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Estimating Shift And Scale In An Exponential Model

Match the first two sample moments to estimate the location shift and scale of a shifted exponential distribution, and identify when the equations have no valid solution.

Intermediate 10 min read

What You'll Learn

  • State the shifted exponential model using its location and scale parameters.
  • Derive its first two raw moments from the exponential model.
  • Match sample raw moments to obtain shift and scale estimates.
  • Check the estimates by substituting them back into the matched moments.
  • Recognize why a constant sample gives no valid scale estimate.
  • Avoid using the sample variance with divisor n minus one in these moment equations.

Why Shifted Exponential Estimation Needs Two Moments

An exponential model describes a nonnegative waiting time measured from a fixed starting point. Some waiting times, however, cannot fall below a positive baseline. A shifted exponential model allows for that baseline: its shift sets the lower endpoint, while its scale controls the amount of variation above that endpoint. Estimating both parameters requires two independent moment equations.

This tutorial derives the method-of-moments estimates from the first two raw moments. The key point is that shifting the distribution changes its mean but not its variance. The sample variance formed with divisor \(n\) therefore identifies the scale, after which the sample mean identifies the shift. These estimates are useful only when the sample has positive variation.

Definition: A shifted exponential random variable has the representation \(X=\theta+Z\), where \(\theta\) is a real-valued shift, \(\beta>0\) is a scale parameter, and \(Z\) has an exponential distribution with scale \(\beta\). Equivalently, its density is \(f_X(x)=\frac{1}{\beta}\exp\!\left(-\frac{x-\theta}{\beta}\right)\) for \(x\geq\theta\), and zero for \(x<\theta\). Thus \(X\) cannot be less than \(\theta\).

Assume \(X_1,\ldots,X_n\) are an independent, identically distributed sample from this model. We use the scale convention: the unshifted exponential variable \(Z\) has \(E[Z]=\beta\) and \(\operatorname{Var}(Z)=\beta^2\). Some sources instead describe an exponential distribution by its rate \(\lambda=1/\beta\); the estimates below first estimate the scale.

The Population Moments

The first two raw moments are the population quantities to be matched. Since \(X=\theta+Z\), the first raw moment follows by taking expectations. For the second, expand the square and use \(E[Z]=\beta\) and \(E[Z^2]=\operatorname{Var}(Z)+(E[Z])^2=2\beta^2\).

$$ E[X]=\theta+\beta $$
$$ E[X^2] =E[(\theta+Z)^2] =\theta^2+2\theta E[Z]+E[Z^2] =\theta^2+2\theta\beta+2\beta^2 $$

The variance gives another useful expression. Subtract the square of the mean from the second raw moment:

$$ \operatorname{Var}(X) =E[X^2]-(E[X])^2 =(\theta^2+2\theta\beta+2\beta^2)-(\theta+\beta)^2 =\beta^2. $$

This cancellation explains why the shift does not affect the variance: adding the constant \(\theta\) moves every possible value by the same amount, without changing the distances among values. The variance depends only on the scale.

Deriving the Method-of-Moments Estimates

For the sample, define the first and second raw moments by \(\overline X=\frac{1}{n}\sum_{j=1}^{n}X_j\) and \(M_2=\frac{1}{n}\sum_{j=1}^{n}X_j^2\). The corresponding sample central moment with divisor \(n\) is \(V_n=M_2-\overline X^2\). Method of moments sets these sample moments equal to the matching population raw moments.

1
Match the first raw moment.
Set \(\overline X=\theta+\beta\). This gives \(\theta=\overline X-\beta\).
2
Match the second raw moment.
Set \(M_2=\theta^2+2\theta\beta+2\beta^2\). The right-hand side is also \((\theta+\beta)^2+\beta^2\).
3
Use the first equation in the second.
Because \(\theta+\beta=\overline X\), the second equation becomes \(M_2=\overline X^2+\beta^2\), so \(\beta^2=M_2-\overline X^2=V_n\).
4
Select the parameter-space solution.
The scale must be positive. If \(V_n>0\), take \(\widehat\beta_{\mathrm{MOM}}=\sqrt{V_n}\), then use the first equation to obtain \(\widehat\theta_{\mathrm{MOM}}=\overline X-\sqrt{V_n}\).
Theorem: If \(V_n=M_2-\overline X^2>0\), the method-of-moments estimates based on the first two raw sample moments are
$$ \widehat\beta_{\mathrm{MOM}}=\sqrt{V_n}, \qquad \widehat\theta_{\mathrm{MOM}}=\overline X-\sqrt{V_n}. $$
If \(V_n=0\), the moment equations have no solution with \(\beta>0\). In particular, a sample of size one or any constant sample does not yield a valid shifted-exponential method-of-moments estimate.

