Pricing of Black-Scholes with dividend

Question

Consider the payoff $g(S_T)$ shown in the figure below. Consider Black-Scholes model for the price of a risky asset with $T = 1$, $r = .04$, and $\sigma = .02$ and dividends are paid quarterly with dividend yield $10\%$. Take $S_0 = 10$, $K_1 = 9$, and $K_2 = 11$. Find the Black-Scholes price, $\Delta$, $\Gamma$, $\rho$, and $\mathcal{V}$ of this option at time $t = 0$. Find $\Theta$ at time $t = 0$ without taking derivatives with respect to $S$.

Solution: The payoff is, $$g(S_t) = (S_t - K_1)_{+} - 2(S_t - \frac{(K_1 + K_2)}{2})_{+} + (S_t - K_2)_{+}$$ The Black-Scholes formula with dividend gives \begin{align*} V(t = 0,S) &= e^{-r\tau}\hat{\mathbb{E}}[g(\tilde{d}S_T)]\\ &= \tilde{d}\left(BS_{call}(\frac{K_1}{\tilde{d}}) - 2BS_{call}(\frac{K_1+K_2}{2\tilde{d}}) + BS_{call}(\frac{K_2}{\tilde{d}})\right) \end{align*} where $$\tilde{d} = \left(1 - \frac{d}{4} \right)^{4} = .9037$$ So, $$V(t = 0,S) = e^{-r\tau}\hat{\mathbb{E}}[g(\tilde{d}S_T)] = (.9037)((0) - 2(0) + (0)) \approx 0 $$ For the Greeks we have $$\Delta = \partial_S V(t = 0,S) = \tilde{d}\left[\Phi(d_1(\frac{K_1}{\tilde{d}})) + \Phi(d_1(\frac{K_1+K_2}{2\tilde{d}})) + N(d_1(\frac{K_2}{\tilde{d}})) \right] \approx 0$$ $$\Gamma = \partial_{SS}V(t = 0, S) = 0$$ $$\rho = \partial_r V(t = 0,S) = \left( e^{-rt}(\frac{K_1}{\tilde{d}})(t)\Phi(d_2) + e^{-rt}(\frac{K_1+K_2}{2\tilde{d}})(t)\Phi(d_2) + e^{-rt}(\frac{K_2}{\tilde{d}})(t)\Phi(d_2)\right) \approx 0$$ $$\mathcal{V} = (S\sqrt{t}\Phi(d_1) + S\sqrt{t}\Phi(d_1) + S\sqrt{t}\Phi(d_1)) \approx 0$$

Answer 1

In your answer, you don't include dividend. I am sorry to say it is wrong.

Payoff function is $$ g(S_T) = (S_T - K_1)_+ - 2(S_T - \frac{K_1+K_2}{2})_+ + (S_T - K_2)_+ $$

BS pricing formula with dividend gives $$ V(t=0,S) = e^{-r}E(g(\tilde{d}S_T)) = \tilde{d} \left(BS_{call}\left(\frac{K_1}{\tilde{d}}\right) - 2BS_{call}\left(\frac{K_1+K_2}{2 \tilde{d}}\right) + BS_{call}\left(\frac{K_2}{\tilde{d}}\right)\right) $$

Where $$ \tilde{d} = (1-\frac{d}{4})^4 = 0.9037 $$ Plug in all numbers in your question, I get 0.3905 (double check it yourself).

As to the greeks,

$$ \Delta = \frac{\partial V(t=0,S)}{\partial S} = \tilde{d} \left[ N(d_1(K_1/\tilde{d})) - 2N(d_1((K_1+K_2)/(2\tilde{d}))) + N(d_1(K_2/\tilde{d})) \right] $$ where $N$ is the normal c.d.f. and $d_1(K)=\frac{log(S/K)+(r+\frac{1}{2}\sigma)\tau}{sigma\sqrt{\tau}}$.

Rest of the problem is pretty trivial because all greeks are just linear combination of original BS greeks. You just need change the strike price in the original BS greeks (I guess?). I won't go through all the calculation here.

Let me know if anything is not clear.