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Suppose IBM stock is trading at $100 per share and a trader purchases one IBM $100 call option for $2.00 per contract.

Scenario 1

IBM is $105 at expiration. The buyer's profit would be $300 (+ $5 - $2)

Scenario 2

IBM is at $101 at expiration. The buyer's loss would be $100 (+ $1 -$2)

Scenario 3

IBM is below $100 at expiration and the call expires and is worthless. The loss is 100% of the premium paid ( - $200).

In all three scenarios, what would be the payoff for the seller? And why?

Wouldn't the seller have a payoff of the premium paid for the contract by the buyer in each and every scenario, meaning that in all scenario's he would have a gain of $200?

Can someone please explain what happens to the seller's payoff at expiration when the price of the asset is higher than the strike price and what happens to the seller's payoff when price of the underlying is lower than the strike price?

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The call seller has the obligation to sell IBM at $100 if it is over $100 at expiration. If it is, his gain or loss will be the premium received less the intrinsic value of the call. The intrinsic value is the in-the-money amount.

1) At $105, the intrinsic value is $5 so the loss is - $300 (+ $2 - $5)

2) At $101, the intrinsic value is $1 so the gain is + $100 (+ $2 - $1)

3) Below $100 the call expires worthless and the gain is the full + $200

Note that the call seller's P&L is the additive inverse of the call buyer's gain. Whatever the call seller makes, the put buyer loses, and vice versa.

Take note that a naked call seller must either BTC his position by expiration or BTO the underlying so that he is flat after assignment. If he does not, his short call will be automatically exercised by the OCC (Exercise By Exception) and on Monday morning, he will find himself short 100 shares per call sold.

  • i know that the profit for the long call buyer is max(ST - K, 0) - c, and being a max function implies that if st-k is negative, to just use 0. But then the profit for the short call seller is min(k-st)+c. What does min imply in this case ? Cause from your answer which i presume are correct it woulndt imply to use a 0 in the brackets if the value is negative. – GGGG Nov 18 at 15:48
  • practially what i mean is , if the profit function for the short call is min(k-st,0)+c, if the value inside the brackets is negative then should i use a 0 or just my normal k and st imputs? Where k=strike price and st=price of the asset at maturity – GGGG Nov 18 at 15:57
  • Try IF(ST>K,K-ST,0) + C – Bob Baerker Nov 18 at 16:19
  • im sorry what does that mean? – GGGG Nov 18 at 16:23
  • If the stock price (ST) is greater than the strike price (K) then the call seller's intrinsic value loss is the strike price minus the stock price (K - ST) otherwise it's zero. Since the call was sold for a premium of (C), add that to the equation. So at $95, the net gain is $2. It's flat at $102 and at $105 the call seller's loss is -$3 – Bob Baerker Nov 18 at 16:35
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The call option writer gets the option premium. If the option is exercised, the option writer gets the agreed price but bears their own cost of acquiring the underlying share. There may be transaction costs for each part of the process.

So if the option is not called, the payoff for the option writer is:

  • Payoff = OptionPremium - TransactionCosts

And if the option is called, the payoff for the option writer is:

  • Payoff = OptionPremium + SellPrice - BuyPrice - TransactionCosts

If the option writer bought IBM at $90 and wrote a $95 option at an OptionPremium of $2 when the market was trading at $95, then that share represents a $2 profit if the option was not called, or a $2 + ($95-$90) = $7 profit if it was called (ignoring transaction costs), regardless of the price at the time the option was exercised.

Now, if the buy price had been $100 with everything else the same, then the option writer loses $3 if the option is exercised.

The big danger is if the option writer doesn’t have any IBM shares and writes what is known as a ‘naked option’. Then if the share price shoots up and the option is exercised, the option writer faces unlimited losses since their buy price is whatever the share price shot up to, but the sell price is fixed at $95.

If the option writer acquires the share by short selling, the transaction costs can be significant. The buy price would be whatever the price is at the time the borrowed share is returned, so potential losses are still unlimited.

Disclaimer: This is not financial advice. Options trading is risky. Please consult a suitable financial services professional if you plan to do options trading.

  • i know that the profit for the long call buyer is max(ST - K, 0) - c, and being a max function implies that if st-k is negative, to just use 0. But then the profit for the short call seller is min(k-st,0)+c. What does min imply in this case ? Cause from your answers which are correct it woulnd't imply to use a 0 in the brackets if the value is negative for a short call position – GGGG Nov 18 at 16:12
  • practially what i mean is , if the profit function for the short call is min(k-st,0)+c, if the value inside the brackets is negative then should i use a 0 or just my normal k and st imputs? Where k=strike price and st=price of the asset at maturity – GGGG Nov 18 at 16:13
  • There's nothing wrong with your answer but it could use more clarity, particularly if explaining to a noob. As is, it requires a lot of dot connecting. Yes, losses are theoretically unlimited but I have yet to see a stock go to infinity :->) – Bob Baerker Nov 18 at 16:31
  • @GGGG What I was trying to say is that the option writer’s payoff includes another variable: the price at which the option writer bought the underlying shares. – Lawrence Nov 18 at 22:46
  • @BobBaerker Thanks, though the OP seems quite well-informed already. – Lawrence Nov 19 at 4:07

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