The nonnegative square root is required because \(\beta\) is a scale and the model requires \(\beta>0\). The condition \(V_n>0\) is therefore part of the result, not an optional technical detail. If the rate parameterization is desired after estimating the scale, the corresponding rate estimate is \(1/\widehat\beta_{\mathrm{MOM}}\), provided the scale estimate is positive.

Worked Applications

Worked Example: Estimating a Baseline and Scale for Processing Times

A hypothetical processing system records four independent times, in minutes: \(2,2,6,6\). Assume the times follow a shifted exponential distribution with shift \(\theta\) minutes and scale \(\beta>0\) minutes. Estimate both parameters by matching the first two raw moments.

Method. Compute \(\overline X\) and \(M_2\), form \(V_n=M_2-\overline X^2\), then apply the moment estimates.

First raw sample moment.

$$ \overline X=\frac{2+2+6+6}{4}=\frac{16}{4}=4\text{ minutes}. $$

Second raw sample moment.

$$ M_2=\frac{2^2+2^2+6^2+6^2}{4} =\frac{4+4+36+36}{4} =\frac{80}{4}=20\text{ minutes}^2. $$

Sample central moment and estimates.

$$ V_n=20-4^2=20-16=4\text{ minutes}^2, \qquad \widehat\beta_{\mathrm{MOM}}=\sqrt{4}=2\text{ minutes}, $$ $$ \widehat\theta_{\mathrm{MOM}}=4-2=2\text{ minutes}. $$

Check. The fitted population mean is \(2+2=4\) minutes. Its second raw moment is \(2^2+2(2)(2)+2(2^2)=4+8+8=20\text{ minutes}^2\). Both match the sample moments. The estimated lower endpoint is also the smallest observed time.

Common mistake avoided: Do not treat the shift as the sample mean. The sample mean estimates the sum \(\theta+\beta\); the second moment is needed to separate its two components.

Worked Example: Using the Raw Moments for Inspection Durations

A hypothetical inspection process produces four independent durations, in hours: \(1,1,1,5\). Assume a shifted exponential model with shift \(\theta\) hours and scale \(\beta>0\) hours. Find the method-of-moments estimates, rounding to four decimal places.

First raw sample moment.

$$ \overline X=\frac{1+1+1+5}{4}=\frac{8}{4}=2\text{ hours}. $$

Second raw sample moment.

$$ M_2=\frac{1^2+1^2+1^2+5^2}{4} =\frac{1+1+1+25}{4} =\frac{28}{4}=7\text{ hours}^2. $$

Variance-based moment and parameter estimates.

$$ V_n=7-2^2=3\text{ hours}^2, \qquad \widehat\beta_{\mathrm{MOM}}=\sqrt{3}\approx1.7321\text{ hours}, $$ $$ \widehat\theta_{\mathrm{MOM}}=2-\sqrt{3}\approx0.2679\text{ hours}. $$

Check. The fitted mean is \((2-\sqrt{3})+\sqrt{3}=2\) hours. Using \(E[X^2]=(\theta+\beta)^2+\beta^2\), the fitted second raw moment is \(2^2+(\sqrt{3})^2=4+3=7\text{ hours}^2\). These equal the two sample raw moments.

Common mistake avoided: Do not round \(\sqrt{3}\) before computing the shift. Using the unrounded value gives \(\widehat\theta=2-\sqrt{3}\); rounding both estimates only at the final step preserves the moment match.

Worked Example: Checking the Lower Endpoint in a Delay Model

A hypothetical reporting system records four independent delays, in days: \(3,4,7,12\). Assume a shifted exponential distribution with shift \(\theta\) days and scale \(\beta>0\) days. Estimate the parameters and verify the estimates using the fitted moments.

First raw sample moment.

$$ \overline X=\frac{3+4+7+12}{4}=\frac{26}{4}=6.5\text{ days}. $$

Second raw sample moment.

$$ M_2=\frac{3^2+4^2+7^2+12^2}{4} =\frac{9+16+49+144}{4} =\frac{218}{4}=54.5\text{ days}^2. $$

Compute the estimates.

$$ V_n=54.5-(6.5)^2=54.5-42.25=12.25\text{ days}^2, $$ $$ \widehat\beta_{\mathrm{MOM}}=\sqrt{12.25}=3.5\text{ days}, \qquad \widehat\theta_{\mathrm{MOM}}=6.5-3.5=3\text{ days}. $$

Check. The fitted mean is \(3+3.5=6.5\) days. The fitted second raw moment is \(3^2+2(3)(3.5)+2(3.5)^2=9+21+24.5=54.5\text{ days}^2\). The estimates reproduce both sample moments, and the estimated shift equals the smallest observed delay.

