Replicating a Forward Contract to Derive Its Value
Summary
The document asks about an arbitrage proof for the value of a long forward contract and whether borrowing the contract’s value creates a repayment obligation that changes the proof’s final balance. The answer recasts the problem as replication under a constant risk-free rate and no dividends. It gives the forward value as the spot price less the discounted delivery price, then describes a portfolio holding one unit of the underlying asset and a fixed position in the money market account. This portfolio is self-financing and produces the forward payoff at maturity.
The answer also notes that when the initial contract value is negative, the replicating position reverses the stock holding and invests the cash surplus. This gives a replication-based valuation argument, but it does not directly walk through the book’s proposed short-forward and new-long-forward arbitrage or explicitly settle the borrower’s sign question. Its formulas rely on the stated assumptions and would need adjustment for dividends or other carry costs.
Key ideas
- With no dividends and a constant risk-free rate, a forward can be replicated using the asset and a money market position.
- The forward value equals spot less the discounted delivery price.
- The replicating portfolio is self-financing when its asset and cash positions remain constant.
- A negative initial value reverses the stock position and leaves cash to invest.
- The argument depends on the assumptions of no dividends and a constant risk-free rate.
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Full text
# Value of a forward contract proof
# Value of a forward contract proof
I am reading the book - "Mathematics for Finance: An Introduction to Financial Engineering" by Marek Capinski and Tomasz Zastawniak. I am going through the proof of Theorem 6.4 - For any $t$ such that $0 ≤ t ≤ T$ the time $t$ value of a long forward contract with forward price $F(0, T)$ is given by $V(t) = [F(t, T) − F(0, T)]e^{−r(T−t)}$.
I understood the case when the authors proved that $V(t)$ cannot be less than $[F(t, T) − F(0,T)]e^{−r(T−t)}$ by building an arbitrage strategy.
However, I am not able to understand the second case (Exercise 6.6) where the authors proved that $V(t)$ cannot be greater than $[F(t, T) − F(0, T)]e^{−r(T−t)}$ by building an arbitrage strategy at the end of the book. Their proof goes like this:
At time $t$ • borrow and pay (or receive and invest, if negative) the amount $V(t)$ to acquire a short forward contract with forward price $F(0, T)$ and delivery date $T$, • initiate a new long forward contact with forward price $F(t, T)$ at no cost.
Then at time $T$ • close out both forward contracts receiving (or paying, if negative) the amounts $S(T) − F(0, T)$ and $S(T) − F(t, T)$, respectively; • collect $V(t)e^{r(T−t)}$ from the risk-free investment, with interest. The final balance $V(t)e^{r(T−t)} − [F(t, T) − F(0, T)] > 0$ will be your arbitrage profit.
My question is if $V(t)$ is positive and if at time $t$, we are borrowing $V(t)$ to acquire a short forward contract with forward price $F(0,T)$, won't we have to return $V(t)e^{r(T−t)}$ to the bank? In that case in the final balance won't we have $-V(t)e^{r(T−t)} − [F(t, T) − F(0, T)]$ (a minus symbol is coming before $V(t)$ since you have to clear the loan with interest at time $T$)??
## Answer by Kurt G. (score 1)
https://quant.stackexchange.com/a/76615
The question is not self contained and hard to answer if one does not know that book. With the notation
- $S_t$ asset price, $K$ forward price, $T$ maturity, $r$ riskless rate
and assuming zero dividends the PV of the forward contract is $$ V_t=S_t-e^{-r(T-t)}K\,. $$ When we sell this initially for $V_0$ we can use that cash to set up a self-financing trading strategy that gives us the payoff $$ H_T=S_T-K $$ at time $T\,.$ To do so we hold one unit of the stock at all times and $$ \beta_t=\frac{V_t-S_t}{e^{rt}}=-e^{-rT}K $$ units of the money market account $B_t=e^{rt}\,.$ To see that this is self-financing is totally simple:
Since both portfolio weights are constant: $$ dV_t=dS_t+d(\beta_tB_t)=dS_t+\beta_tdB_t\,. $$ Therefore this strategy is self-financing. From this relation it also follows that $$ V_T-V_0=S_T-S_0-e^{-rT}K(B_T-B_0)=S_t-e^{-rT}K-V_0 $$ that is: $V_T=H_T\,.$ The strategy replicates the payoff.
When $V_0$ was initially negative we short one unit of $S_t$ and invest the cash surplus into $B_t\,.$ The formulas stay the same.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.