# Delta Hedging: Clarification example of the book “Hull, Options, Futures, and Other Derivatives” [closed]

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

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? ## closed as off-topic by amdopt, LocalVolatility, msitt, chollida, QuantupleMay 3 '17 at 11:54 This question appears to be off-topic. The users who voted to close gave this specific reason: • "Basic financial questions are off-topic as they are assumed to be common knowledge for those studying or working in the field of quantitative finance." – amdopt, LocalVolatility, msitt, chollida, Quantuple If this question can be reworded to fit the rules in the help center, please edit the question. • One call option is worth a 100 equity shares in the example – nimbus3000 Apr 25 '17 at 13:20 • 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 – noob2 Apr 25 '17 at 13:31 • 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? – Mike9 Apr 25 '17 at 13:46 • 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. – noob2 Apr 25 '17 at 14:18 • 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$ – Mike9 Apr 25 '17 at 14:47

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 $\Delta = \$1, the shorted option position will loss \begin{align*} \frac{\partial C}{\partial S_0} \Delta = 0.6 \times \1 = \0.60. \end{align*} Then the whole option position loss is2,000 \times \$0.60 = \$1,200$. • 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? – Mike9 Apr 25 '17 at 22:09
• I'm so sorry but I don't understand why the option position loss is $2,000 \times 0,60 = \$ 1,200\$ ? – Mike9 Apr 25 '17 at 22:24