Doubt for Put Call Forward parity

CFA Level I Derivatives
#1

Can someone explain this question to me? I tried rearranging the Put-Call-Forward Parity and got B (C-P = F/(1+rf)T - X/(1+rf)T) , but A is the correct answer.

Q. Under put–call–forward parity, which of the following transactions is risk free?

A. Short call, long put, long forward contract, long risk-free bond

B. Long call, short put, long forward contract, short risk-free bond

C. Long call, long put, short forward contract, short risk-free bond

Can someone please explain me the answer for this question, I tried to understand the same from 2-3 CFA aspirates, but honestly they are not able to explain me the reasoning. I would be thankful if someone can please help me in understanding the concept behind this.

Thanks

#2

First, it’s a weird question.

In essence, they’re (first) asking you to solve the put-call forward parity equation for . . . wait for it! . . . zero!

Note that zero is risk-free.

P_0\ + \frac{F_T}{\left(1 + r_f\right)^T} = C_0\ + \frac{X}{\left(1 + r_f\right)^T}
P_0\ - C_0\ + \frac{F_T}{\left(1 + r_f\right)^T} - \frac{X}{\left(1 + r_f\right)^T} = 0

Note that the put and the forward have the same sign: you have to be long both or else short both. So B can’t possibly be correct, nor can C.

But how did they get to A? My equation has a short bond, not a long bond.

Here’s the subtle part: if you add a risk-free asset to a risk-free position, the position remains risk-free.

Instead of adding a long bond to my equation, I’ll add two: one to cancel my short bond and one to give me a long bond:

P_0\ - C_0\ + \frac{F_T}{\left(1 + r_f\right)^T} + \frac{X}{\left(1 + r_f\right)^T} = 2\frac{X}{\left(1 + r_f\right)^T}

So transaction A is equivalent to buying two risk-free bonds.

More than a little sneaky. (However, as it’s easy to eliminate B and C, maybe sneaky, but easy.)