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I have seen the question here and have gone through the answer, but I still don't fully understand why the approach below, based on no-arbitrage, yields a different answer.

To summarize:

  • At time $t_0$, I borrow $S_0$ cash and I immediately spend it to buy one unit of stock

  • Whilst holding the stock, dividends are continuously compounded at a constant rate $q$ and reinvested into the stock until time $t$, at which point in time I will stop reinvesting and just take the cash, the value of which would be $D_t$

  • A time $t$, I need to repay the loaned money $S_0$, which has accumulated a continuously compounded constant interest rate $r$, i.e. $S_0e^{rt}$. I will also receive the cash from the counterparty for the forward (i.e. $F(t_0,t)$, which is the forward price agreed upon at time $t_0$), and I will need to deliver the 1 unit of stock to the counterparty, which I have held the whole time until $t$.

These transactions are summarized in the table below:

enter image description here

Clearly for there to be no opportunity to make cash out of thin air, i.e. for the forward $F(t_0,t)$ to generate no arbitrage, we must trivially have at time $t$ that:

$$-S_0e^{rt}+F(t_0,t)+D_t=0$$

Now let's elaborate on the value of $D_t$, since it seems to have various formulas assigned in different answers on this page (apologies if this is over-laboring the point). First, assume that the dividend is discrete and paid at time $t$ at the forward maturity: clearly, the value of $D_t$ would be $qS_{t}$.

Now if we split the time domain into two equal parts, and assume that the dividend rate would be $\frac{q}{2}$ paid at the mid-point, re-invested into the stock and then another dividend at rate $\frac{q}{2}$ paid at maturity, we would get the following transactions (for notation simplicity, in the table below, $t_1$ is the midway point in time, with $t_{1_{-}}$ and $t_{1_{+}}$ being the infinitesimal points in time just before and just after $t_1$. The maturity is then denoted $t_2$. I assume that once the dividend gets paid, it gets immediately reinvested):

enter image description here

Clearly, if we keep splitting the time domain into increasingly larger number of $n$ parts and take the limit $n\to\infty$, the formula in the table converges to $S_te^{q}$, (since $\lim_{n\to\infty}\left(1+\frac{q}{n}\right)^n=e^q$.) We assumed that the time domain was 1 unit of time.

Generalizing, clearly the value of the continuously compounded and simultaneously reinvested dividends, INCLUDING the value of the 1-unit of stock that has been held the whole time, would be $S_te^{qt}$.

At maturity, value of the left-over dividends would then be: $$D_t=S_te^{qt}-S_t=S_t(e^{qt}-1)$$

Going back to the no-arbitrage equation, since $D_t$ is stochastic, we need to take an expectation (as rightfully pointed out in the comments):

$$-S_0e^{rt}+F(t_0,t)+\mathbb{E}^Q_{t_0}[D_t]=0$$

i.e.

$$F(t_0,t)=S_0e^{rt}-\mathbb{E}^Q_{t_0}[D_t]=\\=S_0e^{rt}-\mathbb{E}^Q_{t_0}[S_t(e^{qt}-1)]=\\=S_0e^{rt}(2-e^{qt})$$

This answer is obviously different to the one given in the linked question, and also different to the answer below. If possible, please point out where the difference might be coming from?

EDIT: As per Kurt's comment, the solution is trivial. Instead of borrowing $S_0$ money at $t_0$ abnd buying a whole 1 unit of stock, it's enough to borrow just $S_0e^{-qt}$ to buy $e^{-qt}$ units of stock, and use the dividends to grow this to 1 unit at maturity, as per the table below:

enter image description here

Trivially, at maturity, we get:

