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Answer: Good choice. You earned $10. Poor choice. You lost $10.

Using the Black-Scholes-Merton formula, first find $d_1$ and $d_2$:

$$\begin{aligned} d_1 &= \dfrac{\ln⁡[S_0/K]+(r+\sigma^2/2).T}{\sigma.\sqrt{T}} \\ &= \dfrac{\ln⁡[5/4]+(0.1+0.47^2/2) \times 1}{0.47 \times \sqrt{1}} \\ &= \dfrac{0.223143551 + 0.21045}{0.47} \\ &= 0.922539471 \\ \end{aligned}$$ $$\begin{aligned} d_2 &= d_1-\sigma.\sqrt{T}=\dfrac{\ln⁡[S_0/K]+(r-\sigma^2/2).T}{\sigma.\sqrt{T}} \\ &= 0.922539471 - 0.47 \times \sqrt{1} \\ &= 0.452539471 \\ \end{aligned}$$

Now we can find the European style call option price now $(c_0)$ using the Black-Scholes-Merton formula:

$$\begin{aligned} c_0 &= S_0.N[d_1] - K_T.e^{-r.T}.N[d_2] \\ &= 5 \times N[0.922539471] - 4 \times e^{-0.1 \times 1} \times N[0.452539471] \\ &= 5 \times 0.821876374 - 4 \times e^{-0.1 \times 1} \times 0.674559804 \\ &= 4.10938187 - 2.441467805 \\ &= 1.66791407 \\ \end{aligned}$$ Comments

The Delta of the call option is $N[d_1]$. This is the option's gradient at $S_0$ on the call option price ($c_0$ on the y-axis) versus underlying asset price ($S_0$ on the x-axis) graph.

$$\text{CallOptionDelta} = N[d_1] = N[0.922539471] = 0.821876374$$

So a one cent ($0.01) increase in the underlying stock price from $5 to $5.01 will result in a 0.8219 cent ($0.08219) increase in the option price.

The risk-neutral probability of the call option maturing in-the-money $(S_T > K_T)$ is $N[d_2]$.

$$\text{RiskNeutralProbabilityOfCallOptionMaturingInTheMoney} \\ = N[d_2] = N[0.452539471] = 0.674559804$$

Answer: Good choice. You earned $10. Poor choice. You lost $10.

Using the Black-Scholes-Merton formula, first find $d_1$ and $d_2$:

$$\begin{aligned} d_1 &= \dfrac{\ln⁡[S_0/K]+(r+\sigma^2/2).T}{\sigma.\sqrt{T}} \\ &= \dfrac{\ln⁡[5/4]+(0.1+0.47^2/2) \times 1}{0.47 \times \sqrt{1}} \\ &= \dfrac{0.223143551 + 0.21045}{0.47} \\ &= 0.922539471 \\ \end{aligned}$$ $$\begin{aligned} d_2 &= d_1-\sigma.\sqrt{T}=\dfrac{\ln⁡[S_0/K]+(r-\sigma^2/2).T}{\sigma.\sqrt{T}} \\ &= 0.922539471 - 0.47 \times \sqrt{1} \\ &= 0.452539471 \\ \end{aligned}$$

Now we can find the European style put option price now $(p_0)$ using the Black-Scholes-Merton formula:

$$\begin{aligned} p_0 &= -S_0.N[-d_1] + K_T.e^{-r.T}.N[-d_2] \\ &= -5 \times N[-0.922539471] + 4 \times e^{-0.1 \times 1} \times N[-0.452539471] \\ &= -5 \times 0.178123626 + 4 \times e^{-0.1 \times 1} \times 0.325440196 \\ &= -0.89061813 + 1.177881868 \\ &= 0.287263739 \\ \end{aligned}$$ Comments

The Delta of the put option is $-N[-d_1]$. This is the option's gradient at $S_0$ on the put option price ($p_0$ on the y-axis) versus underlying asset price ($S_0$ on the x-axis) graph.

$$\text{PutOptionDelta} = -N[-d_1] = -N[-0.922539471] = -0.178123626$$

So a one cent ($0.01) increase in the underlying stock price from $5 to $5.01 will result in a 0.178 cent ($0.00178) decrease in the option price.

