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Converting Exponential Rates Across Time Units

Convert exponential rates across time units and verify that probabilities describe the same waiting-time distribution.

Beginner 10 min read

What You'll Learn

  • Relate exponential rates to the time unit in which they are expressed
  • Convert rates when changing to a larger or smaller time unit
  • Rewrite a waiting time and its rate consistently in a probability calculation
  • Verify a conversion by comparing survival probabilities
  • Identify and correct common time-unit conversion errors

Why Exponential Rates Depend on the Time Unit

An exponential rate describes how quickly events occur relative to a stated unit of time. The waiting-time distribution does not change when we switch from minutes to hours, but the numerical value of its rate does. A calculation can therefore be mathematically correct in form and still be wrong if its rate and time are expressed in incompatible units.

The key is to track two conversions separately: how the numerical waiting time changes, and how the numerical rate changes. If a new time unit is longer, a fixed duration has a smaller numerical value in that unit, while the rate has a larger numerical value per new unit. For example, one hour is 60 minutes, so a rate of 0.05 per minute is 3 per hour, not 0.05 divided by 60.

Definition: Let \(T_A\) be an exponential waiting time measured in units of time \(A\), with rate \(\lambda_A>0\) per unit \(A\). Its survival function is \(P(T_A>t)=e^{-\lambda_A t}\) for \(t\geq 0\). If one unit \(B\) is \(c\) units \(A\), where \(c>0\), then the same waiting time has numerical value \(T_B=T_A/c\), and its rate per unit \(B\) is \(\lambda_B=c\lambda_A\).

The phrase “\(c\) units \(A\)” refers to the length of the new unit, not to the size of the numerical time value. For instance, an hour is 60 minutes, so when changing from minutes to hours, \(c=60\). The rate in hours is 60 times the rate in minutes. In the reverse conversion, \(c=1/60\), so the rate in minutes is the rate in hours divided by 60.

The Conversion Rule and Its Justification

Suppose unit \(B\) is \(c\) times as long as unit \(A\). A duration of \(t\) units \(B\) is therefore \(ct\) units \(A\). The probability that the waiting time exceeds that duration must be identical whichever unit is used. Applying the exponential survival formula gives the conversion rule directly:

$$ P(T_B>t)=P(T_A>ct)=e^{-\lambda_A ct}=e^{-(c\lambda_A)t}. $$

The last expression is the survival function in unit \(B\), so its rate is \(\lambda_B=c\lambda_A\). This also shows why the time value and rate must be converted together: their product, the exponent \(\lambda t\), stays unchanged. If a time unit is enlarged by a factor of \(c\), its numerical rate is multiplied by \(c\). If the same duration is instead written using a unit \(c\) times smaller, its numerical time value is multiplied by \(c\), and its rate is divided by \(c\).

Conversion rule: If one unit of the new measure is \(c\) units of the old measure, then \(t_{\text{new}}=t_{\text{old}}/c\) and \(\lambda_{\text{new}}=c\lambda_{\text{old}}\). In every exponential expression, check that rate and time use matching units so that \(\lambda t\) is dimensionless.

This rule applies to expressing the same exponential waiting-time model in a different time unit. It does not change the event being modeled or the underlying probability distribution. No interest rate or mortality assumption is involved in the examples below; each states its exponential distribution and parameter explicitly.

Worked Conversions

Worked Example: Converting a Per-Minute Rate to a Per-Hour Rate

A hypothetical sensor has an exponential time to its next alert, with rate \(0.05\) per minute. Convert the rate to hours and find the probability that the wait exceeds 12 minutes. Check the probability using hours. The model is \(T\sim\operatorname{Exp}(0.05\text{ per minute})\); no interest or mortality assumption is used.

Method. Since one hour is 60 minutes, the new unit (hour) is 60 old units (minutes), so \(c=60\). Multiply the per-minute rate by 60:

$$ \lambda_{\text{hour}}=60(0.05)=3\text{ per hour}. $$

The same 12-minute wait is \(12/60=0.2\) hours. Using minutes, the survival probability is

$$ P(T>12\text{ minutes})=e^{-(0.05)(12)}=e^{-0.6}\approx 0.5488. $$

Using hours gives the same result:

$$ P(T>0.2\text{ hours})=e^{-(3)(0.2)}=e^{-0.6}\approx 0.5488. $$

The check succeeds because both rate-time products equal \(0.6\). The result is \(0.5488\) to four decimal places. A common mistake is to divide \(0.05\) by 60 when converting to a per-hour rate. That would give \(0.0008333\) per hour and exponent \(0.0008333(0.2)\), which does not preserve the original model. Dividing by 60 is appropriate for converting a per-hour rate to a per-minute rate, not the reverse.

Worked Example: Converting a Per-Hour Rate to a Per-Minute Rate

A hypothetical queueing system is modeled with an exponential wait \(T\sim\operatorname{Exp}(0.18\text{ per hour})\). Find the probability that the wait is between 20 minutes and 90 minutes, first using hours and then using minutes. No interest or mortality assumption is used.

Method. Convert the rate to minutes. One minute is \(1/60\) of an hour, so the new unit is \(c=1/60\) old units. Thus,

$$ \lambda_{\text{minute}}=\frac{1}{60}(0.18)=0.003\text{ per minute}. $$

Convert the endpoints to hours: 20 minutes is \(20/60=1/3\) hour, and 90 minutes is \(90/60=1.5\) hours. For an exponential waiting time, the probability of falling between \(a\) and \(b\), with \(0\leq a<b\), is \(P(T>a)-P(T>b)\). Therefore, using hours:

$$ P(20\text{ min}<T<90\text{ min}) =e^{-(0.18)(1/3)}-e^{-(0.18)(1.5)} =e^{-0.06}-e^{-0.27}. $$

Numerically, \(e^{-0.06}\approx0.9417645\) and \(e^{-0.27}\approx0.7633795\), so the probability is approximately \(0.9417645-0.7633795=0.1783850\), or \(0.1784\) to four decimal places.

