# Put Options and Arbitrage

I came across the following problem on put options:

A European put with strike price $100$ expiring in $1$ year has premium $\$ 1$and a European put with strike price$K$expiring in$1$year has premium$ \$2$. The continuously compounded risk free interest rate is $r>1$. What is the full range of values of $K$ that results in an arbitrage opportunity.

Why do we assume that we buy the $\$100$put option and sell the$K$put option? In other words our position is the following: $$(\max(0, 100-S_1)-1e^{r})- (\max(0, K-S_1)-2e^{r}) >0$$ which means that$K < 100+e^r$for arbitrage. - It would help most readers of this site if you could explain what a "European put" is. – Mark Bennet Jul 10 '11 at 21:44 add comment ## 2 Answers You raise a good point. Suppose$K=300$, for example. Then we could sell three$100$-put options and buy one$300$-put option, netting$1$unit at time$t=0$. Our wealth at time$t=1$will be $$e^r + \max(0,300-S_1) - 3\max(0,100-S_1).$$ If$S_1\le100$, then this reduces to $$e^r + 300 - S_1 - 300 + 3S_1 = e^r + 2S_1 \ge e^r.$$ If$100<S_1\le 300$, then this reduces to $$e^r + 300 - S_1 \ge e^r.$$ And if$S_1 > 300$, then this reduces to$e^r$. So it appears we have arbitrage at$K=300$, and the answer$K<100+e^r$is incomplete. - Mathematically this is correct, but it is extremely unlikely that a put at$300$would only cost$2$. – Ross Millikan Jul 10 '11 at 23:48 Since we can achieve any rational ratio between the numbers of options we buy and sell,$K>200$is already enough. That it's extremely unlikely isn't really an argument, since arbitrage opportunities are inherently unlikely to actually occur. The low premium of$\$1$ indicates that the current price is probably far above 100; if it's, say, 300, it's plausible that someone might assess the value of a 100 put option at $\$1$and someone else might assess that of a 200 put option at$\$2$. –  joriki Jul 11 '11 at 0:06
As the premium of the $K$ option is higher than the premium of the $100$ option, $K \gt 100$. If the price stays above $K$, both options expire worthless and we keep the dollar, worth $e^r$ at the end of the year. If the price falls below $100$, both are exercised and we have the $e^r+100-K$. If the price is between $100$ and $K$, say $P$, the $K$ option is exercised and we sell in the market, ending with $e^r+P-K$.If you were to buy the $K$ and sell the $100$, you would be out if the price stayed high.
@Damien: because it is more valuable to be able to sell something at $110$ than to be able to sell it at $100$, so the premium will be higher. Similarly, the premium on a call goes the opposite direction from the strike price. –  Ross Millikan Jul 10 '11 at 23:34