Why Poisson Rate Tests Reduce to Count Thresholds
A Poisson model is used for counts of events occurring over a fixed exposure, such as hypothetical equipment alarms per operating hour or claims per policy-month. When comparing two specified rates, the observed count contains the evidence for deciding between the hypotheses. The Neyman–Pearson likelihood ratio shows exactly how that evidence is ordered: if the alternative rate is higher, larger counts favor it; if the alternative rate is lower, smaller counts favor it.
The construction is aligned with the SOA VEE Mathematical Statistics syllabus. The hypotheses here are simple: each specifies a single Poisson rate. The exposure is fixed and known, and the Poisson model is assumed to apply under both hypotheses. No interest-rate or mortality assumption is involved in these count examples.
A test can be represented by a function \(\phi(n)\), the probability of rejecting \(H_0\) after observing count \(n\). A nonrandomized test has \(\phi(n)\) equal to either zero or one. A randomized test may assign a probability strictly between zero and one at a boundary count. The size is the probability of rejection under \(H_0\), and the power is the probability of rejection under \(H_1\). For these simple hypotheses, the Neyman–Pearson lemma says that a test rejecting for the largest values of the likelihood ratio is most powerful among tests with the same size.
The Likelihood Ratio Gives the Direction
Write \(\mu_0=t\lambda_0\) and \(\mu_1=t\lambda_1\). For an observed count \(n\), the likelihood ratio in favor of the alternative is \(L(n)=P_{\lambda_1}(N=n)/P_{\lambda_0}(N=n)\). Substitution of the two Poisson probabilities gives
The factor \(e^{-(\mu_1-\mu_0)}\) does not depend on the observed count. The remaining factor changes monotonically with \(n\): it increases when \(\mu_1>\mu_0\), and decreases when \(\mu_1<\mu_0\). Because exposure \(t\) is the same under both hypotheses, the ordering of the rates and the ordering of the means are the same.
Here is the reason the ordering leads to a most powerful test. The Neyman–Pearson lemma selects outcomes with the largest values of \(L(n)\), subject to the required probability of rejection under \(H_0\). If the ratio increases with \(n\), those outcomes are the largest counts; if it decreases, they are the smallest counts. Since Poisson counts are discrete, a threshold can fall between attainable sizes. Randomizing at the boundary count can fill the remaining probability exactly.
For fixed exposure \(t\), set \(\mu_0=t\lambda_0\) and \(\mu_1=t\lambda_1\).
If \(\lambda_1>\lambda_0\), use an upper-count region; if \(\lambda_1<\lambda_0\), use a lower-count region.
Sum the null probabilities over the rejection region. If the target size lies between two attainable sizes, randomize at the boundary.
Use the same rejection rule, but sum probabilities using mean \(\mu_1\), including the boundary randomization if present.
Exact Size and Power for Threshold Tests
For a nonrandomized upper-tail test that rejects when \(N\geq c\), its size and power are
For a lower-tail test that rejects when \(N\leq c\), replace each upper-tail probability by the corresponding lower-tail probability. These are exact probabilities under the stated Poisson models; no normal approximation is needed. A Poisson tail can be computed as a sum of point probabilities or by subtracting the complementary tail from one.
Suppose an upper-tail test must have a specified size \(\alpha\), but no nonrandomized threshold gives exactly that size. Choose a boundary count \(c\) for which \(P_{\mu_0}(N>c)\leq\alpha\leq P_{\mu_0}(N\geq c)\). Reject whenever \(N>c\), and reject with probability \(\gamma\) when \(N=c\). The null rejection probability is then
For a lower-tail test, use \(P_{\mu_0}(N<c)\leq\alpha\leq P_{\mu_0}(N\leq c)\), reject for \(N<c\), and randomize at \(c\) with \(\gamma=[\alpha-P_{\mu_0}(N<c)]/P_{\mu_0}(N=c)\). In either direction, the same \(\gamma\) is used to calculate power, but the boundary probability is evaluated under the alternative.
