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From Wikipedia:

a short position in a futures contract or similar derivative means that the holder of the position will profit if the price of the futures contract or derivative goes down.

Is the price of the futures contract or derivative the cost of buying the derivative? If yes, the price of a derivative is known when buying the derivative, so its price will not change, which is contrary to "the price of the futures contract or derivative goes down" in the quote. So I wonder if the quote actually means the price of the underlying instrument, instead of the price of the derivative?

Thanks!

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@Tim - in this case, a futures contract isn't like an options contract. It's simply a method of entering into an agreement for delivery at a future date.

While the speculators appear to have taken over, there are practical examples of use of the futures market.

I am a gold miner and I see that my cost is $1200/oz given my quality of ore. I see the price of gold at $1600 and instead of worrying that if it goes too low, I run at a loss, I take advantage and sell contracts to match my production for the next year (or as long as the contracts go, I forget how far out gold futures are).

Of course I give up the higher price if gold goes higher, but this scenarion isn't speculation, it's a business decision.

The bread maker, on the other hand, might buy wheat futures to guarantee his prices for the next year.

  • Thanks! What do you mean by "as long as the contracts go, I forget how far out gold futures are"? – Tim Sep 26 '11 at 1:21
  • A given contract has a date, the seller must buy it back or deliver the asset on this date. The dates can go out over a year, but I don't know if they go 2 years or 3. – JTP - Apologise to Monica Sep 26 '11 at 1:36
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No, it means what it says. Prices change, hence price of the derivative can go down even if the price of the underlying doesn't change (e.g. theta decay in options).

  • Thanks! Does the quote mean that when the premium for buying the derivative goes down, the profit of the buyer of the derivative will increase? – Tim Sep 26 '11 at 1:20

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