# Value at maturity of long position in money market

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 $$V_t(P_1)=S_t\ \forall t\in[0,T]$$

• 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. Jan 14 at 10:34
• I guess I intuitively thought money loses value when it sits in my pocket Jan 14 at 10:53

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