Likelihood Ratio Tests For An Exponential Rate

Why Test an Exponential Rate?

An exponential model is often used for nonnegative waiting times when the event rate is treated as constant over time. A hypothesis about that rate might represent a specified event frequency in a hypothetical system, or a proposed failure rate in a hypothetical reliability study. The likelihood ratio test compares how well that specified rate explains the observations with how well the best-fitting rate explains them.

The central calculation is compact: for independent exponential observations, the likelihood depends on the data through their sum. This tutorial derives the resulting statistic and explains its large-sample reference distribution. The method is aligned with the SOA VEE Mathematical Statistics syllabus.

Model and assumptions: Let \(X_1,\ldots,X_n\) be independent observations from an exponential distribution with common rate \(\lambda>0\), with density \(f(x;\lambda)=\lambda e^{-\lambda x}\) for \(x>0\). The observations are complete, measured in the same time unit, and have no censoring. The null hypothesis specifies a fixed interior value \(\lambda_0>0\): \(H_0:\lambda=\lambda_0\). The unrestricted alternative allows any \(\lambda>0\), including values either smaller or larger than \(\lambda_0\).

Here \(\lambda\) is a rate per unit time, so its units are the reciprocal of the units used for each \(X_i\). The observed sample sum is \(T=\sum_{i=1}^n X_i\). The assumptions matter: for example, observations with censoring or unequal exposure times require a likelihood that reflects those features rather than the complete-observation likelihood used here.

Likelihood and the Sample Sum

By independence, the joint likelihood is the product of the individual densities. Taking its logarithm gives the log-likelihood:

L(\lambda)=\lambda^n e^{-\lambda T}, \qquad \ell(\lambda)=n\log\lambda-\lambda T.

The data enter this expression only through \(T\). To maximize over all positive rates, differentiate with respect to \(\lambda\):

\frac{d\ell}{d\lambda}=\frac{n}{\lambda}-T, \qquad \frac{d^2\ell}{d\lambda^2}=-\frac{n}{\lambda^2}<0.

Setting the first derivative to zero gives the unrestricted maximum likelihood estimate \(\widehat{\lambda}=n/T\). The negative second derivative confirms that this stationary point is a maximum.

Definition: The likelihood ratio for this test is the largest likelihood allowed by the null hypothesis divided by the largest likelihood allowed by the unrestricted model: \(\Lambda=L(\lambda_0)/L(\widehat{\lambda})\). Thus \(0<\Lambda\leq 1\). The likelihood ratio test statistic is \(W=-2\log\Lambda\), so larger values indicate greater disagreement between the specified rate and the unrestricted fit.

Substituting the two likelihood values produces the statistic in terms of the sample sum:

\Lambda =\frac{\lambda_0^n e^{-\lambda_0T}} {(n/T)^n e^{-n}}, \qquad W=2\left[n\log\left(\frac{n}{\lambda_0T}\right)-n+\lambda_0T\right].

It is often convenient to set \(z=\lambda_0T/n\), the ratio of the sample mean to its null expectation \(1/\lambda_0\). Then

W=2n\left(z-1-\log z\right).

Why This Formula Measures Departure

The statistic is the difference between the maximized log-likelihood and the log-likelihood under the null, multiplied by two. In particular, \(\ell(\widehat{\lambda})-\ell(\lambda_0)\geq 0\), because the unrestricted model includes \(\lambda_0\) among its possible values. This proves that \(W\) cannot be negative.

The expression \(z-1-\log z\) is nonnegative for \(z>0\). To see this, define \(g(z)=z-1-\log z\). Its derivative is \(g'(z)=1-1/z\), which is negative for \(0<z<1\) and positive for \(z>1\). Its minimum is therefore at \(z=1\), where \(g(1)=0\). Thus \(W=0\) when the sample mean equals its null expectation, and \(W>0\) otherwise. Departures in either direction increase the statistic.

This also explains why the two-sided alternative does not correspond to a single one-sided threshold on \(T\). Values of \(T\) well below \(n/\lambda_0\) suggest a rate greater than \(\lambda_0\); values well above it suggest a rate less than \(\lambda_0\). Either can produce a large likelihood ratio statistic.

