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By "Hull, Options, Futures, and Other Derivatives":

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Suppose that, in figure,the stock price is \$100 and the option price is \$10. Imagine an investor who has sold 20 call option contracts—that is, options on 2,000 shares. The investor’s position could be hedged by buying $0.6 \times 2,000 =1,200 $ shares. The gain (loss) on the stock position would then tend to offset the loss (gain) on the option position.

For example, if the stock price goes up by \$1 (producing a gain of \$1,200 on the shares purchased), the option price will tend to go up by $0.6 \times \$1 = \$0.60$ (producing a loss of \$1,200 on the options written);

Why the option price that will tend to go up by \$ 0.60, produce a loss of \$1,200?

If strike price $K= \$ 50$, we have that investor loss:

$$(\$ 100 - \$ 50 ) \times 2,000 = \$ 100.000$$

If the stock price goes up by $1, the investor loss:

$$(\$ 101 - \$ 50 ) \times 2,000 = \$ 102.000$$

so, if the stock price goes up by \$1, the option contract produce a loss of \$ 2.000

Why?

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closed as off-topic by amdopt, LocalVolatility, msitt, chollida, Quantuple May 3 '17 at 11:54

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    $\begingroup$ One call option is worth a 100 equity shares in the example $\endgroup$ – nimbus3000 Apr 25 '17 at 13:20
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    $\begingroup$ The investor has SOLD options on 2000 shares. He loses if the price goes up. He loses 2000*0.6 = 1200 if the price of options goes up by 0.6 $\endgroup$ – noob2 Apr 25 '17 at 13:31
  • $\begingroup$ If strike price $K= \$ 50$, we have that investor loss: $$(\$ 100 - \$ 50 ) \times 2,000 = \$ 100.000$$ If the stock price goes up by \$1, the investor loss: $$(\$ 101 - \$ 50 ) \times 2,000 = \$ 102.000$$ so, if the stock price goes up by \$1, the option contract produce a loss of \$ 2.000 It is not correct. Why? $\endgroup$ – Mike9 Apr 25 '17 at 13:46
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    $\begingroup$ You are using the formula $2000(S-K)^+ $ which is the value at expiration, the book is talking about the change in option value on a day before expiration. The option cannot be exercised yet, so it's market value is not given by your formula. $\endgroup$ – noob2 Apr 25 '17 at 14:18
  • $\begingroup$ Sorry if I insist and tank you for your comment but I'm a bit confused. If the call option price is \$ 10 and the investor sells 20 option contracts, the investor earns from the sale of options: $20 \times \$ 10 = \$ 200 $ If the price of the option goes up by \$ 0.6, the investor could earn $20 \times \$ 10.6 = \$ 212 $. So the difference is $ \$ 12$ $\endgroup$ – Mike9 Apr 25 '17 at 14:47
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We denote by $C(S_0, K)$ the price for a call option with payoff $(S_T-K)^+$ at the option maturity $T.$ Here $S_0=100$ is the spot stock price. Generally, \begin{align*} C(S_0, K) \ne (S_0-K)^+. \end{align*} Moreover, \begin{align*} C(S_0+\Delta, K)-C(S_0, K) \approx \frac{\partial C}{\partial S_0} \Delta, \end{align*} where $\frac{\partial C}{\partial S_0}=0.6$ is the delta hedge ratio. If the stock price go up by $\Delta = \$1$, the shorted option position will loss \begin{align*} \frac{\partial C}{\partial S_0} \Delta = 0.6 \times \$1 = \$0.60. \end{align*} Then the whole option position loss is $2,000 \times \$0.60 = \$1,200$.

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  • $\begingroup$ My confusion stems from the fact that the value of 1 option contract is \$ 10 written on 100 shares ($2.000 / 20$ option contracts). $$ C (S_0, K) = \$ 10 $$ So $$20 option contracts \times \$ 10 = \$ 200 $$ Could you clarify this aspect? $\endgroup$ – Mike9 Apr 25 '17 at 22:09
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    $\begingroup$ After the option is sold, that sold option price does not matter any more. Here, we consider the option price change as the underlying stock changes. $\endgroup$ – Gordon Apr 25 '17 at 22:15
  • $\begingroup$ I'm so sorry but I don't understand why the option position loss is $2,000 \times 0,60 = \$ 1,200$ ? $\endgroup$ – Mike9 Apr 25 '17 at 22:24
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    $\begingroup$ The option notional is 2,000: each option has notional of 100, and you sold 20 options; that is, options on 2,000 shares, as you described. $\endgroup$ – Gordon Apr 25 '17 at 23:26
  • $\begingroup$ OK! I got it. Thank you so much for clarification and for patience $\endgroup$ – Mike9 Apr 26 '17 at 7:48

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