The risk-neutral probability of the put option maturing in-the-money $(S_T < K_T)$ is $N[-d_2]$.

$$\text{RiskNeutralProbabilityOfPutOptionMaturingInTheMoney} \\ = N[-d_2] = N[-0.452539471] = 0.325440196$$

Answer: Good choice. You earned $10. Poor choice. You lost $10.

Using the Black-Scholes-Merton formula, first find $d_1$ and $d_2$. Remember to replace all instances of $S_0$ with $S_0.e^{-q.T}$ since the stock index pays a continuously compounded dividend yield $q$ pa.

$$\begin{aligned} d_1 &= \dfrac{\ln⁡[S_0.e^{-q.T}/K]+(r+\sigma^2/2).T}{\sigma.\sqrt{T}} \\ &= \dfrac{\ln⁡[2700 \times e^{-0.02 \times 0.5}/2800]+(0.05+0.25^2/2) \times 0.5}{0.25 \times \sqrt{0.5}} \\ &= \dfrac{\ln⁡[2673.134551/2800]+0.040625}{0.176776695} \\ &= \dfrac{-0.046367644+0.040625}{0.176776695} \\ &= -0.032485301 \\ \end{aligned}$$ $$\begin{aligned} d_2 &= d_1-\sigma.\sqrt{T}=\dfrac{\ln⁡[S_0.e^{-q.T}/K]+(r-\sigma^2/2).T)}{\sigma.\sqrt{T}} \\ &= -0.032485301 - 0.25 \times \sqrt{0.5} \\ &= -0.209261996 \\ \end{aligned}$$

We can find the European style call option price now $(c_0)$ using the Black-Scholes-Merton formula, again replacing $S_0$ with $S_0.e^{-q.T}$ to take the continuously compounded dividend yield pa $q$ into account:

$$\begin{aligned} c_0 &= S_0.e^{-q.T}.N[d_1] - K_T.e^{-r.T}.N[d_2] \\ &= 2700 \times e^{-0.02 \times 0.5} \times N[-0.032485301] - 2800 \times e^{-0.05 \times 0.5} \times N[-0.209261996] \\ &= 2700 \times e^{-0.02 \times 0.5} \times 0.487042519 - 2800 \times e^{-0.05 \times 0.5} \times 0.417121859 \\ &= 1301.930185 - 1139.104633 \\ &= 162.8255519 \\ \end{aligned}$$

Answer: Good choice. You earned $10. Poor choice. You lost $10.

Using the Black-Scholes-Merton formula, first find $d_1$ and $d_2$. Remember to replace all instances of $S_0$ with $F_{0,T}.e^{-r.T}$. This is because $F_{0,T}$, the current $(t=0)$ futures price that matures at time $T$, takes into account the market's estimate of the underlying stock index's continuously compounded dividend yield which is not given in the question (see the comment at the end of this question for an alternative method):

$$\begin{aligned} d_1 &= \dfrac{\ln⁡[F_{0,T}.e^{-r.T}/K]+(r+\sigma^2/2).T}{\sigma.\sqrt{T}} \\ &= \dfrac{\ln⁡[2740.805274 \times e^{-0.05 \times 0.5}/2800]+(0.05+0.25^2/2) \times 0.5}{0.25 \times \sqrt{0.5}} \\ &= \dfrac{\ln⁡[2673.134551/2800]+0.040625}{0.176776695} \\ &= \dfrac{-0.046367644+0.040625}{0.176776695} \\ &= -0.032485301 \\ \end{aligned}$$ $$\begin{aligned} d_2 &= d_1-\sigma.\sqrt{T}=\dfrac{\ln⁡[F_{0,T}.e^{-r.T}/K]+(r-\sigma^2/2).T}{\sigma.\sqrt{T}} \\ &= -0.032485301 - 0.25 \times \sqrt{0.5} \\ &= -0.209261996 \\ \end{aligned}$$