Worked Example: A Sample Sum Near Its Null Expectation

A hypothetical reliability study records \(n=10\) complete waiting times, in hours, assumed independent and identically distributed as exponential with rate \(\lambda\) per hour. There is no censoring. Test \(H_0:\lambda=0.20\) per hour against the unrestricted alternative using the asymptotic 5% likelihood ratio test. No interest rate or mortality assumption is relevant to this waiting-time model.

Method. Calculate the unrestricted estimate and \(W\), then compare \(W\) with the 5% critical value from a chi-square distribution with one degree of freedom.

The observed total is \(T=40\) hours. The unrestricted estimate is \(\widehat{\lambda}=n/T=10/40=0.25\) per hour. Under the null, the expected total is \(n/\lambda_0=10/0.20=50\) hours, so the observed total is below its null expectation.

Compute \(\lambda_0T=0.20(40)=8\) and \(n/(\lambda_0T)=10/8=1.25\). Therefore

W=2\left[10\log(1.25)-10+8\right] =2\left[2.2314355-2\right] =0.462871\ldots

Rounded to four decimal places, \(W=0.4629\). The 5% critical value is approximately \(3.841\), so \(0.4629<3.841\): do not reject \(H_0\). The calculation is consistent with the observed sample mean, \(40/10=4\) hours, being reasonably near the null mean of \(1/0.20=5\) hours. A common mistake is to treat a point estimate above the null value as automatic evidence of rejection; the test instead measures the likelihood improvement and compares it with a calibrated threshold.

Asymptotic Calibration

Under the null, the rate \(\lambda_0\) is a positive interior parameter, and the independent exponential model is regular. The null fixes one parameter while the unrestricted model estimates one parameter. Wilks' theorem therefore gives the large-sample reference distribution

W\ \xrightarrow{d}\ \chi^2_1 \qquad\text{under }H_0.

For an asymptotic test of level \(\alpha\), reject when \(W>\chi^2_{1,1-\alpha}\), where \(\chi^2_{1,1-\alpha}\) is the \(1-\alpha\) quantile of a chi-square distribution with one degree of freedom. Equivalently, the approximate p-value is the upper-tail probability \(P(\chi^2_1\geq W)\). At the 5% level, the critical value is about \(3.841\).

The degrees of freedom are one because the unrestricted model has one more freely estimated parameter than the null model. This is an asymptotic calibration, not a claim that \(W\) has exactly a chi-square distribution for every finite sample size. Under \(H_0\), the exact distribution of \(2\lambda_0T\) is chi-square with \(2n\) degrees of freedom. That fact can support exact finite-sample procedures, but the statistic \(W\) itself is generally not exactly chi-square with one degree of freedom.

Worked Example: An Observed Total Below the Null Expectation

A hypothetical monitoring process produces \(n=8\) independent, complete exponential waiting times in days with common rate \(\lambda\) per day and no censoring. The observed sum is \(T=48\) days. Test \(H_0:\lambda=0.25\) per day against an unrestricted rate at the asymptotic 5% level. There is no interest rate or mortality assumption in this statistical model.

Method. Use the sample-sum likelihood ratio statistic and the one-degree-of-freedom asymptotic critical value.

Under the null, \(\lambda_0T=0.25(48)=12\), while \(n=8\). Hence \(n/(\lambda_0T)=8/12=2/3\), and

W=2\left[8\log(2/3)-8+12\right] =2\left[4+8(-0.4054651)\right] =2(0.756279) =1.512558\ldots

Rounded to four decimal places, \(W=1.5126\). Since \(1.5126<3.841\), do not reject the specified rate at the 5% level. The unrestricted estimate is \(8/48=1/6\), or about \(0.1667\) per day, below the null rate, as expected from a total \(T=48\) that exceeds the null expected total of \(8/0.25=32\) days. The common error of rounding the logarithm too early can change the reported statistic; retain adequate precision until the final rounding.

Worked Example: A Likelihood Ratio Large Enough to Reject

A hypothetical study records \(n=12\) independent, uncensored exponential waiting times measured in months, with common unknown rate \(\lambda\) per month. The observed sum is \(T=15\) months. Test \(H_0:\lambda=0.40\) per month against an unrestricted rate at the asymptotic 5% level. The observations have no interest rate or mortality assumption.

