Simple and Composite Hypotheses

Why the Number of Allowed Distributions Matters

A hypothesis is a statement about the probability model that generated the observations. Some hypotheses specify exactly one distribution. Others allow a range of parameter values and therefore a range of possible distributions. This distinction matters because a conclusion based on one distribution need not describe the other distributions permitted by the same hypothesis.

For example, a statement that a population proportion equals \(0.25\) specifies one value of the proportion. A statement that it is at most \(0.25\) permits many values. In a model where the proportion determines the distribution, the first statement is simple and the second is composite. The same idea applies to means, rates, variances, and parameter vectors.

Definition: A simple hypothesis specifies exactly one probability distribution for the observations. A composite hypothesis specifies more than one possible probability distribution. In an identifiable model indexed by a parameter \(\theta\), a hypothesis is simple when its parameter set contains one value of \(\theta\), and composite when that set contains multiple values. More generally, the distribution—not merely the number of parameter labels—is decisive.

Let \(\Theta\) be the parameter space of a model and let \(\Theta_0\) and \(\Theta_1\) denote the parameter values allowed by the null and alternative hypotheses. For example, \(H_0:\theta=\theta_0\) restricts \(\theta\) to one value, while \(H_0:\theta\leq\theta_0\) allows a set of values. In an identifiable model, the first is simple and the second is composite. An alternative can be simple or composite too; the labels do not depend on whether the hypothesis is null or alternative.

H_0:\theta=\theta_0 \quad\Longrightarrow\quad \Theta_0=\{\theta_0\} \qquad\text{(simple, if the distribution is thereby fully specified),}

H_0:\theta\leq\theta_0 \quad\Longrightarrow\quad \Theta_0=\{\theta:\theta\leq\theta_0\} \qquad\text{(composite when this set gives multiple distributions).}

A hypothesis may restrict one component of a parameter while leaving another unspecified. In that case, a single equality does not necessarily specify a single distribution. A model with an unknown mean and unknown variance illustrates this point: fixing the mean alone still allows different variances, and hence different normal distributions.

Classifying Hypotheses by Their Parameter Sets

The classification follows a consistent process. First state the model and its full parameter space. Then translate the hypothesis into the set of permitted parameter values. Finally, check whether that set corresponds to one distribution or several. This avoids classifying a hypothesis merely by noticing an equality sign or an inequality sign.

State the model.
Specify the distributional family, the parameter or parameters, and any assumptions such as independence or known variance.

Translate each hypothesis into a parameter set.
Write down every value permitted under the null and under the alternative, including any parameters left unspecified.

Count the distinct distributions.
One permitted distribution makes the hypothesis simple; multiple permitted distributions make it composite.

Worked Example: A Proportion Hypothesis in a Bernoulli Model

A hypothetical review examines 12 independent policies. For each policy, let \(X_i=1\) if a specified feature is present and \(X_i=0\) otherwise. Assume \(X_i\sim\operatorname{Bernoulli}(p)\), independently, where \(p\in[0,1]\). Classify \(H_0:p=0.25\) and \(H_1:p\neq0.25\). Then compare the probability of observing zero features at \(p=0.25\) and \(p=0.40\).

Step 1: Identify the parameter space. The model is fully described by \(p\), with \(0\leq p\leq1\). Under \(H_0\), the only permitted value is \(p=0.25\), so the null is simple. Under \(H_1\), every value in \([0,1]\) except \(0.25\) is permitted, so the alternative is composite.

Step 2: Calculate the event probability under the null. Independence gives the probability of zero features as the probability that all 12 observations are zero:

P(X_1=0,\ldots,X_{12}=0\mid p=0.25) =(1-0.25)^{12} =0.75^{12} =0.031676352\ldots

Step 3: Calculate it at another parameter value. At \(p=0.40\), which is permitted under \(H_1\), the same event has probability

P(X_1=0,\ldots,X_{12}=0\mid p=0.40) =(1-0.40)^{12} =0.60^{12} =0.002176782336.

The event probability differs because the two parameter values imply different distributions. To four decimal places, the probabilities are \(0.0317\) and \(0.0022\), respectively. The method is to identify the distributions allowed before calculating; the common mistake avoided here is treating the alternative as one distribution just because it is written as one statement.

Why a Composite Hypothesis Requires Attention to Every Value

For an event \(A\) and a model indexed by \(\theta\), its probability is a function of the parameter: \(P_\theta(A)\). If a hypothesis permits several values of \(\theta\), the probability of \(A\) can vary across those values. Thus, calculating \(P_\theta(A)\) at one convenient value does not establish what happens throughout a composite hypothesis.

