Hello, I read the explanation but still not quite get it. Where does F(0, T) - X come from? I thought put-call parity is P = C + S - (X - FP)/(1 + Rf)^T. Why would F > X be critical to this question? If we’re taking the long position for the forward, why do we still have to take the long for the zero coupon bond?
If F > X , you short the bond (the PDF answer is wrong).
I had the same question. So it would be long the future and short the bond, right?
??? I am not sure. But I think I got it now. The put-call for forward/furture: (I typed it wrong) P = C - (X - FP)/(1 + Rf)^T At time 0: (and long the forward, since it costs nothing, we don’t add it to the formula) P0 = C0 - (X - FP)/(1+Rf)^T = C0 + (FP - X)/(1+Rf)^T At time t: Pt = Ct + (FP - X) + FP, where FP - X is the face value of the zero-coupon bond and FP is the forward price.
folks, literally this very moment grappling with the same issue IS this the confirmation that: Co = P - (X-F)/(( 1 + r) ^T Therefore the synthetic call is created being: Long the Put, Long the forward, and Short the bond?