This should be easy, but for some reason I am struggling with it.

Say you have a long position in the money market (you hold dollars), say you own a quantity of $Ke^{-r(T-t)}$ dollars at time $t$, where $K$ is some positive real number, $T$ is some future time $>t$ and $r>0$ is the risk-free interest rate earned by the money market account.

The question is: what is the value at time $T$ of that portfolio consisting of just some dollars?

I would think it is however much you own times $e^{-r(T-t)}$, which produces $Ke^{-2r(T-t)}$ but in my lecture slides it is actually $K$. As if the values at $T$ was computed by multiplying by $e^{r(T-t)}$. But this doesn't make sense to me since $e^{-r(T-t)}<1$.

Am I wrong or are the slides wrong?

For context, this was in a slide about the Law of one price, where portfolios $P1$ and $P2$ are created so that their value at $T$ is the same.

P1: long Ke−r(T−t) euros and one call, short one put. $V_T(P_1)=K−\max(K−S_T,0)+\max(S_T −K,0)=S_T =V_T(P_2)$

P2: long one share, $V_t(P_2) = S_t$

From this we can conclude that $V_t(P_1)=S_t\ \forall t\in[0,T]$

  • 1
    $\begingroup$ Here's your mistake: "I would think it is however much you own times $e^{-r(T-t)}$", whilst in fact it should be "$e^{+r(T-t)}$" (why? Because your money market account compounds interest, so $r$ is a positive quantity. Why would you put minus sign in front of the $r$? You only do that when you discount. $\endgroup$ Jan 14 at 10:34
  • $\begingroup$ I guess I intuitively thought money loses value when it sits in my pocket $\endgroup$
    – H. Walter
    Jan 14 at 10:53

The value at time $t$ will be $K e^{-r(T-t)}$ times $e^{+r(T-t)}$ that is $K e^{-r(T-t)+r(T-t)}=K$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.