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From Terence Tao Blog the price of a put option at time $t_0$ cannot exceed the strike price P at time $t_1$. The reason is that otherwise there would be an arbitrage opportunity. Everyone in the market could build a put option Simply putting aside a certain amount of cash for paying the strike price P. In this way could one make risk-free profit and how?

  • What do you mean? Just sell the put and keep $P$ out of the proceeds to pay for the asset if the put is exercised. Even if that asset you are forced to buy is worthless, you keep the excess premium. – lulu Oct 07 '22 at 16:41
  • Unless you are worried about technicalities...I mean, it is technically possible for an asset to trade at a negative price. It happens. Tax residual CMO's, for instance, can carry excess tax liability so, sometimes, people need to be paid to accept the assets. But, from the general context, I am assuming that you were speaking of ordinary assets which can certainly trade for $0$, but not lower. – lulu Oct 07 '22 at 16:44

1 Answers1

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You are the seller, I am the buyer.

At $t_0$, I pay you $x$ to purchase a put option for $S$ at strike price $P$ at time $t_1$, meaning that at $t_1$ I can choose to force you to buy $S$ at price $P$.

Generally I will exercise this option if the market price of $S$ is far less than $P$ at $t_1$, I can buy $S$ off the market and force you to pay extra for it. (as the buyer of a put option, I am basically betting that the price of $S$ will decrease in the future)

If $x > P$, then you simply set aside $P$ and pocket $x-P$. Then at time $t_1$ even if $S$ is worthless and I force you to buy it you still come out ahead.

Of course, as a buyer I would never accept such a deal because I am guaranteed to lose money, so it's not really an arbitrage opportunity because no one would buy this option. Similarly "You give me \$2 and I will give you between \$0 and \$1 back" is also "risk-free profit" if you can find anyone gullible enough to take your offer.

BaronVT
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  • you come out ahead because you earn x-P? –  Oct 07 '22 at 16:56
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    @Pirofotulis The words "you" and "me" are confusing. BaronVT's answer can be rephrased as: if the option buyer pays $x>P$ for the put option then the seller will always earn at least $x-P$ which is an unjustified arbitrage profit. – Kurt G. Oct 07 '22 at 17:45
  • @Kurt G. Yes i see. So the price at time $t_0$ cannot exceed P –  Oct 07 '22 at 18:04