Controlling Error Across a Composite Null

Why a Composite Null Requires More Than One Check

For a simple null hypothesis, the null specifies one probability model. For a composite null, it permits a set of parameter values, so the probability of rejecting the null can vary across those values. A test that has a small Type I error probability at one null parameter value may have a much larger one elsewhere in the null parameter space.

The relevant control is therefore not an error probability calculated at an arbitrarily selected null value. It is the largest Type I error probability permitted by the null. This is the central idea behind test size and is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Let $\Theta_0$ be the set of parameter values specified by the null hypothesis, and let $R$ be the rejection region of a test. At a particular $\theta\in\Theta_0$, the Type I error probability is $P_\theta(X\in R)$. The size of the test is $$ \sup_{\theta\in\Theta_0} P_\theta(X\in R). $$ If the supremum is attained, it is the maximum Type I error probability over the null. If it is not attained, the supremum is still the least upper bound. A test has level $\alpha$ if its size is at most $\alpha$.

The notation $P_\theta$ means that probabilities are calculated under the model with parameter value $\theta$. The same rejection rule can consequently have different rejection probabilities for different values of $\theta$. The size is the supremum over the null values only; rejection probabilities at parameter values in the alternative do not contribute to size.

A nominal significance level and a test’s size are related but not identical concepts. A claimed level $\alpha$ is a guarantee that the test’s size is no greater than $\alpha$. It is not enough that the Type I error probability at one selected null value be no greater than $\alpha$.

Deriving the Control Requirement

The reason for taking a supremum follows directly from what it means to control Type I error across a composite null. To guarantee a bound for every null value, the rejection probability must satisfy the bound at each member of $\Theta_0$. This is equivalent to requiring that its supremum satisfy the bound.

Write the pointwise error probability.
For each $\theta\in\Theta_0$, let $a(\theta)=P_\theta(X\in R)$. This is the Type I error probability if that particular null value is true.

Require control at every null value.
A level-$\alpha$ guarantee requires $a(\theta)\leq\alpha$ for every $\theta\in\Theta_0$, not merely for one chosen value.

Express the uniform bound as a supremum.
The inequalities $a(\theta)\leq\alpha$ for all $\theta\in\Theta_0$ hold exactly when $\sup_{\theta\in\Theta_0}a(\theta)\leq\alpha$. That supremum is the test’s size.

Worked Examples

Worked Example: A Normal Upper-Tail Test

Suppose one observation $X$ has distribution $N(\theta,1)$, with known variance $1$. Consider $H_0:\theta\leq 0$ against $H_1:\theta>0$. The test rejects when $X>1.645$. Find its size and check the error probability at $\theta=-1$. State whether the test has level $0.05$.

Method. Calculate the rejection probability as a function of the null parameter, then maximize it over all $\theta\leq 0$. The common mistake to avoid is evaluating only at a convenient interior null value and treating that as the size.

For any $\theta\leq0$, standardizing under $X\sim N(\theta,1)$ gives $$ P_\theta(X>1.645) =P\left(\frac{X-\theta}{1}>1.645-\theta\right) =1-\Phi(1.645-\theta), $$ where $\Phi$ is the standard normal cumulative distribution function. As $\theta$ increases toward $0$, $1.645-\theta$ decreases, so the upper-tail probability increases. Its largest value over $\theta\leq0$ is therefore at $\theta=0$: $$ \sup_{\theta\leq0}P_\theta(X>1.645) =1-\Phi(1.645) \approx 1-0.9500 =0.0500. $$ Thus the test has size approximately $0.05$ and is level $0.05$, subject to the rounding in the stated critical value.

At $\theta=-1$, the standardized cutoff is $1.645-(-1)=2.645$, so $$ P_{-1}(X>1.645)=1-\Phi(2.645)\approx0.0041. $$ This smaller pointwise error probability does not determine the size: the value at $\theta=0$ is larger. The check confirms that considering just $\theta=-1$ would understate the largest Type I error probability.

Check. The null includes $\theta=0$, and its rejection probability there is about $0.05$; all smaller null values have no greater rejection probability. This agrees with the calculated size.

Worked Example: A Binomial Test That Fails a One-Value Check

Let $X\sim\operatorname{Binomial}(2,p)$, and test $H_0:0\leq p\leq0.4$ using the rule “reject if $X\geq1$.” A reviewer checks the Type I error probability only at $p=0.2$ and finds it no greater than $0.4$. Determine whether that check establishes level $0.4$, and find the actual size.

Method. Express the rejection probability as a function of $p$ and maximize it on the entire null interval. The trap is to mistake a satisfactory check at one null value for a bound that holds throughout the composite null.

The test rejects unless both trials are failures. Thus, for $p\in[0,0.4]$, $$ P_p(X\geq1)=1-P_p(X=0)=1-(1-p)^2. $$ This probability increases as $p$ increases on the null interval: if $p$ increases, $1-p$ decreases, so $(1-p)^2$ decreases and $1-(1-p)^2$ increases. The largest value is consequently at $p=0.4$: $$ \sup_{0\leq p\leq0.4}P_p(X\geq1) =1-(1-0.4)^2 =1-(0.6)^2 =1-0.36 =0.64. $$ The size is $0.64$, so the test is not level $0.4$.

At the reviewer’s selected value $p=0.2$, the calculation is $$ P_{0.2}(X\geq1)=1-(1-0.2)^2 =1-(0.8)^2 =1-0.64 =0.36. $$ This pointwise error probability is indeed at most $0.4$, but the value $0.64$ at $p=0.4$ exceeds $0.4$. The proposed check therefore does not establish the claimed error control.

