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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.

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  • $\begingroup$ 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. $\endgroup$
    – emot
    Jul 29 at 14:47
  • $\begingroup$ 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). $\endgroup$
    – emot
    Jul 29 at 15:31
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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)

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  • $\begingroup$ 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? $\endgroup$
    – emot
    Jul 30 at 12:36
  • $\begingroup$ 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. $\endgroup$
    – Quantuple
    Aug 2 at 8:50
  • $\begingroup$ Would you care to look at my question quant.stackexchange.com/q/67909/6686? Thank you. $\endgroup$
    – Hans
    Sep 14 at 0:29
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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.

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    $\begingroup$ I now better see what you mean. Agreed! $\endgroup$
    – Quantuple
    Aug 2 at 11:28

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