The question does not state the precise intervals or yields so the $H_{0}$ hypothesis must be conservative with infinite intervals and yields with pessimist approximations, .*logic's suggestion won't qualify. Confidence interval not calculated. So:
$lim_{ m \rightarrow \infty } \left[ 1+\frac{r}{m} \right] ^{mt}=e^{rt}$,
where r is the rate p.a. and t is the time. The sum is $\sum_{k=1}^{n} x_{k} e^{rt}$. If you have data of different signs, you must calculate positive numbers to one sum and negative numbers to one sum, this way you get proper upper/lower bounds. The exponent is $r_{k} MINUS(timestamp_{1}, timestamp_{2})/365$, the MINUS -function returns days between the timestamps and $r_{k}$ is the rate. The terms FV and PV are:
$FV = x_{0} \left( 1+r \right)^{n} + x_{1} \left(1+r \right)^{n-1}+...+x_{n}$
$PV = x_{0} + \frac{x_{1}}{1+r} +...+ \frac{x_{n}}{\left( 1+r\right)}^{n}$
so $FV$ is with the sum -formula, while you just take reciprocal with $PV$.
Example
1 01/1/1980
2 01/2/1999
3 03/12/2000
-1 03/6/2005
-5 07/07/2007
Work in progress.
Trying to find one-liner to operate over the data: $if (term_{k} > 0) : sum (positive_{k}e^{MINUS(t_{k},t_{0})/365})$