Now we can find the European style call option price now $(c_0)$ using the Black-Scholes-Merton formula, again replacing $S_0$ with $F_{0,T}.e^{-r.T}$:

$$\begin{aligned} c_0 &= F_{0,T}.e^{-r.T}.N[d_1] - K_T.e^{-r.T}.N[d_2] \\ &= 2740.805274 \times e^{-0.05 \times 0.5} \times N[-0.032485301] - 2800 \times e^{-0.05 \times 0.5} \times N[-0.209261996] \\ &= 2740.805274 \times e^{-0.05 \times 0.5} \times 0.487042519 - 2800 \times e^{-0.05 \times 0.5} \times 0.417121859 \\ &= 1301.930185 - 1139.104633 \\ &= 162.8255519 \\ \end{aligned}$$

Commentary

Note that this question is almost identical to Question 903. This is because the futures price $(F_{0,T})$ of 2740.805274 points was chosen to match a 2% pa continuously compounded dividend yield $(q)$ in the underlying index:

$$F_{0,T} = S_0.e^{(r-q).T}$$ $$2740.805274 = 2700 \times e^{(0.05-q) \times 0.5}$$ $$q = 0.05 - ln[2740.805274/2700]/0.5 = 0.02$$

So another way to solve this question is to work out $q$ (which equals $r - ln[F_{0,T}/S_0]/T$) and then replace all instances of $S_0$ with $S_0.e^{-q.T}$ similarly to Question 903.

Answer: Good choice. You earned $10. Poor choice. You lost $10.

No explanation provided.

Answer: Good choice. You earned $10. Poor choice. You lost $10.

No explanation provided.

Answer: Good choice. You earned $10. Poor choice. You lost $10.

No explanation provided.

Answer: Good choice. You earned $10. Poor choice. You lost $10.

No explanation provided.

Answer: Good choice. You earned $10. Poor choice. You lost $10.

To Delta and Gamma hedge the 1000 long put options, short 2078.4226 call options and long 2397.9617 stocks. This is the correct answer to option E about Gamma and Delta hedging.

To Gamma hedge the 1000 puts, set the portfolio Gamma equal to zero. Note that the Gamma of the shares is zero since their payoff is a straight line, there's no curvature:

$$\Gamma_\text{portfolio} = n_\text{puts}.\Gamma_\text{put} + n_\text{calls}.\Gamma_\text{call} + n_\text{stocks}.\Gamma_\text{stock}$$ $$0 = 1000 \times 0.098885989 + n_\text{calls} \times 0.047577422 + n_\text{stocks} \times 0$$ $$\begin{aligned} n_\text{calls} &= \dfrac{-1000 \times 0.098885989}{0.047577422} \\ &= -2078.422592 \\ \end{aligned}$$

Since the number of calls to buy is negative, short 2078.422592 calls (sell the call options).

To Delta hedge the long 1000 puts and short 2078.422592 calls, set the portfolio Delta equal to zero. Note that the Delta of the shares is one since their payoff is a diagonal straight line with a gradient of one.

$$\Delta_\text{portfolio} = n_\text{puts}.\Delta_\text{put} + n_\text{calls}.\Delta_\text{call} + n_\text{stocks}.\Delta_\text{stock}$$ $$0 = 1000 \times -0.552034778 + -2078.422592 \times 0.888138405 + n_\text{stocks} \times 1$$ $$\begin{aligned} n_\text{stocks} &= 1000 \times 0.552034778 + 2078.422592 \times 0.888138405 \\ &= 2397.961704 \\ \end{aligned}$$

Since the number of stocks to buy is positive, long 2397.961704 stocks.

To simply Delta hedge the long 1000 puts without Gamma hedging, set the portfolio Delta of puts and stocks equal to zero. This is the answer to option D. Again, the Delta of the shares is one since their payoff is a diagonal straight line with a gradient of one.

$$\Delta_\text{portfolio} = n_\text{puts}.\Delta_\text{put} + n_\text{stocks}.\Delta_\text{stock}$$ $$0 = 1000 \times -0.552034778 + n_\text{stocks} \times 1$$ $$\begin{aligned} n_\text{stocks} &= 1000 \times 0.552034778 \\ &= 552.034778 \\ \end{aligned}$$

Since the number of stocks to buy is positive, long 552.034778 stocks.