Worked Example: A Higher Alternative Rate
A hypothetical monitoring system records event counts over one fixed operating interval. Compare \(H_0:\lambda=2\) events per interval with \(H_1:\lambda=5\) events per interval. The model is Poisson with exposure \(t=1\) interval, so \(\mu_0=2\) and \(\mu_1=5\). Find the most powerful nonrandomized rule that rejects for counts of at least 4, then calculate its exact size and power.
Method and direction. The likelihood ratio is \(L(n)=e^{-3}(5/2)^n\), which increases with \(n\). Thus an upper-count region is the Neyman–Pearson ordering. The requested rule is to reject when \(N\geq4\).
Size under \(H_0\). The nonrejection counts are 0, 1, 2, and 3. Their probabilities are \(e^{-2}\), \(2e^{-2}\), \(2^2e^{-2}/2!=2e^{-2}\), and \(2^3e^{-2}/3!=(4/3)e^{-2}\). Therefore
Power under \(H_1\). Under mean 5, the probabilities of counts 0 through 3 sum to \(e^{-5}(1+5+25/2+125/6)=e^{-5}(118/3)\). Hence
The size is about 0.1429, and the power is about 0.7350. The method uses the alternative-to-null likelihood ratio and then sums the probabilities over the stated rejection region. A common mistake is to reverse the ratio: \(P_{\mu_0}(N=n)/P_{\mu_1}(N=n)\) decreases with \(n\), but that is evidence against the alternative ordering, not the ratio used to select outcomes favoring \(H_1\). The final results are checked by noting that the upper-tail probability is larger under mean 5 than under mean 2.
Worked Example: A Lower Alternative and Exact Target Size
A hypothetical laboratory counts a rare signal in a fixed observation window. Compare \(H_0:\lambda=4\) signals per window with \(H_1:\lambda=1\) signal per window, using a Poisson model with \(t=1\) window. Construct a most powerful test of size exactly 0.10, allowing randomization if needed, and calculate its power.
Method and direction. The means are \(\mu_0=4\) and \(\mu_1=1\). The likelihood ratio is \(L(n)=e^3(1/4)^n\), which decreases with \(n\). Small counts therefore favor \(H_1\), so begin with the lower tail.
Under \(H_0\), rejecting for \(N\leq1\) has probability \(P_{\mu_0}(N\leq1)=e^{-4}(1+4)=0.0915782\), which is below 0.10. The next boundary count is 2, with probability \(P_{\mu_0}(N=2)=e^{-4}4^2/2!=8e^{-4}=0.1465251\). Reject for \(N\leq1\), and reject with probability \(\gamma\) when \(N=2\), where
The size is \(5e^{-4}+\gamma(8e^{-4})=0.1000\), as required. Under \(H_1\), the probability of rejecting for \(N\leq1\) is \(e^{-1}(1+1)=2e^{-1}\), and the probability of the boundary count is \(P_{\mu_1}(N=2)=e^{-1}/2\). Thus
The method avoids rounding the size up to the nonrandomized value 0.0916 or jumping to the larger value for \(N\leq2\). The boundary randomization gives exactly 0.10 under the null. As a check, the alternative mean is smaller, so the lower-tail rejection rule has substantially greater probability under the alternative than under the null.
Worked Example: Exposure, an Upper Tail, and Randomization
A hypothetical sensor records alarms over 50 operating days. Compare a rate of \(0.04\) alarms per day under \(H_0\) with \(0.07\) alarms per day under \(H_1\). Assume a Poisson count over the fixed 50-day exposure and construct a most powerful test of size exactly 0.10. Find the boundary randomization probability and the power.
Method and direction. Convert each rate to a mean count over the actual exposure: \(\mu_0=50(0.04)=2\) and \(\mu_1=50(0.07)=3.5\). Since the alternative rate is higher, the likelihood ratio increases with the observed alarm count, so use an upper tail.
