# put call parity?

i thought it was: C + x/(1+r) = S + P And just algebraically rearranging terms to solve for whatever you need in some schweser questions, i’m seeing: Put = Call + Xe−rt − S what’s with the “e”? are you supposed to assume continuous compounding?

They are probably using continuous compounding. I doubt it will have a major effect on your answer choice.

Black Scholes - uses Continuous Compounding. As bpdulog said - it should not make too much of a difference to your answer.

yes, you are correct, answer choices are usually about .35-.45 cents off, so it hasnt impacted the choice (and yes, it was a black scholes problem). WOuldnt surprise me if CFAI put the non-continously compounded answer as an alternative choice on the test

If CFAI puts both continuous and non-continuous as choices on the test that would make it easy if you just know discrete formula. solve the discrete formula - you now know that answer is wrong and the correct answer is the one .35-.45 cents away. the third answer I would think would be something completely different. This is all very theoretical and trying to guess what the CFAI will do hasn’t worked out for me in most cases so take this as you will.