Why Derive the Moments?
For an exponential model, the rate parameter determines both the average waiting time and the variability of that waiting time. The formulas are compact, but they are easy to misread: the mean is the reciprocal of the rate, while the variance is the reciprocal of the rate squared. Deriving them from the density makes the distinction clear and provides a check on units and calculations.
Here, \(X\) represents a nonnegative continuous measurement, such as a waiting time. The model assumes a fixed positive rate \(\lambda\) and uses the exponential density on the nonnegative real line. The derivation below uses density integrals; it does not require assumptions about insurance benefits, interest, or mortality.
The density is properly normalized because \[ \int_0^\infty \lambda e^{-\lambda x}\,dx =\left[-e^{-\lambda x}\right]_0^\infty =0-(-1)=1. \] The condition \(\lambda>0\) ensures that the exponential term tends to zero as \(x\) tends to infinity. All the moment integrals used below are finite under this condition.
Deriving the First Two Moments
For a continuous random variable with density \(f_X\), the \(r\)th raw moment, when it exists, is \(E[X^r]=\int_{-\infty}^{\infty}x^r f_X(x)\,dx\). Since the exponential density is zero for negative \(x\), the integrals for \(X\) begin at zero. We first evaluate the integrals needed for the mean and second raw moment.
For the exponential density, \(E[X]=\lambda\int_0^\infty x e^{-\lambda x}\,dx\) and \(E[X^2]=\lambda\int_0^\infty x^2 e^{-\lambda x}\,dx\).
Choose the polynomial factor as \(u=x^n\) and the exponential factor as \(dv=e^{-\lambda x}dx\). Then \(du=nx^{n-1}dx\) and \(v=-e^{-\lambda x}/\lambda\).
For \(n=1\) and \(n=2\), the boundary term at zero is zero because it contains \(x^n\). At infinity it is zero because \(x^n e^{-\lambda x}\) tends to zero for \(\lambda>0\).
More explicitly, for \(n=1,2\), integration by parts gives
The remaining base integral is \(\int_0^\infty e^{-\lambda x}\,dx=1/\lambda\). Thus the first integral is \(1/\lambda^2\), and the second is \(2/\lambda^3\). Multiplying each by the density's leading factor \(\lambda\) gives the raw moments.
The variance is the second raw moment minus the square of the mean. Substituting the expressions just derived yields
Interpreting the Rate and Its Units
The rate is not itself an average waiting time. It has units reciprocal to those of \(X\), whereas the mean has the same units as \(X\). If \(X\) is measured in days, then \(1/\lambda\) is in days and \(1/\lambda^2\) is in days squared. A variance cannot be compared directly with a mean as though they had the same units.
The formulas also describe how changes in the rate affect the distribution's moments. A larger \(\lambda\) gives a smaller mean and a smaller variance; a smaller \(\lambda\) gives a larger mean and a larger variance. In particular, doubling the rate halves the mean and reduces the variance to one quarter of its former value. These are consequences of the formulas, not separate assumptions about the model.
| Quantity | Formula | Units when \(X\) is measured in time |
|---|---|---|
| Rate | \(\lambda\) | Inverse time |
| Mean | \(1/\lambda\) | Time |
| Second raw moment | \(2/\lambda^2\) | Time squared |
| Variance | \(1/\lambda^2\) | Time squared |
| Standard deviation | \(1/\lambda\) | Time |
Worked Applications
Worked Example: Mean and Variance of a Waiting Time
In a hypothetical service system, let \(X\) be the time in hours from submitting a request until a technician begins work. Assume \(X\) has an exponential distribution with rate \(\lambda=0.25\) per hour. Find the mean, second raw moment, variance, and standard deviation.
Method. Apply the density-integral results for an exponential rate parameter. The density is \(f_X(x)=0.25e^{-0.25x}\) for \(x\geq0\), with \(x\) in hours.
Mean. The reciprocal of the rate is \[ E[X]=\frac{1}{0.25}=4\text{ hours}. \]
Second raw moment. Since \(\lambda^2=(0.25)^2=0.0625\), \[ E[X^2]=\frac{2}{(0.25)^2}=\frac{2}{0.0625}=32\text{ hours}^2. \]
Variance. Subtract the squared mean from the second raw moment: \[ \operatorname{Var}(X)=32-(4)^2=32-16=16\text{ hours}^2. \] The standard deviation is \(\sqrt{16}=4\) hours.
Check. Directly from the variance formula, \(1/(0.25)^2=1/0.0625=16\text{ hours}^2\), agreeing with the calculation from raw moments. The common mistake to avoid is reporting \(0.25\) hours as the mean: \(0.25\) is a rate in inverse hours, so its reciprocal gives the mean.
Worked Example: Computing Moments From a Density
A hypothetical laboratory models the time \(T\), in days, for an automated calibration cycle as exponential with density \(f_T(t)=2e^{-2t}\) for \(t\geq0\). Find \(E[T]\), \(E[T^2]\), and \(\operatorname{Var}(T)\), showing how the density integrals produce the results.