Now work entirely in minutes. The endpoints are 20 and 90, and the rate is \(0.003\) per minute:

$$ e^{-(0.003)(20)}-e^{-(0.003)(90)} =e^{-0.06}-e^{-0.27} \approx0.1784. $$

The two calculations agree. The common trap is to convert the endpoints but leave the rate at \(0.18\) per hour in the minute-based expression. That would produce exponents \(0.18(20)\) and \(0.18(90)\), incorrectly treating minutes as hours. Match the rate unit to the time unit before evaluating.

Worked Example: Converting a Per-Day Rate to a Per-Week Rate

A hypothetical monitoring process has an exponential time to a signal, \(T\sim\operatorname{Exp}(0.14\text{ per day})\). Convert the rate to weeks and calculate the probability that the signal occurs within three days. Verify the probability in weeks. No interest or mortality assumption is used.

Method. One week is seven days, so \(c=7\) when converting from days to weeks. The rate per week is

$$ \lambda_{\text{week}}=7(0.14)=0.98\text{ per week}. $$

The probability of occurring within three days is one minus the probability of waiting more than three days. In days:

$$ P(T\leq3\text{ days})=1-e^{-(0.14)(3)} =1-e^{-0.42} \approx1-0.6570468 =0.3429532. $$

Thus the probability is \(0.3430\) to four decimal places. In weeks, three days is \(3/7\) week, so

$$ P(T\leq 3/7\text{ week}) =1-e^{-(0.98)(3/7)} =1-e^{-0.42} \approx0.3430. $$

The equality follows because \(0.98(3/7)=0.42\), the same rate-time product as \(0.14(3)\). A plausible but incorrect conversion is \(0.14/7=0.02\) per week. This reverses the direction: a week is a larger unit than a day, so the rate per week must be larger, not smaller. The correct rate is \(0.98\) per week.

A Reliable Unit-Conversion Procedure

For an exam calculation, it is useful to write the unit relationship before substituting any values. The procedure below keeps the direction of conversion explicit.

1
Name the current rate unit.
Record whether the rate is per minute, per hour, per day, or another unit. Do not use a bare numerical rate without its unit.
2
State the unit-size relationship.
For example, one hour is 60 minutes. Specify which unit is new and which is old before choosing the conversion factor.
3
Convert the rate and time consistently.
If the new unit is \(c\) old units long, multiply the old rate by \(c\), and divide the numerical time by \(c\).
4
Check the exponent.
Compute the rate-times-time product in both units. It must be identical, since the waiting-time probability has not changed.

A quick directional check can catch many errors. A per-hour rate should be 60 times a per-minute rate, because an hour contains 60 minutes. Conversely, a per-minute rate should be one-sixtieth of a per-hour rate. Likewise, a per-week rate is seven times a per-day rate, and a per-day rate is one-seventh of a per-week rate.

Common Mistakes and Exam Checks

Most mistakes come from thinking of “per unit” as a directionless label. The number of events expected over a fixed duration is represented by the product of the rate and the duration. Changing units alters each numerical value in opposite ways, leaving that product unchanged.

  • Reversing the rate conversion. When moving from a smaller unit to a larger one, multiply the numerical rate by the number of smaller units in the larger unit. For example, per minute to per hour means multiplying by 60.
  • Changing only one quantity. If a time is changed from minutes to hours, either convert the rate to per hour as well or keep both quantities in minutes.
  • Using a conversion factor without naming its direction. Write “one hour equals 60 minutes” and identify the old and new units before multiplying or dividing.
  • Forgetting what the exponential exponent represents. The product \(\lambda t\) must use compatible units and remain unchanged under conversion.
  • Rounding too early. Keep exact conversion factors such as 60 or 7 through the calculation, and round the final probability only at the end.
Key takeaway: If the new time unit is \(c\) times as long as the old unit, multiply the rate by \(c\) and divide the numerical time by \(c\). Confirm that the product \(\lambda t\) is unchanged before interpreting the probability.

Check Your Understanding

For each problem, keep the distribution and time units explicit. Probabilities in the answers are rounded to four decimal places.

  1. A hypothetical device has an exponential waiting time with rate \(0.08\) per minute. Convert the rate to hours and find the probability that the wait exceeds 10 minutes. (Answer: \(4.8\) per hour; \(P(T>10\text{ min})=e^{-0.8}\approx0.4493\))
  2. A hypothetical service time is exponential with rate \(0.25\) per hour. Convert the rate to minutes and find the probability the service ends within 12 minutes. (Answer: \(0.0041667\) per minute; \(1-e^{-0.05}\approx0.0488\))
  3. A hypothetical process has an exponential time to completion with rate \(0.09\) per day. Convert the rate to weeks and find the probability completion occurs between two and four days. (Answer: \(0.63\) per week; \(e^{-0.18}-e^{-0.36}\approx0.1376\))
  4. A hypothetical waiting time is exponential with rate \(1.2\) per hour. Express the rate per 15-minute unit, then find the probability the wait exceeds 45 minutes. (Answer: \(0.3\) per 15-minute unit; \(e^{-0.9}\approx0.4066\))
  5. A hypothetical monitoring time is exponential with rate \(0.004\) per second. Convert the rate to minutes and find the probability the wait is less than two minutes. (Answer: \(0.24\) per minute; \(1-e^{-0.48}\approx0.3812\))

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