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.


1 Answer 1


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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