Method. The coefficient of \(t\) in the exponent identifies the rate as \(\lambda=2\) per day. The density is zero for \(t<0\). For the mean, \[ E[T]=\int_0^\infty t(2e^{-2t})\,dt =2\int_0^\infty te^{-2t}\,dt. \] Integration by parts with \(u=t\) and \(dv=e^{-2t}dt\) gives \(du=dt\) and \(v=-e^{-2t}/2\). The boundary term \([-te^{-2t}/2]_0^\infty\) is zero, leaving \[ E[T]=2\left(\frac{1}{2}\int_0^\infty e^{-2t}\,dt\right) =2\left(\frac{1}{2}\cdot\frac{1}{2}\right) =\frac{1}{2}\text{ day}. \]
For the second raw moment, \[ E[T^2]=2\int_0^\infty t^2e^{-2t}\,dt. \] Integration by parts with \(u=t^2\) and \(dv=e^{-2t}dt\) gives a zero boundary term and \[ E[T^2]=2\left(\frac{2}{2}\int_0^\infty te^{-2t}\,dt\right). \] The remaining integral is \(1/2^2=1/4\), as found from the first integration-by-parts step. Therefore \[ E[T^2]=2\left(1\cdot\frac{1}{4}\right)=\frac{1}{2}\text{ day}^2. \]
Variance. Subtract the squared mean: \[ \operatorname{Var}(T)=\frac{1}{2}-\left(\frac{1}{2}\right)^2 =\frac{1}{2}-\frac{1}{4} =\frac{1}{4}\text{ day}^2. \] Check. The rate formulas give \(E[T]=1/2\) day and \(\operatorname{Var}(T)=1/2^2=1/4\) day squared. The common mistake to avoid is calling \(E[T^2]\) the variance; it must first be reduced by \((E[T])^2\).
Worked Example: Interpreting Rate, Mean, and Variability
For a hypothetical monitoring process, let \(W\) be the time in minutes between recorded events, with exponential density \(f_W(w)=0.5e^{-0.5w}\) for \(w\geq0\). A report states that the model's mean is 2 minutes and its variance is 2 minutes squared. Determine whether each statement is correct, and give the standard deviation.
Method. Read the rate from the density: \(\lambda=0.5\) per minute. The rate's units are inverse minutes, so use its reciprocal for the mean: \[ E[W]=\frac{1}{0.5}=2\text{ minutes}. \] The report's mean is correct.
For the variance, square the rate before taking the reciprocal: \[ \operatorname{Var}(W)=\frac{1}{(0.5)^2} =\frac{1}{0.25}=4\text{ minutes}^2. \] Thus the report's variance is incorrect; it is \(4\), not \(2\), minutes squared. The standard deviation is \(\sqrt{4}=2\) minutes.
Check. The second raw moment is \(E[W^2]=2/(0.5)^2=2/0.25=8\text{ minutes}^2\), and \(8-(2)^2=4\text{ minutes}^2\), matching the variance formula. The common mistake is to use \(1/\lambda\) for the variance as well as for the mean; the variance requires \(1/\lambda^2\).
Common Errors and Practical Checks
The derivation supplies a short checklist for calculations. Before substituting values, confirm that the parameter is the rate in the density, not a scale parameter written in another convention. Then check that the result's units are appropriate and, if using raw moments, subtract the squared mean to obtain the variance.
- Confusing rate with mean. If the density is \(\lambda e^{-\lambda x}\), then \(\lambda\) has inverse units and \(E[X]=1/\lambda\), not \(\lambda\).
- Using the mean formula for variance. The variance is \(1/\lambda^2\); it has squared units. The standard deviation, not the variance, is \(1/\lambda\).
- Stopping at the second raw moment. The integral gives \(E[X^2]=2/\lambda^2\). The variance is \(E[X^2]-(E[X])^2=1/\lambda^2\), not \(2/\lambda^2\).
- Ignoring the support. The density integral starts at zero because the exponential random variable is nonnegative. Integrating over the full real line as if the density had the same formula for negative values is not valid.
These moments summarize the scale of variation in the model; they do not establish that the exponential assumption is appropriate for a particular process. If that assumption is not justified, the formulas need not describe the process's actual mean or variance. The results here are aligned with the SOA VEE Mathematical Statistics syllabus.
Check Your Understanding
For each problem, identify the exponential rate and use the density-based moment results. Include units in your calculations.
- A hypothetical dispatch center models the time \(X\), in minutes, until a callback with exponential rate \(0.5\) per minute. Find the mean, variance, and standard deviation. (Answer: mean \(2\) minutes; variance \(4\) minutes squared; standard deviation \(2\) minutes.)
- A hypothetical equipment check has density \(f_T(t)=4e^{-4t}\) for \(t\geq0\), where \(t\) is in hours. Find \(E[T]\), \(E[T^2]\), and \(\operatorname{Var}(T)\). (Answer: \(E[T]=0.25\) hour; \(E[T^2]=0.125\) hour squared; \(\operatorname{Var}(T)=0.0625\) hour squared.)
- A hypothetical process has exponentially distributed completion time with mean \(3\) hours. Determine its rate in inverse hours and its variance in hours squared. (Answer: \(\lambda=1/3\) per hour; variance \(9\) hours squared.)
- Two hypothetical queues have exponential waiting times measured in days. Queue A has rate \(0.2\) per day and Queue B has rate \(0.5\) per day. Find each mean and variance, then identify which has greater variability by variance. (Answer: A has mean \(5\) days and variance \(25\) days squared; B has mean \(2\) days and variance \(4\) days squared; A has greater variance.)
- A hypothetical exponential waiting time \(Z\), measured in minutes, has variance \(16\) minutes squared. Find its positive rate and mean, and verify the variance using the rate. (Answer: \(\lambda=0.25\) per minute; mean \(4\) minutes; \(1/(0.25)^2=16\) minutes squared.)