I'm asking about etymology, and not what these options are. The answers beneath don't feel convincing; can't 'call' and 'put' be interchanged in them? I'll abbreviate Call Option to CO, and Put Option to PO. From Quant SE:

Not as much on origin of this, but some analogy that I find helpful:

Call - think of an open auction when bidders call out their prices for an auctioned item. So with a call option you've got a privilege to call the strike price, effectively having a right to buy the underlying asset.

Put - think that you own a privilege to put the asset for sale at the strike price, effectively having a right to sell the underlying asset.

Mind that for an option buyer this is optional to exercise it, yet for the option writer it's an obligation to transact, should the buyer opt to exercise it.

But can't I be said to:

  1. for a CO, "put" (to the CO seller) to buy the security at the strike price (for a CO)?

  2. for a PO, "call" (to the CO seller) to sell the security at the strike price?

From Quora:

When you buy a call and exercise it you are receiving stock that you have "called" up from the person that sold you the call, the right to buy. When you buy a put you have purchased the right to sell the stock, or "put it" to the person who sold you the put.

Again, can't I say that a:

  1. CO is when a CO seller "puts" the call option to the CO buyer?

  2. PO is when a PO buyer is "calling" the PO seller to buy the security from the buyer?


closed as off-topic by EnergyNumbers, Adam Bailey, denesp, BB King, Herr K. Jan 14 at 15:44

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    $\begingroup$ I'm voting to close this question as off-topic because this is not an etymology Q&A site. $\endgroup$ – EnergyNumbers Jan 11 at 7:36
  • $\begingroup$ I’ve always understood a call option as an option to call on the option seller for the underlying (at a specified price), while a put option is an option to put the underlying to the option seller (at a specified price). That certainly seems to be the simplest explanation, though I don’t have a source for you. $\endgroup$ – dismalscience Jan 12 at 13:55

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