Discounting and continuous compounding

Tez4715 · November 7, 2018, 7:43am

I know there is a relationship between discounting by (1+R_f)^T and e^-R(f)xT but for the life of me can’t figure it out.

Any of you math wizards able to clarify this relationship for me so I can understand it?

Thanks in advance.

S2000magician · November 7, 2018, 8:16am

First, multiplying by e⁻^R(f)T is the same as dividing by e^R(f)T, just as discounting by (1 + R_f)^T is accomplished by dividing by (1 + R_f)^T.

Second, if R(f) is the continuously compounded rate, and R_f is the annual effective rate, then

e^R(f) = 1 + R_f

Finally,

e^R(f)T = [e^R(f)]^T

so,

e^R(f)T = [e^R(f)]^T = (1 + R_f)^T

Therefore, multiplying by e⁻^R(f)T is the same as dividing by e^R(f)T, which is the same as dividing by (1 + R_f)^T.

Voilà!

Tez4715 · November 7, 2018, 5:59pm

Brilliant, thank you Sir. That makes interpreting the BSM much easier!

S2000magician · November 7, 2018, 6:01pm

You’re quite welcome.