Reading Bond Holding Returns with Time and Maturity Indices

Summary

The discussion explains the notation for a one-period holding return on a zero-coupon bond. The return compares the bond’s log price at time t, when it has n periods remaining, with its price at time t+1, when one period has passed and only n−1 periods remain. The maturity index measures time remaining, while the time index identifies when the price is observed.

The apparent zero return comes from treating these two indices as if they were calendar dates that can be added or subtracted interchangeably. They are distinct dimensions: the bond’s maturity date stays fixed as the observation date advances and remaining maturity declines. Thus the expression compares prices at different observation times, and their difference need not be zero. The answer clarifies the indexing convention but does not derive further return calculations or address coupon-paying bonds.

Key ideas

• The time index identifies when a bond price is observed, while the maturity index gives periods remaining.
• After one period passes, a bond with n periods remaining has n−1 periods to maturity.
• The bond’s calendar maturity remains fixed as observation time advances.
• The two price terms in the holding-return expression refer to different observation dates, so their difference is not automatically zero.

Tags

# Bond one period holding return definition


# Bond one period holding return definition












In the literature I am reading Crump & Gospodinov Deconstructing the yield curve, Federal Reserve Bk of NY, Staff Report 884 (2019), I came across the definition for a one period holding return of a bond as:

The one-period holding return on a bond of maturity $n$ from time t to t + 1 is defined as

$$ r^{(n)}_{t,t+1} \equiv p^{(n−1)}_{t+1} − p^{(n)}_{t} $$

The idea is that the price is defined as $p^{(n)}_t = $ the time t log price of a zero-coupon bond which pays $1 at time t + n.

Given this, the one period return from time t to t+1 is

$ r^{(n)}_{t,t+1} = p^{(n−1)}_{t+1} − p^{(n)}_{t} $

$ =$ price at time:(t+1+n-1) - price at time:(t+n)

$ =$ price at time:(t+n) - price at time:(t+n)

$=$ p(n)t − p(n)t

$ = $ 0

I am wondering if I am missing something to understanding the return? because this would always result in 0.

## Answer by David Duarte (score 2, accepted)

https://quant.stackexchange.com/a/51125

You can't really add/subtract the n and t indices. The fact that you didn't use Latex in your question makes it even more confusing :)

The $t$ refers to the time of the price and $n$ refers to the time to maturity of the bond, the bond has $n$ periods remaining and matures at ($t+n$).

If today's price is $p_t^{(n)}$ then tomorrow's price is $p_{t+1}^{(n-1)}$ because one day has passed ($t+1$) and there's one less day to maturity ($n-1$).

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