Common mistake avoided: The second raw moment is the average of the squared observations, not the square of their average. Here those quantities are \(54.5\) and \(42.25\), respectively.

Common Exam Trap: Using the Wrong Variance Divisor

Method of moments matches the sample raw moments defined with divisor \(n\). Consequently, the central moment in the equations is \(V_n=M_2-\overline X^2=\frac{1}{n}\sum_{j=1}^{n}(X_j-\overline X)^2\), not the usual unbiased sample variance with divisor \(n-1\). Using the latter changes the scale estimate and generally fails to reproduce the second raw moment.

Worked Example: Correcting a Divisor Error

Use the inspection durations \(1,1,1,5\) hours from the earlier calculation. A candidate incorrectly uses the variance with divisor \(n-1\), then estimates the scale by taking its square root. Identify the error and give the correct estimates.

The deviations from the sample mean \(2\) are \(-1,-1,-1,3\) hours. Their squared sum is \(1+1+1+9=12\text{ hours}^2\). The variance with divisor \(n-1=3\) is \(12/3=4\text{ hours}^2\), so the incorrect calculation gives scale \(2\) hours and shift \(2-2=0\) hours.

For method of moments, the required central moment uses divisor \(n=4\):

$$ V_n=\frac{12}{4}=3\text{ hours}^2, \qquad \widehat\beta_{\mathrm{MOM}}=\sqrt{3}\text{ hours}, \qquad \widehat\theta_{\mathrm{MOM}}=2-\sqrt{3}\text{ hours}. $$

The incorrect pair would imply a second raw moment of \((0+2)^2+2^2=8\text{ hours}^2\), not the observed \(M_2=7\text{ hours}^2\). The corrected pair implies \(2^2+3=7\text{ hours}^2\), as required.

Interpreting the Estimates

The method uses the sample mean to estimate the total location \(\theta+\beta\), and the sample central moment to estimate \(\beta^2\). Once the scale is obtained, subtracting it from the mean yields the shift. This is why both equations are needed: the first alone cannot distinguish shift from scale.

A zero sample central moment requires special care. If all recorded values are identical, including when \(n=1\), then \(V_n=0\). Setting \(\widehat\beta=0\) would violate the model condition \(\beta>0\). The correct conclusion is that there is no solution to these moment equations within the shifted exponential parameter space, not that the model has a valid zero-scale estimate.

The method-of-moments calculations are aligned with the SOA VEE Mathematical Statistics syllabus. For a duration or delay application, the units also matter: \(\theta\) and \(\beta\) have the same units as the observations, while \(V_n\) has squared units. Taking its square root returns a scale in the original units.

Check Your Understanding

For each nonconstant sample, use raw sample moments with divisor \(n\). Report parameter estimates to four decimal places when needed.

  1. A hypothetical sensor records \(2,2,6,6\) seconds. Find \(M_2\), \(V_n\), and the shift and scale estimates. (Answer: \(M_2=20\text{ seconds}^2\), \(V_n=4\text{ seconds}^2\), \(\widehat\theta_{\mathrm{MOM}}=2\) seconds, \(\widehat\beta_{\mathrm{MOM}}=2\) seconds.)
  2. A hypothetical service has completion times \(1,1,1,5\) hours. Find the shift and scale estimates. (Answer: \(\widehat\theta_{\mathrm{MOM}}=0.2679\) hours, \(\widehat\beta_{\mathrm{MOM}}=1.7321\) hours.)
  3. A hypothetical delivery process records delays \(3,4,7,12\) days. Calculate the two raw sample moments, then estimate shift and scale. (Answer: \(\overline X=6.5\) days, \(M_2=54.5\text{ days}^2\), \(\widehat\theta_{\mathrm{MOM}}=3\) days, \(\widehat\beta_{\mathrm{MOM}}=3.5\) days.)
  4. A hypothetical machine records operating times \(5,8,8,14,25\) hours. Compute the method-of-moments estimates and state whether \(V_n>0\). (Answer: \(\overline X=12\) hours, \(M_2=194.8\text{ hours}^2\), \(V_n=50.8\text{ hours}^2>0\), \(\widehat\beta_{\mathrm{MOM}}\approx7.1274\) hours, \(\widehat\theta_{\mathrm{MOM}}\approx4.8726\) hours.)
  5. A one-observation sample contains the value \(9\) days. Do the first two moment equations give a valid shifted exponential estimate? Explain using \(V_n\). (Answer: No. \(V_n=0\), which would require \(\beta=0\), contrary to the model requirement \(\beta>0\).)

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