If I start with Rs 100 and it becomes Rs 200 in a year, why is the continuous compounding rate obtained by doing Ln(200/100)? Could someone please help me?
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V_t = V_0e^{r_{cont}t}
e^{r_{cont}t} = \frac{V_t}{V_0}
r_{cont}t = ln\left(\frac{V_t}{V_0}\right)
r_{cont} = \frac1tln\left(\frac{V_t}{V_0}\right)
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I can’t understand this. Could you please simplify it?
It can’t be simplified.
I can explain it.
V_0 is the value at time t = 0; in your case, that’s Rs 100.
V_t is the value at some time t in the future; in your case, that’s Rs 200 at time t = 1.
The first equation says that the value V_t at any time in the future is the value at time 0 increased at the continuous compounding rate r_{cont} for the amount of time t. We know t = 1, V_0 = 100, and V_1 = 200; we’re trying to solve for r_{cont}.
First, we divide each side by V_0.
Then, we take the natural logarithm of each side, remembering that ln\left(e^x\right) = x.
Finally, we divide each side by t.
Using the values we know,
r_{cont} = \frac1tln\left(\frac{V_t}{V_0}\right) = \frac11ln\left(\frac{200}{100}\right) = ln\ 2 = 0.693147 = 69.3147\%
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Thank you!
My pleasure.
Hey, if i may add something. You say we are solving for “rcount” but arent you actually soving for “r”? Why u refer to it as “rcount”? THe rate only becomes compound after you have raised it on “e”. Am I wrong?
The rate depends on how often it’s compounded.
A continuously compounded rate of 4% is equivalent to an effective annual rate of 4.0811%, or a semiannually compounded (bond equivalent) rate of 4.0403%.
I agree, i am only referring to the terms you are using cuz it confused me. The 4% only becomes continuously compunded once u have e^0,04. Otherwise the 0,04 are just the “r” and not “rcount”? Am I right?
Thanks for the answer btw!