# formula for pricing bond-futures

Is anybody able to help me understanding why does $$P_t(S)$$ appear in the solution to the following problem; deriving the price of bond forward contracts?
Thank you
Given:

• $$r_t$$, the instantaneous rate process
• $$P_t(T)$$, the price of a zero-coupon bond at $$t$$ and expiring at $$T$$
• $$F_t(S,T)$$, the price at $$t$$ of a forward expiring at $$S\leq T$$
• $$B_S(T)$$, the price at $$S$$ of a coupon-bearing bond expiring at $$T>t$$

The payoff for going long the forward is $$B_S(T)-F_t(S,T)$$, and the position is costless at inception such that, in the absence of arbitrage, the forward satisfies
$$F_t(S,T)=\frac{1}{P_t(S)}\mathbb{E}_t\left(e^{-\int_t^Sr_\tau d\tau} B_S(T)\right)$$
where $$\mathbb{E}_t$$ denotes the expectation

• What is $F_t(S,T)$? You say is the forward expiring... you mean the forward rate? or is $F_t(S,T)$ the forward price of the coupon-bearing bond $B_S(T)$? Nov 25 '21 at 11:54
• $F_t(S, T)$ is the price at $t$ of the forward (or futures) contract Nov 25 '21 at 13:47
• What contract? The underlying asset is the coupon-bearing bond, right? Nov 25 '21 at 13:52
• Yes, the contract $F_t(S, T)$ is a forward contract in which the underlying asset $B_S(T)$ is a coupon-bearing bond. Nov 25 '21 at 14:23
• Is the conditional expectation taken under the risk-neutral measure or under the forward measure? When is the inception, $t=0$ or just $t$? Nov 25 '21 at 16:58

I assume the forward contract matures at time $$S$$ and $$0 \leq t < S < T.$$ Let $$F_t(S,T)$$ be the price that makes the forward contract be worth zero at time $$t$$. Then by the risk-neutral valuation formula we obtain $$0= \mathbb E^{\mathbb Q} \left[(B_S(T) - F_t(S,T))e^{-\int_t^S r(u)du} | \mathcal F_t \right].$$ Since $$F_t(S,T))e^{-\int_t^S r(u)du}$$ is $$\mathcal F_t$$-measurable, $$F_t(S,T)=e^{\int_t^S r(u)du} \mathbb E^{\mathbb Q} \left[B_S(T)e^{-\int_t^S r(u)du} | \mathcal F_t \right].$$ Under the risk-neutral probability measure $$\mathbb Q$$ the discounted bearing-bond is a martingale, so $$F_t(S,T)=e^{\int_t^S r(u)du} B_t(T)=\frac{1}{P_t(S)}B_t(T).$$
• I guess there's a small typo in the first equation. It should $F_t(S,T)$ instead of $F_S(S,T)$ Nov 26 '21 at 7:36