Check. The null parameter $p=0.4$ is allowed under $H_0$, and at that value the rejection probability is $0.64$. The test fails the level-$0.4$ requirement, consistent with its size.

Worked Example: The Largest Error Can Occur Inside the Null

Suppose $X\sim N(\theta,1)$ and $H_0:-1\leq\theta\leq1$. A test rejects when $-0.5<X<0.5$. Find the size and compare the rejection probability at $\theta=0$ and at the endpoint $\theta=1$. Explain why checking only the endpoints would be inadequate.

Method. Write the probability of the rejection interval as a function of $\theta$, then locate its maximum. The common mistake is to assume that a maximum over a parameter interval must occur at an endpoint.

For any $\theta$, standardization gives $$ g(\theta)=P_\theta(-0.5<X<0.5) =\Phi(0.5-\theta)-\Phi(-0.5-\theta). $$ At $\theta=0$, using standard normal CDF values with enough digits to show the subtraction, $$ g(0)=\Phi(0.5)-\Phi(-0.5) \approx 0.69146246-0.30853754 =0.38292492 \approx0.3829. $$ At $\theta=1$, $$ g(1)=\Phi(-0.5)-\Phi(-1.5) \approx0.30853754-0.06680720 =0.24173034 \approx0.2417. $$ By symmetry, $g(-1)=g(1)$. These endpoint values are both smaller than $g(0)$.

To verify that $\theta=0$ gives the maximum on the whole null interval, differentiate: $$ g'(\theta)=-\phi(0.5-\theta)+\phi(-0.5-\theta), $$ where $\phi$ is the standard normal density. For $\theta>0$, $|-0.5-\theta|>|0.5-\theta|$. Since the standard normal density decreases as the absolute value of its argument increases, $\phi(-0.5-\theta)<\phi(0.5-\theta)$, so $g'(\theta)<0$. For $\theta<0$, the reverse comparison gives $g'(\theta)>0$. Therefore $g$ increases up to $0$ and decreases after $0$; its maximum on $[-1,1]$ occurs at the interior point $\theta=0$.

The size is thus $g(0)\approx0.3829$. Looking only at $\theta=-1$ and $\theta=1$ would yield approximately $0.2417$, which misses the larger rejection probability at an interior null value.

Check. The rejection interval is centered at the mean when $\theta=0$, giving it the greatest probability under this normal model. The derivative calculation confirms that no other null value produces a larger probability.

Finding and Reporting the Size

The calculation of size has two distinct parts: finding the rejection probability under each possible null parameter and then taking its supremum over the null set. A useful analysis should make both parts visible. Reporting only a numerical answer can conceal whether the null parameter space was handled correctly.

Write the null parameter set. For example, $H_0:\theta\leq0$ means $\Theta_0=(-\infty,0]$, while $H_0:0.1\leq p\leq0.6$ means $\Theta_0=[0.1,0.6]$.
Calculate the pointwise rejection probability. Express $P_\theta(X\in R)$ in terms of the parameter. Use the distribution specified by the model under each parameter value.
Find the supremum over the full null set. Monotonicity may place it at an endpoint, symmetry may identify equivalent values, or calculus may show that it occurs in the interior. Do not assume an endpoint maximum without justification.
Compare the size with the target level. The test is level $\alpha$ only when the supremum is at most $\alpha$.

In some models the rejection probability is monotone over the null parameter space, making a boundary value the least favorable null value—the value that produces the largest Type I error probability. In other models the probability is not monotone and the least favorable value may be interior. The normal interval example illustrates why “check the boundary” is not a universal rule. What is universal is to maximize over the whole null set.

Common Errors and Their Corrections

Incorrect approach	Why it fails	Correct approach
Calculate the rejection probability at one null parameter and call it the size.	Another permitted null value may have a larger rejection probability.	Take the supremum over every parameter value in the null set.
Assume the maximum occurs at a boundary.	The rejection probability can attain its maximum at an interior value.	Establish where the maximum occurs using monotonicity, symmetry, differentiation, or a direct comparison.
Check that one pointwise error probability is below $\alpha$ and claim level $\alpha$.	A level guarantee must hold uniformly over the null parameter space.	Verify $\sup_{\theta\in\Theta_0}P_\theta(X\in R)\leq\alpha$.
Include alternative parameter values in the size calculation.	Size concerns Type I error when the null is true.	Take the supremum over the null set only; study rejection probabilities under alternatives separately.

Key takeaway: For a composite null, the test’s size is the largest rejection probability over all parameter values allowed by the null. A single pointwise calculation is not enough to establish error control; the entire null parameter space must be considered.

Check Your Understanding

For each problem, determine the size by considering every parameter value in the stated null.

Let $X\sim N(\theta,4)$, with known variance $4$. The null is $H_0:\theta\leq3$, and the test rejects if $X>6$. Find the size. (Answer: $1-\Phi(1.5)\approx0.0668$.)
Let $X\sim\operatorname{Binomial}(3,p)$, with $H_0:0\leq p\leq0.2$. A test rejects if $X\geq2$. Find its size. (Answer: $3(0.2)^2(0.8)+(0.2)^3=0.1040$.)
Let $X\sim\operatorname{Bernoulli}(p)$, with $H_0:0.1\leq p\leq0.6$. The test rejects if $X=0$. Find its size and identify the null value where it occurs. (Answer: size $0.9000$, attained at $p=0.1$.)
Let $X\sim N(\theta,1)$, with $H_0:-1\leq\theta\leq1$. A test rejects if $X>2$ or $X<-2$. Find the size and the null parameter value or values where it occurs. (Answer: at $\theta=\pm1$, size $1-\Phi(1)+\Phi(-3)\approx0.1600$.)

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