$$F(t_0,t)=S_0e^{rt-qt}$$

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  • $\begingroup$ The basic difference is that your formula has a term in it that is not known at $t_0$ (The integral) so it can’t be correct. Essentially the other answer has replaced that term by its expectation. $\endgroup$
    – dm63
    Nov 8, 2022 at 14:57
  • $\begingroup$ @dm63: thank you. I have taken the expectation and I still get the wrong answer. $\endgroup$
    – Conductor
    Nov 8, 2022 at 15:28
  • $\begingroup$ Smart ! You are trying to set up a portfolio in which, instead of investing the paid dividends into the money market account $e^{rt}$, they get invested into buying more shares of the stock $S_t$. This is not uncommon in practice. I thought about this recently and found it quite complicated to set up an equation that reflects the fact that every newly bought piece of share $qS_t$ immediately pays its own dividends that must be reinvested also. It seems easier to do this with a discrete dividend that is just paid once a year. $\endgroup$
    – Kurt G.
    Nov 8, 2022 at 18:28
  • $\begingroup$ @KurtG.: say I would invest the dividends into the money market account, not the stock. The transaction table above would still be valid: wouldn't I still get a different result to yours? $\endgroup$
    – Conductor
    Nov 8, 2022 at 18:42
  • $\begingroup$ Your $D$ should then be $D_t=\int_0^tqS_u e^{-ru}\,du$. Then you should get the same forward. $\endgroup$
    – Kurt G.
    Nov 8, 2022 at 18:51

1 Answer 1

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It is in fact simple to constructing the portfolio $\Pi_t$ in which dividends are reinvested into buying more shares of the stock, instead of putting that cash into the money market account $e^{\int_0^t r(s)\,ds}\,,$ as it was done alternatively in the linked answer: Namely, when the time interval is divided into steps $\Delta t$ we have $$\tag{A} \Pi_t=S_t\prod_{k=1}^{\lfloor t/\Delta t\rfloor}\Big(1+q\,\Delta t\Big) $$ which reflects the fact that at each dividend date $t=k\Delta t$ the portfolio value increases by $\Pi_t\,q\,\Delta t$ which is the value of the newly bought shares. This formula also reflects the fact that newly bought shares themselves pay dividends that get reinvested. In the limit $\Delta t\to 0$ we get $$\tag{B} \Pi_t=S_t\,e^{qt}\,. $$ In the Black-Scholes case this is $\Pi_t=S_0\,e^{rt+\sigma W_t-\frac{\sigma^2 t}{2}}\,,$ the GBM of the non dividend paying stock. From no arbitrage it follows that $$\tag{C} e^{-\int_0^tr(s)\,ds}\,\Pi_t $$ must be a martingale. Therefore, $$\tag{D} \Pi_0=S_0=\textstyle\mathbb E\Big[e^{-\int_0^tr(s)\,ds}\,S_t\Big]\,e^{qt}\,. $$ Because the forward $F_t$ of the stock is -as always- defined by $$ \tag{E} \mathbb E\left[e^{-\int_0^tr(s)\,ds}\right]F_t-\mathbb E\left[e^{-\int_0^tr(s)\,ds}S_t\right]=0\, $$ it follows from (D) that it is the same as in the linked answer, namely $$ \tag{F} \boxed{F_t=\frac{S_0\,e^{-q t}}{p_t}\,} $$ where $p_t=\mathbb E[e^{-\int_0^tr(s)\,ds}]\,.$

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  • $\begingroup$ Thank you, Kurt. Painfully, I still get a different answer, even after changing the $D_t$ value to $S_te^{qt}$. I think maybe the difference comes from the fact that I assume that the arbitrage must not happen between two discrete points in time, whilst somehow you assume it must not occur between two points infenitesimally close to each other? Also, I don't discount the value of everything to PV, but that is Ok if rates are deterministic and shouldn't make any difference. $\endgroup$
    – Conductor
    Nov 9, 2022 at 15:26
  • $\begingroup$ It looks like you are assuming that initially you have to borrow $S_0$ to buy the full stock. This is unnecessary. We only have to buy $S_0e^{-qt}$ because this will earn us dividends and grow to $S_t$ at $t$. Pay back $F_t=S_0e^{-qt+rt}$ at $t$ gives you the correct forward. $\endgroup$
    – Kurt G.
    Nov 9, 2022 at 18:06
  • $\begingroup$ OMG, that was so simple!!! :D :D :D Thank you. $\endgroup$
    – Conductor
    Nov 9, 2022 at 20:00
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    $\begingroup$ It was more than worth getting to the bottom of it. I learned a lot as well. $\endgroup$
    – Kurt G.
    Nov 9, 2022 at 20:02

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