Value of a forward contract proof

Question

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 1

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.