# Why do we need to borrow money in the call-put parity? [closed]

As I understand it, the call put parity is given by

$$c = p + S - \frac{X}{(1 + r)^T}$$

I understand the rationale behind simultaneously buying the call, put and underlying asset for $$S$$, but why is it necessary at $$t=0$$ to borrow $$\frac{X}{(1 + r)^T}$$?

• Assume the call ends in the money. Then it pays $S-X$; the put pays nothing, and you're holding a stock, so the only way to cancel the call's payoff (preserve the call-put parity equation) is to also hold a cash amount of $X$: $S-X=S-X$. Reciprocally, if the put ends in the money, its pays $X-S$, the call pays nothing, hence again you also need a cash amount $X$ in addition to the stock $S$ to cancel payments: $0=(X-S)+S-X$. – Daneel Olivaw Jun 19 '19 at 9:00

Better to understand the call-put parity as,

$$c + \frac{X}{{(1+r)^T}} = p + S$$

You would be equally good at all times, if you hold the LHS or the RHS.

At maturity, if the call is in the money you pay $$X$$ (which is what your cash amount will be worth at $$T$$) and get a stock worth $$S(T)$$.

Under same circumstances (i.e. call is in the money at maturity), the put will expire worthless and your stock will be worth $$S(T)$$.

Similarly, if call were to expire worthless, you would have cash equal to $$X$$ from LHS. And for RHS, you exercise the put by paying the stock and getting cash equal to $$X$$ back. So, equally good in both scenarios.

There is no borrowing of cash or stock involved.