Under \(H_0\), the probability of \(N\geq5\) is \(1-e^{-2}(1+2+2+8/6+16/24)=1-7e^{-2}=0.0526530\). The boundary probability at 4 is \(P_{\mu_0}(N=4)=e^{-2}2^4/4!=(2/3)e^{-2}=0.0902235\). Therefore reject for \(N\geq5\), and reject with probability
when \(N=4\). This produces size \(0.0526530+(0.5248)(0.0902235)\), which rounds to 0.1000.
For power, under mean 3.5 the probability of counts 0 through 4 is \(e^{-3.5}(1+3.5+6.125+7.1458333+6.2526042)=0.7254450\). Therefore \(P_{\mu_1}(N\geq5)=1-0.7254450=0.2745550\), and \(P_{\mu_1}(N=4)=e^{-3.5}(3.5)^4/4!=0.1888123\). The power is
The final check is that the boundary probability is between zero and one and that the randomized null rejection probability is the target size. A common mistake is to use the rate \(0.04\) as the Poisson mean; the count is observed over 50 days, so the null mean is 2, not 0.04.
Common Exam Traps and Interpretation
A Poisson rate test is short to construct, but several details affect the answer:
- Using the wrong likelihood-ratio direction. For the Neyman–Pearson ordering in favor of \(H_1\), use \(P_{\lambda_1}(N=n)/P_{\lambda_0}(N=n)\). Reversing it reverses the ordering.
- Using a lower tail for every test. The direction follows from the specified alternative: a higher alternative rate gives an upper tail, while a lower alternative rate gives a lower tail.
- Forgetting the exposure. The Poisson mean is \(t\lambda\), not the rate \(\lambda\) alone, unless the exposure is one unit.
- Claiming a discrete threshold has exactly the target size. A nonrandomized test may only have size below or above the target. If exact size is required, check whether a boundary randomization is needed.
- Using the null mean to calculate power. Size is calculated under \(H_0\); power is calculated under \(H_1\), with the same rejection rule.
The size is the probability of a false rejection when the null rate is true; the power is the probability of rejection when the specified alternative rate is true. A high observed count can support a higher rate, but the test does not establish that the alternative is true. Its error probabilities are conditional on the two specified models. If the rate is not fixed at one of these two values, this simple-versus-simple Neyman–Pearson construction does not by itself describe a test for that broader hypothesis.
Check Your Understanding
For each problem, assume the stated Poisson model and use exact Poisson probabilities. Answers are rounded to four decimal places.
- A hypothetical counter observes one interval of events. Test \(H_0:\lambda=1.5\) against \(H_1:\lambda=3\) using the rule \(N\geq3\). Determine the direction of the likelihood-ratio ordering, the size, and the power. (Answer: Upper-tail ordering; size \(0.1912\), power \(0.5768\).)
- A hypothetical device produces counts over one fixed window. Test \(H_0:\lambda=3\) against \(H_1:\lambda=1.5\) with the nonrandomized rule \(N\leq2\). Calculate the size and power. (Answer: Lower-tail ordering; size \(0.4232\), power \(0.8088\).)
- For one observation with \(N\sim\operatorname{Poisson}(\lambda)\), test \(H_0:\lambda=1\) against \(H_1:\lambda=2\) at size exactly 0.10. Use an upper-tail test and randomize at the boundary if necessary. Give the boundary count, randomization probability, and power. (Answer: Reject for \(N\geq3\), randomize at \(N=2\) with probability \(0.1071\); power \(0.3523\).)
- A hypothetical monitoring system runs for 20 hours. Test a rate of \(0.05\) events per hour against \(0.10\) events per hour using the rule \(N\geq2\). Convert the rates to Poisson means, then calculate size and power. (Answer: Means \(1\) and \(2\); upper-tail ordering; size \(0.2642\), power \(0.5940\).)
- A hypothetical counter observes one interval. Test \(H_0:\lambda=4\) against \(H_1:\lambda=6\) using the nonrandomized rule \(N\geq5\). Calculate the size and power, and state whether the rule has exact size 0.10. (Answer: Upper-tail ordering; size \(0.3712\), power \(0.7149\); no, its size is not \(0.10\).)