Deriving Stock Forward Prices with Continuous Dividends by No Arbitrage

# How to derive forward price on stock with continuous dividend


# 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.

## Answer by Kurt G. (score 9, accepted)

https://quant.stackexchange.com/a/68520

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.

## Answer by user34971 (score 2)

https://quant.stackexchange.com/a/68521

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 $e^{q(T-t)} \times S_T e^{-q(T-t)} = S_T$.

## Answer by ming yuan (score 0)

https://quant.stackexchange.com/a/81971

Replicating portfolio: At t = 0, Buy $e^{-qT}$ units of stock and short a bond with face value K. let N(t),s(t) denote the number of shares and price of shares respectively at time t. Since under perfect market one can buy fractional shares and continuously reinvest, we then have the following separable ODE: \begin{align*} &\frac{dN(t)}{dt} = \frac{s(t)N(t)q}{s(t)} = N(t)q \\ \implies&\int\frac{1}{N(t)}dN(t) = \int qdt \\ \implies &\ln|N(t)| = qt+C \iff N(t) = Ae^{qt}, A\in\mathbb{R} \\ &\text{with initial condition:}N(0) = A =e^{-qT},\text{we have}:\\ &N(t) = e^{-q(T-t)},\text{In particular} N(T)=1 \end{align*}

This implies that by buying $e^{-qT}$ units of stock at time 0 and continuously reinvesting at each time t, the replicating portfolio would contain 1 unit of stock only at the expiration date, taking into account the bond position at time T, the payoff of the replicating portfolio is S(T) - k which is identical to that of derivative security, hence by non-arbitrage pricing we must have: \begin{equation*} V(s(t),T-t) = s(t)e^{-q(T-t)} - ke^{-r(T-t)} \end{equation*} The forward price is the stike price K that makes the above equation = 0,hence the result.