# A question about exercise from "Paul Wilmott introduces Quantitative Finance"

I am new on this forum and i have just begun my adventure with finances, so please be patient.

I was solving exercises from "Paul Wilmot introduces Quantitative Finance" and i came across the following task(exercise 6, chapter 1):

A particular forward contract costs nothing to enter into at time $$t$$ and obliges the holder to buy the asset for an amount $$F$$ at expiry, $$T$$ . The asset pays a dividend $$DS$$ at time $$t_d$$ , where $$0\le D\le 1$$ and $$t \le t_d \le T$$ . Use an arbitrage argument to find the forward price, $$F (t)$$ .

And there was also a hint:

Hint: Consider the point of view of the writer of the contract when the dividend is re-invested immediately in the asset.

As far i understand, because there is no arbitrage opportunity, i should earn nothing. So it implies that all the profit i have got from dividends or interest rate from my bank account is equal to $$F$$.

Here is my plan how to evaluate the $$F(t)$$

1. In order to get the dividends from particular stock, firstly i must buy their share. ($$-S(t)$$)
2. I get the dividends ($$+D S(t_d)$$)
3. I sell the share ($$+S(t_d)$$)
4. Assuming that $$(D+1)S(t_d)-S(t)>0$$ I can put it in my bank account with interest rate $$r$$, so i do this and after time $$T-t_d$$ i have earned $$[(D+1)S(t_d)-S(t)]e^{r(T-t_d)}$$
5. This is reduced by the costs $$F$$, so: $$F(t) = [(D+1)S(t_d)-S(t)]e^{r(T-t_d)}$$

However official answer(https://www.wiley.com/legacy/wileychi/pwiqf2/supp/c01.pdf) is $$F(t) = (1-D)S(t)e^{r(T-t)}$$ and i don't understand his explanation.

I would be grateful if someone explain me why i am wrong.

• And what is the value $S(t_d)$? This value is stochastic and we don't know it at $t=0$, so you don't know what is F in your equation. You assumed you that you need 1 unit of the stock in step 1, but actually you can use dividends to buy more shares to end up with 1 unit of share in the end. Therefore to hedge it in step 1 you have to buy (1-D) units of S not 1, then reinvest dividend back to stock.
– emot
Jul 29 at 14:47
• Also in step 1 you have to borrow cash to buy stock, so you have to pay interest on that, so your position at step 1 is Stock - cash borrowing (worth 0).
– emot
Jul 29 at 15:31

You ought to compare the $$t$$-values of two self-financing strategies, under the assumption that there exists a risk-free money market account and that the dividend is deterministic but proportional to the random stock price.

Strategy 1 - Entering a forward contract

• At inception ($$t=0$$), you do not pay anything by definition, $$\Pi_1(0)=0$$
• At maturity ($$t=T)$$, you pay the forward price and receive the stock (whether cash/physical settlement): $$\Pi_1(T)=-F(0,T)+S(T)$$

Strategy 2 - Cash & carry, assuming proportional dividend

• At inception ($$t=0$$), you borrow cash and purchase the stock, $$\Pi_2(0)=-S(0) + S(0) = 0$$
• At dividend ex-date ($$t=t_d$$), you receive $$DS(t_d)$$ as an extra cash proceed. Your current cash balance is then $$\Pi_2(t_d) = -S_0 e^{rt_d} + D S(t_d) + S(t_d)$$, the first reflecting what you need to give back to your lender (borrow), the second the cash proceed from the dividend, the last being your long stock pose.
• At maturity ($$t=T$$) you are left with $$\Pi_2(T) = -S(0) e^{rT} + D S(t_d) e^{r(T-t_d)} + S(T)$$ It's the same idea as at $$t=t_d$$ except all cash has grown at the risk-free rate.

