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VEE Mathematical Statistics · Estimators and Moments Methods · Tutorial 441 of 880

Uniform Endpoint Estimation From The Mean

Learn how matching the sample mean to the uniform model’s population mean gives an estimate of its unknown upper endpoint.

Intermediate 10 min read

What You'll Learn

  • State the support and assumptions for a uniform model from zero to an unknown endpoint.
  • Derive the population mean of that model from its density.
  • Match the population and sample first raw moments to obtain the endpoint estimator.
  • Calculate an endpoint estimate from observed sample values.
  • Distinguish the method-of-moments estimate from the sample maximum.
  • Recognize when a method-of-moments estimate falls below an observed value.

Estimating a Uniform Upper Endpoint

Suppose a measurement is modeled as uniformly distributed between zero and an unknown positive endpoint. The endpoint determines the range of possible values and also determines the population mean. Matching that mean to the observed sample mean gives a direct method-of-moments estimate.

The model is intentionally specific: the lower endpoint is known to be zero, and only the upper endpoint is unknown. The result below does not estimate two unknown endpoints or apply automatically to a uniform distribution on some other interval.

Definition: Let \(X_1,\ldots,X_n\) be independent and identically distributed observations from \(U(0,\theta)\), where the unknown parameter satisfies \(\theta>0\). The density is \(f_X(x;\theta)=1/\theta\) for \(0\leq x\leq\theta\), and \(f_X(x;\theta)=0\) otherwise. The method-of-moments estimator of \(\theta\) matches the population first raw moment \(E[X]\) to the sample first raw moment \(\bar X=(1/n)\sum_{i=1}^n X_i\).

Deriving the Estimator

The population first raw moment is the expected value of one observation. Using the density specified by the model,

$$ E[X]=\int_0^\theta x f_X(x;\theta)\,dx =\int_0^\theta x\frac{1}{\theta}\,dx =\frac{1}{\theta}\left[\frac{x^2}{2}\right]_0^\theta =\frac{\theta}{2}. $$

The sample first raw moment is the arithmetic mean of the observations. The method of moments equates these two quantities and solves for the unknown parameter:

$$ \bar X=\frac{\theta}{2} \quad\Longrightarrow\quad \hat{\theta}_{\mathrm{MM}}=2\bar X. $$

Thus, to calculate the estimate, add the observations, divide by the sample size, and double the resulting mean. If the observed data are \(x_1,\ldots,x_n\), the numerical estimate is \(\hat{\theta}_{\mathrm{MM}}=2\bar{x}\), where \(\bar{x}=(1/n)\sum_{i=1}^n x_i\).

Result: Under the independent, identically distributed \(U(0,\theta)\) model with known lower endpoint zero and \(\theta>0\), the method-of-moments estimator is \(2\bar X\). The estimate uses the sample mean; it is not defined by taking the largest observation.

Worked Calculations

Worked Example: Estimating a Range of Hypothetical Processing Times

A hypothetical audit records four processing times, in hours: \(3, 4, 6,\) and \(7\). Assume the times are independent observations from \(U(0,\theta)\), with unknown upper endpoint \(\theta\) hours. Find the method-of-moments estimate and check it against the observations.

Method. Use the first-moment equation \(E[X]=\theta/2\) and match it to the sample mean.

Step 1: Calculate the sample mean. The sum is \(3+4+6+7=20\) hours, and the sample size is \(n=4\). Therefore, \(\bar{x}=20/4=5\) hours.

Step 2: Double the sample mean. The estimator gives \(\hat{\theta}_{\mathrm{MM}}=2\bar{x}=2(5)=10\) hours.

Step 3: Check the calculation. Under the estimated model, the implied population mean is \(10/2=5\) hours, equal to the observed sample mean. Each observation is between zero and the estimated endpoint of 10 hours.

Common mistake avoided. The largest observation is 7 hours, but the method-of-moments estimate is 10 hours. Substituting the sample maximum for the endpoint would not match the model mean to the sample mean.

Worked Example: Estimating a Hypothetical Sensor Range

A hypothetical sensor records five positive response times, in seconds: \(18, 21, 24, 27,\) and \(30\). Assume the observations are independent and identically distributed as \(U(0,\theta)\), where \(\theta\) is an unknown upper endpoint in seconds. Calculate the method-of-moments estimate.

Method. Calculate the first sample raw moment and solve the first-moment equation for \(\theta\).

Step 1: Add the values. Their sum is \(18+21+24+27+30=120\) seconds.

Step 2: Find the sample mean. There are five observations, so \(\bar{x}=120/5=24\) seconds.

Step 3: Apply the estimator. The estimated endpoint is \(\hat{\theta}_{\mathrm{MM}}=2(24)=48\) seconds.

Step 4: Check the result. The model with endpoint 48 seconds has mean \(48/2=24\) seconds, which agrees with the sample mean. The maximum observed time, 30 seconds, is below the estimated endpoint.

Common mistake avoided. Do not divide the sample mean by two. The model mean is half the endpoint, so solving \(\bar{x}=\theta/2\) for the endpoint requires multiplying the sample mean by two.

Worked Example: When the Estimate Is Below an Observation

A hypothetical study records four measurements, in millimeters: \(2, 2, 2,\) and \(14\). Suppose the model assumption is that these are independent observations from \(U(0,\theta)\), with unknown \(\theta>0\). Find the method-of-moments estimate and assess what the observed maximum implies.

