# Pricing of Black-Scholes with dividend

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 $$\Delta = \partial_S V(t = 0,S) = \tilde{d}\left[\Phi(d_1(\frac{K_1}{\tilde{d}})) + \Phi(d_1(\frac{K_1+K_2}{2\tilde{d}})) + N(d_1(\frac{K_2}{\tilde{d}})) \right] \approx 0$$ $$\Gamma = \partial_{SS}V(t = 0, S) = 0$$ $$\rho = \partial_r V(t = 0,S) = \left( e^{-rt}(\frac{K_1}{\tilde{d}})(t)\Phi(d_2) + e^{-rt}(\frac{K_1+K_2}{2\tilde{d}})(t)\Phi(d_2) + e^{-rt}(\frac{K_2}{\tilde{d}})(t)\Phi(d_2)\right) \approx 0$$ $$\mathcal{V} = (S\sqrt{t}\Phi(d_1) + S\sqrt{t}\Phi(d_1) + S\sqrt{t}\Phi(d_1)) \approx 0$$

• Hint: read the question again; it is asking about price and greeks at t=0, not t=1. Apr 5, 2016 at 0:56
• @onlyvix.blogspot.com How is my solution now? Apr 6, 2016 at 23:11

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.

• Just so I understand I need to calculate the greeks gamma, rho, and vega as well? Apr 5, 2016 at 19:39
• Also, if you don't mind could you post the $BS_{call}$ formula for dividend. Apr 5, 2016 at 19:49
• As in $BS_{call}(K_1/\tilde{d})$ If you can show me the formula you are using that would be great as I am not sure how to apply your way with what I have in my notes Apr 5, 2016 at 20:05
• Calculate all greeks except $\Theta$, then you find $\Theta$ via BS PDE. Please refer to this note (econ-pol.unisi.it/fm10/greeksBS.pdf). $BS_{call}$ is exact the same as C in this note.You need to change strike price $K$ to $K/\tilde{d}$. Apr 5, 2016 at 20:29
• Sorry to abuse the notation $BS_{call}(K_1/\tilde{d})$. All I want to say is to change the strike price to $K/\tilde{d}$ in the BS call option pricing formula. Apr 5, 2016 at 20:30