Method. Compare the maximized unrestricted likelihood with the likelihood at the specified null rate by calculating \(W\).

The unrestricted estimate is \(\widehat{\lambda}=12/15=0.80\) per month. Under the null, the expected sum is \(n/\lambda_0=12/0.40=30\) months, twice the observed total. Also, \(\lambda_0T=0.40(15)=6\), and \(n/(\lambda_0T)=12/6=2\). Thus

W=2\left[12\log 2-12+6\right] =2\left[8.317766-6\right] =4.635532\ldots

Rounded to four decimal places, \(W=4.6355\). Since \(4.6355>3.841\), reject \(H_0\) at the asymptotic 5% level. The result reflects a substantial likelihood improvement when the rate is allowed to rise from \(0.40\) to \(0.80\) per month. A common mistake is to compare \(W\) with a standard normal critical value: \(W\) is calibrated against \(\chi^2_1\), not directly against a standard normal variable.

Procedure and Common Traps

Check the model and units.
Confirm independent, complete exponential observations with a common rate, and express all waiting times in a consistent unit.

Find the sufficient data summary and estimate.
Calculate \(T=\sum X_i\) and \(\widehat{\lambda}=n/T\). The sample mean is \(T/n\).

Calculate the likelihood ratio statistic.
Use \(W=2[n\log\{n/(\lambda_0T)\}-n+\lambda_0T]\), keeping the rate and time units consistent.

Calibrate and state a contextual conclusion.
For the asymptotic test, compare \(W\) with the specified chi-square critical value and say whether the evidence warrants rejection.

Several errors are especially costly. First, the ratio inside the logarithm is \(n/(\lambda_0T)\), not its reciprocal, in this version of the formula. Second, the terms \(-n+\lambda_0T\) must remain in the statistic; omitting them can even produce an incorrect value when the estimate equals the null. Third, a two-sided unrestricted-rate test uses a large-\(W\) rejection region, not only a large- or only a small-\(T\) rejection region. Finally, the test does not prove that the null rate is true when it is not rejected; it says only that the observed likelihood improvement did not exceed the chosen threshold.

A quick check is to verify that \(W\geq0\), that \(W=0\) when \(T=n/\lambda_0\), and that extreme totals in either direction lead to large values. These checks can expose sign errors before a decision is made.

Key takeaway: For independent complete exponential observations with a common positive rate, the sample sum \(T\) determines the likelihood ratio statistic. Under a specified interior rate, the statistic is asymptotically compared with \(\chi^2_1\); the finite-sample statistic is not itself exactly chi-square.

Check Your Understanding

For each problem, use the asymptotic 5% test unless stated otherwise. Round reported test statistics to four decimal places.

Six independent, complete waiting times are modeled as exponential with rate \(\lambda\) per day. For \(H_0:\lambda=0.50\) per day, the observed total is 9 days. Calculate the unrestricted estimate and statistic, then decide. (Answer: \(\widehat{\lambda}=0.6667\) per day, \(W=0.4522\); do not reject.)
Ten independent, uncensored exponential waiting times are measured in hours and have observed total 150 hours. Test the specified rate \(\lambda_0=0.10\) per hour. Find \(W\) and the decision. (Answer: \(W=1.8907\); do not reject.)
Fifteen independent complete exponential observations, measured in weeks, have total \(T=40\) weeks. Test \(H_0:\lambda=0.20\) per week. Calculate the unrestricted rate and determine whether the null is rejected. (Answer: \(\widehat{\lambda}=0.3750\) per week, \(W=4.8583\); reject.)
Twenty-five independent, complete exponential waiting times have a total of 10 years. Test a specified rate of \(2\) per year. State the expected total under the null and the asymptotic decision. (Answer: null expected total \(=12.5\) years, \(W=1.1572\); do not reject.)
Nine independent, uncensored exponential observations in days have total \(T=15\) days. For \(H_0:\lambda=0.30\) per day, calculate \(z=\lambda_0T/n\), \(W\), and the 5% decision. (Answer: \(z=0.5\), \(W=3.4767\); do not reject.)

What You'll Learn