So I am studying Paul Glasserman's book on Monte Carlo in finance and I am currently working through Chapter 3 just to brush up on my risk neutral pricing processes (It's been a while). I have a question regarding Glasserman's derivation of a price process for a future contract. The derivation goes as follows:
Let $\beta(t) = e^{rt}$ be the value a dollar at time t invested at time $0$ with interest rate r. In pricing under the risk-neutral measure, we discount a payoff to be received at time t back to time $0$ by dividing by $\beta(t)$. Suppose the asset S pays no dividends; then, under the risk-neutral measure, the discounted price process $\frac{S(t)}{\beta(t)}$ is a martingale. In other words, $$\frac{S(u)}{\beta(u)} = E[\frac{S(t)}{\beta(t)}|S(t) 0 \leq t \leq u] \quad (3.25)$$
(I understand this part, and skipping forward a bit we have...)
Let $S(t)$ denote the price of the underlying asset and let $F(t,T)$ denote the futures prices at time t for a contract to be settled at a fixed time T in the future. For this contract to have zero value at the inception time t entails, $$0 = e^{-r(T-t)}E[(S(t) - F(t,T))|\mathcal{F}_t] \quad (3.27)$$ where $\mathcal{F}$ is the history of the market prices up to time t. At $t = T$, the spot price and futures prices must agree, so $S(T) = F(T,T)$ and we can write this as $$F(t,T) = E[F(T,T)|\mathcal{F}_t]$$ Thus the futures price is a martingale under the risk neutral measure. It follows that if we choose to model a futures price using geometric brownian motion, we should set its drift parameter to zero: $$\frac{dF(t, T)}{f(t,T)} = \sigma dW(t)$$ Comparison of (3.27) and (3.25) reveals that: $$F(t,T) = e^{(r)(T-t)}S(t)$$
The very last step is what I don't understand. If we need $F(t,T)$ to be a martingale under brownian motion, shouldn't we have $F(t,T) = e^{-r(T-t)}S(t)$? Brownian motion is only a martingale if it's drift parameter is equal to zero, otherwise the concavity of the exponential function imparts a specific curvature. If someone could do this derivation a lot more explicitly then that would be very, very helpful. It has been a while since I have worked with this level of probability theory. Thanks!
Okay, so after taking a break and going over this, the answer is painstakingly obvious. Since we know that the discounted stock price is a martingale, then we use (3.25) with $u = T$ to obtain $$S(T) = \beta(T) \cdot E[\frac{S(t)}{\beta(t)}|\mathcal{F}_t].$$ But $\frac{\beta(T)}{\beta(t)} = \beta(T-t)$. So we know that $$S(T) = \beta(T-t)\cdot E[S(t)|\mathcal{F}_t] = \beta(T-t)S(t)$$ because $S(t)$ is $\mathcal{F}_t$-measurable.
Going back to (3.27), since we know that $F(t,T)$ must be a martingale, then we have $$0 = e^{-r(T-t)}\cdot E[S(T) - F(t,T)|\mathcal{F}_t] =e^{-r(T-t)}\cdot E[S(T)|\mathcal{F}_t] - e^{-r(T-t)}\cdot E[ F(t,T)|\mathcal{F}_t].$$ But $e^{-r(T-t)} = \frac{1}{\beta(T-t)}$ so when we substitute for $S(T)$, we have $$0 = S(t) - e^{-r(T-t)}F(t,T) \leftarrow \text{b/c F(t,T) is a martingale}$$ Finally, solving for $F(t,T)$ gives $$F(t,T) = e^{r(T-t)}S(t)$$