Time Units and Payment-Rate Scaling

Why Time Units Matter

A continuous-annuity value is an integral over time. Its result is unchanged if the calculation is expressed in years or months, but only when the force of interest, payment intensity, and integration limits all use the same time unit. Changing just the limits, or just the force, does not convert the valuation; it changes the mathematical model.

For example, a stream paying $3,600 per year over 1.5 years has the same value whether its present value is integrated over 1.5 years or over 18 months. In the monthly calculation, however, the payment intensity must be $300 per month, not $3,600 per month. This scaling is the central discipline of this topic.

Definition: A payment intensity is the amount paid per unit of time. If $c_y(t)$ is an intensity in dollars per year at time $t$, measured in years, then the corresponding monthly intensity at $u=12t$ is $c_m(u)=c_y(u/12)/12$, in dollars per month. A force of interest is also expressed per unit of time: if the annual force is $\delta_y(t)$, the corresponding monthly force is $\delta_m(u)=\delta_y(u/12)/12$.

The division by 12 in both conversions has different dimensional interpretations. The annual payment intensity is divided by 12 because one month is one-twelfth of a year. The annual force is divided by 12 because the accumulated force over an interval must remain unchanged: an interval of $u$ months is $u/12$ years.

The Conversion Rule

Suppose the annual-time valuation uses a force $\delta_y(t)$ and payment intensity $c_y(t)$. Its present value at time 0 over $T$ years is

\text{PV}=\int_0^T c_y(t)\exp\left(-\int_0^t \delta_y(s)\,ds\right)dt.

Set $u=12t$. Then $t=u/12$, $dt=du/12$, and the monthly interval ends at $12T$. The monthly intensity includes the factor $1/12$, while the monthly force is also the annual force per one-twelfth-year unit. The accumulated force from time 0 to $u$ months is therefore the same in either description.

Theorem: When converting from years to months, use $u=12t$, replace $c_y(t)$ by $c_m(u)=c_y(u/12)/12$, replace $\delta_y(t)$ by $\delta_m(u)=\delta_y(u/12)/12$, and replace the integration limits $0,T$ by $0,12T$. These conversions preserve the present value. The same principle applies to other time-unit conversions: scale the time variable, intensity, force, and limits consistently.

For constant annual force $\delta_y$ and constant annual payment intensity $c_y$, the annual-time formula is

\text{PV}=c_y\int_0^T e^{-\delta_y t}\,dt =c_y\frac{1-e^{-\delta_yT}}{\delta_y},\qquad \delta_y\ne 0.

In monthly units, $\delta_m=\delta_y/12$, $c_m=c_y/12$, and the term is $12T$ months. Thus

\text{PV}=c_m\frac{1-e^{-\delta_m(12T)}}{\delta_m} =\frac{c_y}{12}\frac{1-e^{-\delta_yT}}{\delta_y/12} =c_y\frac{1-e^{-\delta_yT}}{\delta_y}.

This equality is a direct check on the unit conversion. The formula assumes a continuous payment stream, with the stated intensity applying throughout the term, and a constant force in the displayed constant-force case. It does not describe a sequence of separate monthly payments; that is a different cash-flow model.

Worked Conversions

Worked Example: Converting an Annual Stream to Monthly Units

A hypothetical continuous stream pays $3,600 per year for 1.5 years. The force of interest is constant at 0.08 per year. Find the present value at time 0, first in years and then in months. Give the answer to the nearest cent.

Method. Value the same continuous stream in both time units. The annual duration is 1.5 years, so the monthly duration is $1.5(12)=18$ months. The monthly payment intensity is $3,600/12=\$300$ per month, and the monthly force is $0.08/12=0.0066666667$ per month.

In years, the present value is

\begin{aligned} \text{PV} &=3,600\int_0^{1.5}e^{-0.08t}\,dt\\ &=3,600\frac{1-e^{-0.08(1.5)}}{0.08}\\ &=3,600\frac{1-e^{-0.12}}{0.08}. \end{aligned}

Using $e^{-0.12}=0.8869204367$, the numerator is $1-0.8869204367=0.1130795633$. Therefore,

\text{PV}=3,600\left(\frac{0.1130795633}{0.08}\right) =3,600(1.4134945413) =\$5,088.58.

Now calculate in months. The monthly exponent over 18 months is $0.0066666667(18)=0.12$, so

\begin{aligned} \text{PV} &=300\int_0^{18}e^{-0.0066666667u}\,du\\ &=300\frac{1-e^{-0.12}}{0.0066666667}\\ &=300\left(\frac{0.1130795633}{0.0066666667}\right)\\ &=300(16.9619345)=\$5,088.58. \end{aligned}

Both calculations give $5,088.58 to the nearest cent. The common mistake this conversion avoids is carrying the annual intensity of $3,600 into a monthly integral: that would treat the stream as paying $3,600 each month, twelve times the stated intensity.

Worked Example: Converting an Effective Annual Rate Before Monthly Valuation

A hypothetical stream pays continuously at $100 per month for 12 months. The effective annual interest rate is 12%. Find the present value at time 0, to the nearest cent.

Method. Convert the effective annual rate to an annual force first, then express that force per month. The annual force is $\delta_y=\ln(1.12)$, not 0.12. Consequently, the monthly force is $\delta_m=\ln(1.12)/12$.