Method. Apply the first-moment equation, without replacing it by a rule based on the sample maximum.

Step 1: Calculate the sample mean. The sum is \(2+2+2+14=20\) millimeters. With \(n=4\), the mean is \(\bar{x}=20/4=5\) millimeters.

Step 2: Match moments. The estimate is \(\hat{\theta}_{\mathrm{MM}}=2\bar{x}=2(5)=10\) millimeters.

Step 3: Compare with the data. The observed maximum is 14 millimeters, which exceeds the method-of-moments estimate of 10 millimeters. The calculation is still the method-of-moments estimate because it solves the specified moment equation. However, the fitted \(U(0,10)\) model cannot include an observation of 14; a parameter value compatible with all four observations must be at least 14 millimeters.

Common mistake avoided. Setting the method-of-moments estimate equal to 14 millimeters would silently change the estimation rule. Report \(10\) millimeters as the method-of-moments result, and separately note the support mismatch rather than concealing it.

Why the Formula Works and What It Guarantees

The method works because the endpoint controls the first population moment in a simple, one-to-one way: \(E[X]=\theta/2\). Setting that model quantity equal to the observed first moment uses the sample mean as the estimate of the model mean, then algebraically solves for the endpoint. The calculation is not based on a claim that the sample mean is itself an endpoint.

There is also a useful expectation check. Under the stated model, every \(X_i\) has expectation \(\theta/2\). By linearity of expectation,

$$ E[\hat{\theta}_{\mathrm{MM}}] =E[2\bar X] =2E\left[\frac{1}{n}\sum_{i=1}^n X_i\right] =\frac{2}{n}\sum_{i=1}^n E[X_i] =\frac{2}{n}\left(n\frac{\theta}{2}\right) =\theta. $$

Therefore, under this model, the estimator is unbiased: its expected value over repeated samples equals the true endpoint. This fact does not mean that every individual estimate equals the true endpoint, nor does it guarantee that every computed estimate is large enough to contain every observation in its sample.

The support requirement deserves particular attention. Under \(U(0,\theta)\), every possible observation must lie between zero and \(\theta\). A particular sample with maximum \(x_{\max}\) can therefore be compatible with the model only if \(\theta\geq x_{\max}\). Yet \(2\bar{x}\) can be less than \(x_{\max}\), as the third example showed. The method-of-moments equation still gives \(2\bar{x}\); a support conflict should be reported as a diagnostic, not “fixed” by replacing the answer with the maximum.

1
State the model.
Confirm that the observations are independent and identically distributed as \(U(0,\theta)\), with known lower endpoint zero and unknown \(\theta>0\).
2
Write the population moment.
For this model, \(E[X]=\theta/2\).
3
Calculate the sample moment.
Find \(\bar{x}=(1/n)\sum x_i\), keeping the measurement units consistent.
4
Solve and check.
Set \(\bar{x}=\theta/2\) to obtain \(\hat{\theta}_{\mathrm{MM}}=2\bar{x}\). Compare the estimate with the sample maximum and report any support mismatch.

Common Exam Traps

  • Using the maximum instead of matching the mean. The method-of-moments estimate for this model is twice the sample mean, not the sample maximum.
  • Reversing the factor of two. Since the population mean is half the endpoint, the endpoint estimate is twice the sample mean.
  • Forgetting the model’s known lower endpoint. The formula \(2\bar{x}\) is derived for \(U(0,\theta)\); it is not the formula for a model with an unknown lower endpoint.
  • Ignoring the support check. If \(2\bar{x}<x_{\max}\), the method-of-moments calculation remains \(2\bar{x}\), but that fitted endpoint does not cover the observed sample.

When presenting a calculation, state the model and units, show the sample mean, write the moment equation, and solve for the endpoint. This makes clear both why the estimate has the value reported and whether the fitted support is consistent with the observed values. This focused endpoint calculation is aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, assume independent, identically distributed observations from the stated \(U(0,\theta)\) model. Find the method-of-moments estimate and perform any requested support check.

  1. A hypothetical inspection records \(5, 7, 8,\) and \(10\) seconds. Find the estimated upper endpoint in seconds. (Answer: \(15\) seconds.)
  2. Six hypothetical readings, in units, are \(12, 16, 20, 24, 28,\) and \(30\). Find the estimate and compare it with the observed maximum. (Answer: \(\hat{\theta}_{\mathrm{MM}}=43\frac{1}{3}\) units; it exceeds the maximum of 30 units.)
  3. Five hypothetical measurements, in centimeters, are \(0.4, 0.8, 1.2, 1.6,\) and \(2.0\). Calculate the estimate in centimeters. (Answer: \(2.4\) centimeters.)
  4. A hypothetical sample, in units, is \(3, 3, 3,\) and \(15\). Find the method-of-moments estimate and state whether its fitted support covers all observations. (Answer: \(12\) units; no, the estimate is below the observed maximum of 15 units.)
  5. Seven hypothetical readings, in minutes, are \(9, 11, 13, 17, 20, 22,\) and \(28\). Calculate the estimate, rounding only the final result to two decimal places, and check it against the maximum. (Answer: \(34.29\) minutes; it exceeds the maximum of 28 minutes.)

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