Continuous Compounding Links Log Bond Price to Yield

Summary

The note clarifies how the continuously compounded yield of a zero-coupon bond relates to its log price. For a bond paying one unit at maturity after n periods, its price is the exponential of minus yield times maturity. Taking logarithms gives log price equal to minus yield multiplied by maturity, so yield is minus log price divided by maturity.

The apparent discrepancy comes from terminology: a cited convention calls the log yield something different, while the basic bond-pricing identity concerns the continuously compounded yield. The document offers only a brief clarification and no numerical example or broader discussion of yield conventions. The relationship depends on the stated zero-coupon bond setup and should not be generalized to other yield definitions without checking their conventions.

Key ideas

• A zero-coupon bond’s price under continuous compounding is determined by yield and time to maturity.
• Taking logarithms yields log price equal to negative yield times maturity.
• The terminology around log yield can be misleading; the formula describes continuously compounded yield.

Tags

# How to derive the relationship between log yield and log price?


# How to derive the relationship between log yield and log price?












Usually, people write $y_t^{(n)}=-\frac{p_t^{(n)}}{n}$ where $y, p$ and log yield and log price respectively. My question is how do one derive this expression?

Note that $e^{-Y_t^{(n)}\cdot n}=P_t^{(n)}$ if $Y,P$ are the continuous yield and price of the one dollar $n$ period bond respectively at time $t$. Now, if we take logs, we get $$-Y_t^{(n)}\cdot n=p_t^{(n)}.$$ Therefore, we have $Y_t^{(n)}=-\frac{p_t^{(n)}}{n}$ which differs from the equation from the first paragraph. What went wrong?

## Answer by Dom (score 2, accepted)

https://quant.stackexchange.com/a/28436

This is a misnomer by Cochrane and Piazzesi. It should simply be called the continuously compounded yield.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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