The accumulated force over 12 months is $12\delta_m=\ln(1.12)$, so the discount factor at 12 months is $e^{-\ln(1.12)}=1/1.12$. The present value is

\begin{aligned} \text{PV} &=100\frac{1-e^{-12\delta_m}}{\delta_m}\\ &=100\frac{1-e^{-\ln(1.12)}}{\ln(1.12)/12}\\ &=1,200\frac{1-1/1.12}{\ln(1.12)}. \end{aligned}

Here $1-1/1.12=0.1071428571$ and $\ln(1.12)=0.1133286853$. Thus,

\text{PV} =1,200\left(\frac{0.1071428571}{0.1133286853}\right) =1,200(0.9454169) =\$1,134.50.

The method preserves the effective annual accumulation: the 12-month accumulation factor is exactly 1.12. A common error is to use $0.12/12$ as the monthly force. Dividing an effective annual rate by 12 does not generally produce an equivalent monthly force. Here, the force conversion follows from $\delta_y=\ln(1+i)$, where $i=0.12$.

Worked Example: Rescaling a Varying Payment Intensity

A hypothetical stream has annual payment intensity $c_y(t)=1,200(1+t)$ dollars per year, where $t$ is measured in years. Payments continue for two years, and the force is constant at 0.06 per year. Find the present value to the nearest cent, and express the same stream in monthly units.

Method. First value the stream using years. Then substitute $u=12t$ to identify its monthly intensity and confirm that the monthly integral is the same valuation. The annual-time value is

\text{PV}=1,200\int_0^2(1+t)e^{-0.06t}\,dt.

For $a=0.06$ and $T=2$, split the integral into its constant and linear parts:

\int_0^T(1+t)e^{-at}\,dt =\frac{1-e^{-aT}}{a} +\frac{1-e^{-aT}(1+aT)}{a^2}.

This identity follows by integrating $e^{-at}$ directly and integrating $te^{-at}$ by parts. With $aT=0.12$ and $e^{-0.12}=0.8869204367$, the first part is \[ \frac{1-0.8869204367}{0.06} =\frac{0.1130795633}{0.06} =1.8846593883. \] For the second part, $1+aT=1.12$, and

\begin{aligned} e^{-0.12}(1.12)&=0.9933508891,\\ \frac{1-0.9933508891}{0.06^2} &=\frac{0.0066491109}{0.0036} =1.84697525. \end{aligned}

The integral is therefore $1.8846593883+1.84697525=3.7316346383$, giving

\text{PV}=1,200(3.7316346383)=\$4,477.96.

For monthly time, $u=12t$. The intensity becomes $c_m(u)=1,200(1+u/12)/12=100(1+u/12)$ dollars per month, the force becomes $0.06/12=0.005$ per month, and the term is 24 months. Therefore the equivalent monthly calculation is

\text{PV}=\int_0^{24}100(1+u/12)e^{-0.005u}\,du.

Substituting $u=12t$ in this integral gives $du=12\,dt$, $100(1+u/12)du=1,200(1+t)dt$, and $e^{-0.005u}=e^{-0.06t}$. It is exactly the annual-time integral, so it also gives $4,477.96. The common mistake avoided is to change the variable in a varying intensity without changing its argument: $1+t$ becomes $1+u/12$, not $1+u$.

Practical Checks and Exam Traps

A unit conversion should preserve the economic stream, not merely produce an integral with convenient-looking limits. Before calculating, label each rate with its unit and check the horizon:

An annual payment intensity is in dollars per year; a monthly intensity is in dollars per month.
An annual force is per year; a monthly force is per month.
A duration in years must be multiplied by 12 to obtain months.
When an intensity varies with time, its time argument must be converted as well as its units.

For example, in the first worked example, using $3,600 per month with a monthly force of $0.08/12$ over 18 months gives

3,600\frac{1-e^{-0.12}}{0.08/12}=\$61,062.96,

which is twelve times the correct value. The force and term were converted, but the payment intensity was not. Conversely, keeping an annual force of 0.08 while integrating over 18 months treats the exponent as if months were years, so it does not preserve the original discounting.

When the interest input is an effective annual rate, first use $\delta_y=\ln(1+i)$ for a constant force equivalent to that annual accumulation. The corresponding monthly force is $\ln(1+i)/12$, and the equivalent monthly effective rate is $e^{\delta_y/12}-1=(1+i)^{1/12}-1$. Dividing an effective annual rate by 12 is not the same conversion. By contrast, if a quoted rate is already a nominal annual rate convertible monthly, its contractual monthly rate is defined by that nominal-rate convention; do not silently treat it as an effective annual rate.

Consistent units are especially useful when a problem supplies one quantity in months and another in years, or when the payment intensity changes over time. Write the time conversion explicitly before substituting into an integral. This simple step helps prevent errors in scale, discounting, and the shape of a varying stream. These conversion techniques are aligned with the SOA Exam FM (Financial Mathematics) syllabus.

Check Your Understanding

For each problem, keep the force, payment intensity, time variable, and limits in consistent units. Round monetary answers to the nearest cent.

A continuous stream pays $600 per month for 10 months. The constant force is 0.004 per month. Find its present value at time 0. (Answer: $5,881.58)
A continuous stream pays $500 per month for 6 months. The effective annual interest rate is 8%, and the force is constant. Find its present value at time 0. (Answer: $2,943.01)
A continuous stream pays $2,400 per year for one year under a constant force of 0.05 per year. Convert to monthly units and find its present value. (Answer: $2,340.99)
A stream has annual intensity $c_y(t)=800+200t$ dollars per year for one year, where $t$ is measured in years. The constant force is 0.04 per year. State the monthly intensity as a function of months $u$, then find the present value to the nearest cent. (Answer: $c_m(u)=66.6667+1.38889u$ dollars per month; $881.58)

What You'll Learn