Arbitrage-free pricing

Suppose you create a strategy $$\Pi$$ where you implement being long strat 1 and short strat 2 simultaneously. $$\Pi$$ is entered at at zero cost by design. Its payout at $$T$$ should hence be zero in expectation to preclude any arbitrage opportunity: $$\Bbb{E}_0[ \Pi(T) ] = \Bbb{E}_0[ \Pi_1(T) - \Pi_2(T) ] = \Bbb{E}_0[ - F(0,T) + S(T) + S(0) e^{rT} - D S(t_d) e^{r(T-t_d)} - S(T) ] = 0$$ which yields \begin{align} F(0,T) &= \Bbb{E}_0[ S(0)e^{rT} - D S(t_d)e^{-rt_d} e^{rT} ] \\ &= (S(0) - D \Bbb{E}_0[ S(t_d)e^{-rt_d} ]) e^{rT} \\ &= S(0)(1 - D)e^{rT} \end{align} where the last line leverages the fact in between any capital distribution (i.e. here prior to the dividend payment), investing in the stock constitutes a self-financing strategy (hence stock priced expressed in the risk-free numéraire, i.e. dicounted stock prrice, should be martingale).

REM I just saw that in your OP you consider a generic $$t$$ hence time to maturity $$T-t$$, I just gave the example for $$t=0$$ hence time to maturity $$T$$ (generalisation should be straightforward)

• Your point is valid and I agree, but I wanted to add that the payoff only matches in expectation and it doesn't mean that the strategy 2 should be executed. From Strategy 1 the payoff is: $$-S(0)e^{rT}+DS(0)e^{rT}+S(T)$$ From Strategy 2 we have: $$-S(0)e^{rT}+DS(t_d)e^{r(T-t_d)}+S(T)$$ and it only matches when $S(t_d)=Se^{rt_d}$. In order to replicate forward payoff we have to buy less than 1 unit of stock and reinvest dividend it back to stock. Do you agree?
– emot
Jul 30 at 12:36
• They only match in expectation because the dividend (in cash) is stochastic (it's a fixed proportion, but a fixed proportion of a random quantity, $S(t_d)$). I'm not sure how you obtain the above payoff for strategy 1: when you enter a forward contract you are blind to divs, you just contractually pay at $T$ the price you've agreed upon at $0$ and get the stocks in exchange: $-F(0,T)+S(T)$. No div appears in that strat, so no, I don't agree a priori. Aug 2 at 8:50
• Would you care to look at my question quant.stackexchange.com/q/67909/6686? Thank you.
– Hans
Sep 14 at 0:29

Although the answer to this question has been provided, I would like to give another point of view to this problem. It is worth noting that at $$t=0$$ you should hold less than 1 units of stock ($$(1-D)$$ units to be precise) to replicate forward payoff. Below I present replicating strategy.

• At $$t=0$$ you buy $$1-D$$ units of stock worth $$(1-D)S(0)$$, you finance it by borrowing $$(1-D)S(0)$$ from the bank account. Portfolio value at $$t=0$$ is thus: $$\Pi_0=(1-D)S(0)-(1-D)S(0)=0$$
• At dividend date $$t=t_d$$ you receive $$S(t_d)(1-D)D$$ dividend and then reinvest it fully back to stock now worth $$S(t_d^{+})=S(t_d)(1-D)$$ - it's value dropped due to dividend. Therefore from dividend proceeds you buy additional $$\frac{(1-D)S(t_d)D}{S(t_d)(1-D)}$$ units of stock $$S(t_d^{+})$$

Portfolio value is thus: $$\Pi_{t_d^{+}}=(1-D)S(t_d^{+})+(1-D)S(0)e^{rt_d}+\frac{(1-D)S(t_d)D}{S(t_d)(1-D)}S(t_d^{+})$$ where $$t_d^{+}$$ indicates time just after dividend cut-off.

• At time $$t=T$$ portfolio value is therefore: $$\Pi_{T}=(1-D)S(T)+(1-D)S(0)e^{rT}+\frac{(1-D)S(t_d)D}{S(t_d)(1-D)}S(T)=S(T)-(1-D)S(0)e^{rT}$$

Long forward payoff is $$S(T)-F(0,T)$$. The value of $$F(0,T)$$ is the value of the financing cost, therefore $$F(0,T)=(1-D)S(0)e^{rT}$$

Note that all stochasticity i.e. the term $$S(t_d)D$$ dropped from our equation, therefore $$F(0,T)$$ is known.

• I now better see what you mean. Agreed! Aug 2 at 11:28