach you use the no-arbitrage argument right? – Anon Sep 25 '19 at 8:31
• In the second approach, I use the no arbitrage principle explicitly but the first approach, which was outlined by Slade, also relies on no arbitrage since the conditional expectation of the discounted payoff is the (fair) no arbitrage price of a put option. – KeSchn Sep 25 '19 at 8:42