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.


2 Answers 2


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 [1]) 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 $$ \boxed{F_t=S_0e^{(r-q) t}\,.} $$ [1] D. Duffie, Dynamic Asset Pricing Theory. Princeton University Press, 1991.

  • $\begingroup$ 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)? $\endgroup$ Oct 27, 2021 at 2:04
  • $\begingroup$ @BaroqueFreak : exactly . See my edit which I did after a good sleep last night. $\endgroup$
    – Kurt G.
    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 $e^{q(T-t)} \times S_T e^{-q(T-t)} = S_T$.


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