# How to derive forward price on stock with continuous dividend

Let $$F_{t,T}$$ be the forward price of a stock $$S$$ at time $$T$$ and $$t$$ be the current time. The stock pays a proportional continuous dividend at a rate of $$q$$ and the risk-free rate is $$r$$. How can I prove that the price is given by $$F_{t,T} = S_{t}e^{(r-q)(T-t)}$$, preferably with a no arbitrage argument?

In the no dividend case, I know the derivation can be done with a no arbitrage argument: the forward payoff at $$T$$ is $$S_{T}-F_{t, T}$$, so a replicating portfolio consists of one unit of stock with current price $$S_t$$ and $$F_{t, T}e^{-r(T-t)}$$ units of cash. Since $$F_{t,T}$$ is chosen to make the initial value of the forward zero, we have $$F_{t,T} = S_te^{rt}$$.

In the dividend case, I am not sure what the terminal payoff should be. I feel like we would need to subtract the accumulated value of the dividends, but I am not sure what form it should take in order to get $$F_{t,T} = S_{t}e^{(r-q)(T-t)}$$. My initial guess was $$\text{AV}(\text{div})_{T} = \int_{t}^{T}q S_t dt$$, since $$qS_tdt$$ is the dividend payment per share on $$[t, t+dt]$$, but this seems wrong as I do not know how to remove the integral.

Note: If possible, I do not want to reference a risk-neutral measure or the Black-Scholes framework. I believe the equation for $$F_{t,T}$$ should hold as long as there is no arbitrage, but please correct me if I am wrong here.

When the dividend yield $$q$$ is constant one can in fact derive a very simple forward formula under no model assumptions on $$S_t$$ (see (4) below). Only no arbitrage arguments are needed:

The forward price $$F_t$$ with maturity $$t$$ is by definition the solution of the equation $$\tag{1} \mathbb E\left[e^{-\int_0^tr(s)\,ds}\right]F_t-\mathbb E\left[e^{-\int_0^tr(s)\,ds}S_t\right]=0\,$$ where $$\mathbb E$$ is the expectation under the risk-neutral measure. This equation means that the difference of two present values in this equation should be equal. That's a no arbitrage argument saying that today's commitment to buy the stock at time $$t$$ for the fixed price $$F_t$$ should be worth the same as buying it at its price prevailing at time $$t$$. I don't think we can avoid the reference to the risk neutral measure here.

Side Remark: When the stock pays dividends it is not true that the deflated stock price $$e^{-\int_0^tr(s)\,ds}S_t$$ is a martingale. But instead (see ) no arbitrage theory dictates that the process $$\tag{2} M_t:=e^{-\int_0^tr(s)\,ds}S_t+D_t$$ is a martingale where $$D_t$$ is the pathwise present value of all dividends paid until time $$t\,$$: $$D_t=\int_0^tq\,S_ue^{-\int_0^ur(s)\,ds}\,du\,.$$ To understand this a bit better note that the portfolio consisting of the stock plus its past dividends, when they got put into the money market account, is $$\Pi_t=S_t+\int_0^t q\,S_u\,e^{\int_u^tr(s)\,ds}\,du\,.$$ This is an asset that does not pay dividends. Hence $$e^{-\int_0^tr(s)\,ds}\Pi_t$$ must be a martingale, and it obviously equals $$M_t\,.$$

From (1), $$\tag{3} F_t=\frac{\mathbb E\left[e^{-\int_0^tr(s)\,ds}S_t\right]}{\mathbb E\left[e^{-\int_0^tr(s)\,ds}\right]}\,.$$ Let's write $$p_t:=\mathbb E\left[e^{-\int_0^tr(s)\,ds}\right]\,,\quad\tilde F_t:=p_t\,F_t\,.$$ Then from (2) and the fact that $$M_t$$ is a martingale, \begin{align} S_0&=M_0=\mathbb E\left[e^{-\int_0^tr(s)\,ds}S_t\right]+\int_0^t q\,\mathbb E\left[S_u\,e^{-\int_0^ur(s)\,ds}\right]\,du\\ &=p_t\,F_t+\int_0^tq\,p_u\,F_u\,du\,\\ &=\tilde F_t+\int_0^tq\,\tilde F_u\,du\,. \end{align} Differentiation yields $$\frac{d}{dt}\tilde F_t+q\,\tilde F_t=0\,.$$ The solution to this ODE is $$\tilde F_t=\tilde F_0e^{-q t}=F_0e^{-qt}=S_0e^{-q t}\,.$$ In other words: $$\tag{4} \boxed{F_t=\frac{S_0e^{-q t}}{p_t}\,.}$$ The only model assumptions on $$S_t$$ were that the dividend yield $$q$$ was constant.

When the interest rate is constant this simplifies to the known formula $$\boxed{F_t=S_0e^{(r-q) t}\,.}$$  D. Duffie, Dynamic Asset Pricing Theory. Princeton University Press, 1991.

• Thanks for the answer - one quick question, you mentioned that we may not get a formula for $F = \mathbb{E}[S_t]$ even if we assume deterministic interest rate. Can't we just take out the factor $\exp(-\int_{0}^{t} r_s ds)$ as a constant with respect to the expected value in equation (3)? Oct 27, 2021 at 2:04
• @BaroqueFreak : exactly . See my edit which I did after a good sleep last night. Oct 27, 2021 at 8:56

The reason why the forward price is $$S_t e^{-q(T-t)}$$ (let's set $$r=0$$ as that is the easy part) is because the asset pays a continuous dividend rate $$q$$. In other words, if today you purchase $$e^{-q(T-t)}$$ amount of the stock $$S_t$$ by borrowing $$S_t e^{-q(T-t)}$$ from the bank, since a continuous dividend rate is being paid, the 'infinitesimal dividends' received can be continuously reinvested in the asset so that at the end of the road you will have $$e^{q(T-t)} \times S_T e^{-q(T-t)} = S_T$$.