Converting Annual Interest Rates for Black–Scholes Pricing

Summary

The document explains how to convert an annually compounded interest rate into the continuously compounded rate used in the Black–Scholes model. Equating the accumulated value under each convention over the same time horizon gives the conversion: take the natural logarithm of one plus the annual rate. For the reverse conversion, exponentiate the continuous rate and subtract one.

It illustrates the distinction with an annual rate of 3.6%, which converts to approximately 3.53% continuously compounded. The questioner’s expression, one minus an exponential with a negative rate, gives a nearby but different value because the signs are reversed. The explanation is limited to translating between compounding conventions; it does not discuss choosing a risk-free rate, term structures, day-count conventions, or other Black–Scholes assumptions.

Key ideas

• Black–Scholes uses a continuously compounded interest rate.
• Convert an annual effective rate to a continuous rate by taking the logarithm of one plus that rate.
• Convert a continuous rate back to an annual effective rate by exponentiating and subtracting one.
• The two conventions produce close but distinct values for modest rates.

Tags

# Risk free rate in black-scholes model


# Risk free rate in black-scholes model












Currently reading A. Damodaran‘s book Investment Valuation. In chapter 5 in order to value an option using black-scholes model he adjusts risk free rate using the following formula: $1-e^{-r}$ I. E. If given risk free rate is $3,6%$ it becomes $1-e^{(-0,036)} = 3,54\%$ Why is that? I wasn’t able to find any explanation on that

## Answer by Dom (score 1, accepted)

https://quant.stackexchange.com/a/60960

Black Scholes uses a continuously compounded rate $r$. To go from a $T$-year annually compounded rate $\hat{r}$ to a $T$-year continuously compounded $r$ you use the formula

$e^{rT} = (1+\hat{r})^T$

So to solve for the Black-Scholes continuously compounded rate you take logs and simplify which gives

$r = \ln(1+\hat{r})$.

This is what Damodoran quotes on page 132 (chapter 5, page 12) on the second edition of his book (I found a free online version).

So if $\hat{r} = 0.036$ then $r=0.0353$. It's very close to the result you got, but not identical.

If you are converting in the opposite direction in order to solve for $\hat{r}$ the formula is

$\hat{r} = e^r - 1$

which is not the same either.

I don't know where the formula you get comes from. It's almost the correct formula - but the signs of the interest rates are incorrectly reversed.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.