Two-period binomial model with dividends

Consider a two-period binomial model for a risky asset with each period equal to a year and take $S_0 = 1$, $u = 1.15$ and $l = 0.95$. The interest rate is $R = .05$.

a.) If the asset pays 10% of its value as dividend in the first period and 20% in the second period, find the price of the ATM call option.

b.) Consider a more complicated dividend strategy which pays 10% dividend only if the price moves up and no dividend if the price moves down at each period. Find the price of the ATM call option.

Forgive me my professor can have many errors in his problems. I am not sure how to solve this problem and what ATM call option mean. We have never covered dividends in regards to binomial model. I need some help with this any suggestions is greatly appreciated.

• Could you provide a solution I am having a difficult time with your notation are you saying $S_0 u = .9$? – Wolfy Mar 16 '16 at 22:45

(a) First of all, ATM means strike price $K=S_0$. By the end of second period, the risky asset has values (from top to down) $S_0 u^2 (1-d_1) (1-d_2)=0.95$, $S_0 ul (1-d_1) (1-d_2)=0.79$ and $S_0 l^2 (1-d_1) (1-d_2)=0.65$. The risk neutral probabilities are calculated as $\hat{\pi}_u=(1+R-l)/(u-l)=1/2$ and $\hat{\pi}_l=(u-R-1)/(u-l)=1/2$. Hence three stages have probabilities 1/4, 1/2 and 1/4 respectively.

The option is determined by the terminal price $$V(S_T)=(S_T-K)_+$$ However, if $K=S_0=1$, then $V(S_T)=0$ for sure. Then price of the ATM call will be $$C=\frac{1}{(1+R)^2} \hat{E}(V(S_T))=0.$$ If you change $K=0.8$, then $$C=\frac{1}{(1+R)^2} \hat{E}(V(S_T))= \frac{1}{4}(0.95-0.8) + \frac{1}{2}0 + \frac{1}{4}0.$$

(b)

Terminal stages are $S_0 u^2 (1-d)^2=0.93$, $S_0 u l (1-d)=0.98$ and $S_0 l^2=0.9$. Neutral probabilities won't change. Just repeat steps in part (a) you will get zero for $K=S_0$ and a nonzero price for $K=0.8$.

• I wonder if this is really correct... can you have a look at my second attempt of answer? – Quantuple Mar 18 '16 at 18:36
• The idea behind this problem is pretty much the same as Exercise 2.10 in Shreve's book Vol I. Try to get a copy of that volume and do that problem. Especially how the risk-neutral measure is defined for dividend-paying stock. – Pandaaaaaaa Mar 18 '16 at 21:26
• So you are saying that risk neutral probabilities do not involve the div yield? Interesting, I'll see in the Shreve cause this does sound counter intuitive to me. – Quantuple Mar 18 '16 at 21:30
• You are right, risk-neutral measurement does not involve the dividend yield. – Pandaaaaaaa Mar 18 '16 at 21:33
• What about the $q = (exp(r-y)\Delta t - l)/(u-l)$ I can find in many papers/books. See wikipage for instance. Clearly the $y$ figures the div yield. If you look at my answer it makes sense that it should appear... no? – Quantuple Mar 18 '16 at 21:36

I did the calculations, and actually both current answers seem wrong... can someone confirm or point towards any mistake in the following reasoning?

What I claim to show is that: you can use a standard stock price tree (i.e. the terminal values of the stock price do not involve the dividend yield and write $S_0 u^2$, $S_0 ul$ and $S_0 l^2$), but your risk-neutral probabilities will depend on the dividend yield $Y$.

In Exercise A, this can be seen by writing the replicating portfolio: $\Pi = \Delta S + B$, which by the end of a period should satisfy $$\Pi_u = \Delta S_u + \beta (1 + R) + {\color{red}{\Delta S_u Y}} = V_u$$ $$\Pi_d = \Delta S_d + \beta (1 + R) + {\color{red}{\Delta S_d Y}} = V_d$$ Where the additional terms come from the proportional dividend payments that are reinvested in the cash account at the end of each period. These equations are equivalent to $$\Delta S_u (1 + Y) + \beta (1 + R) = V_u$$ $$\Delta S_d (1 + Y)+ \beta (1 + R) = V_d$$ Solving them yields to (it is as if $u$ and $d$ in the standard formulae were replaced by $u(1+Y)$ and $d(1+Y)$ $$q_u = \frac{ \frac{1+R}{1+Y} - d}{u - d}$$ $$q_l = 1 - q_u$$ This result seems consistent with many papers. Even Wikipedia exhibits something similar with $$q_u = \frac{e^{(r-q)\Delta t} - d}{u -d}$$ Plus, it is straightforward to show that these probabilities provide the right forward price if you evaluate $F(0,t)=E[S(t)]=qS_u+(1-q)S_d=S_0(1+R)/(1+Y)$

Yet, because $(1+R)/(1+Y)$ here is smaller than $d$ on the second period, I doubt that this problem is feasible at all.

In Exercise B, using the same rationale I end up with $$\Delta S_u + \beta (1 + R) + {\color{red}{\Delta S_u Y}} = V_u$$ $$\Delta S_d + \beta (1 + R) = V_d$$ and $$q_u = \frac{ 1+R - d}{u(1+Y) - d}$$ $$q_l = 1 - q_u$$ and now the problem indeed seems feasible (no negative probabilities). But this is never a question of the strike contrary to what the OP's professor claims.