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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 $T$)??

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  • $\begingroup$ Please try to use MathJax for formatting going forward. It's much easier to read. $\endgroup$
    – amdopt
    Sep 7, 2023 at 13:23

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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.

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  • $\begingroup$ I have edited my question. Can you please answer my specific query? $\endgroup$
    – anthony
    Sep 8, 2023 at 3:34
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    $\begingroup$ @anthony . Why should we want to follow every "proof" in a literature when we can find our own proofs we understand better? In fact: mine can be drastically simplified. As the seller of a forward contract you are obliged to pay to the buyer $S_T-K$ at maturity. What else would you do to hedge that than holding the stock and lending the amount $Ke^{-rT}$ at time zero to receive $K$ at maturity? Look at those authors' proof from the same angle. I believe their basic idea is similarly simple. $\endgroup$
    – Kurt G.
    Sep 8, 2023 at 4:09
  • $\begingroup$ I agree with @Kurt G. In addition it does seem like the author’s proof should have started with “receive positive V(t) and invest “ . $\endgroup$
    – dm63
    Sep 10, 2023 at 13:37

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