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 $\frac{X}{(1 + r)^T}$?


closed as off-topic by phdstudent, Daneel Olivaw, skoestlmeier, byouness, amdopt Jun 19 at 16:41

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    $\begingroup$ 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: $0=(X-S)+S-X$. $\endgroup$ – Daneel Olivaw Jun 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.


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