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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.

Damien
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2 Answers2

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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.

Ross Millikan
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  • How do we know that higher premiums imply higher strike prices? – Damien Jul 10 '11 at 23:02
  • @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
  • But it is still possible for the person who writes a put option to make an error and have a lower premium for a higher strike price? – Damien Jul 10 '11 at 23:39
  • That makes even a larger arbitrage opportunity. Errors like that are usually not considered. – Ross Millikan Jul 10 '11 at 23:47
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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.

Jason Swanson
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  • 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 00:06