This follows directly from the definition of probability under a parameterized model. In a discrete model with probability mass function \(f_\theta\), the event probability is the sum of the probabilities of its outcomes. Each term can depend on \(\theta\). In a continuous model with density \(f_\theta\), the event probability is the integral of that density over the event, and the density can likewise change with \(\theta\).

P_\theta(A)=\sum_{x\in A}f_\theta(x) \quad\text{(discrete case)},\qquad P_\theta(A)=\int_A f_\theta(x)\,dx \quad\text{(continuous case)}.

When \(\theta\) varies over a composite hypothesis, these expressions generally vary too. A particular event may happen to have the same probability at two parameter values, but that coincidence does not make the distributions identical. The relevant question is whether the hypothesis allows one distribution or more than one.

Worked Example: Event Probabilities Across a Composite Null

Suppose four independent measurements satisfy \(X_i\sim N(\mu,9)\), where the variance \(9\) is known and \(\mu\) is unknown. Consider \(H_0:\mu\leq30\) against \(H_1:\mu>30\). For the event \(\bar X<27\), calculate its probability at \(\mu=27\), \(\mu=30\), and \(\mu=33\), and explain what the comparisons show.

Step 1: Classify the hypotheses. The null permits every \(\mu\leq30\), so it includes multiple normal distributions and is composite. The alternative also permits multiple values, \(\mu>30\), and is composite.

Step 2: Find the distribution of the sample mean. For four independent observations with variance \(9\), the sample mean has variance \(9/4\) and standard deviation \(3/2=1.5\). Therefore, for each specified \(\mu\),

\bar X\sim N\left(\mu,\frac{9}{4}\right), \qquad P_\mu(\bar X<27) =\Phi\left(\frac{27-\mu}{1.5}\right),

where \(\Phi\) is the standard normal cumulative distribution function.

Step 3: Evaluate at \(\mu=27\). The standardized value is \((27-27)/1.5=0\), so

P_{27}(\bar X<27)=\Phi(0)=0.5000.

Step 4: Evaluate at \(\mu=30\). The standardized value is \((27-30)/1.5=-2\), giving

P_{30}(\bar X<27)=\Phi(-2)\approx0.0228.

Step 5: Evaluate at \(\mu=33\). This value is in the alternative, not the null. The standardized value is \((27-33)/1.5=-4\), so

P_{33}(\bar X<27)=\Phi(-4)\approx0.0000317.

To four decimal places, the last probability is \(0.0000\); its more precise value is about \(0.0000317\), so it is not exactly zero. Within the composite null, the event probability at \(\mu=27\) is much larger than at \(\mu=30\). The calculation shows why one null value cannot automatically stand in for all values covered by the null.

A Fixed Mean Does Not Always Specify One Distribution

A hypothesis can look like a point equality and still be composite if the model has other unknown parameters. For a normal model, fixing the mean while leaving the variance unrestricted allows many distributions. The parameter set must include all parameters that determine the distribution, not only the parameter named in the claim.

Worked Example: An Unknown Variance Makes the Null Composite

Let one observation satisfy \(X\sim N(\mu,\sigma^2)\), where \(\mu\) is any real number and \(\sigma>0\) is unknown. Consider \(H_0:\mu=10\), with no restriction on \(\sigma\). For the event \(X\leq8\), calculate the probability under the null when \(\sigma=1\) and when \(\sigma=2\).

Step 1: Classify the null. The null fixes \(\mu=10\), but allows every positive \(\sigma\). It therefore allows, among others, \(N(10,1)\) and \(N(10,4)\), which are distinct distributions. The null is composite.

Step 2: Standardize when \(\sigma=1\). Under \(N(10,1)\), the standard deviation is \(1\), so

P(X\leq8\mid \mu=10,\sigma=1) =\Phi\left(\frac{8-10}{1}\right) =\Phi(-2) \approx0.0228.

Step 3: Standardize when \(\sigma=2\). Under \(N(10,4)\), the standard deviation is \(2\), not \(4\). Thus

P(X\leq8\mid \mu=10,\sigma=2) =\Phi\left(\frac{8-10}{2}\right) =\Phi(-1) \approx0.1587.

The probabilities differ even though the mean restriction is identical. This example avoids the mistake of calling a hypothesis simple merely because one parameter is fixed; all parameters that determine the distribution must be considered.

Common Classification Traps

Equating “equality” with “simple.” An equality fixes one parameter, but other parameters may remain unknown. Check whether the full distribution is specified.
Equating “inequality” with “composite” without checking the model. An inequality usually allows multiple values in standard continuous parameter models, but classification is about the number of distinct distributions actually permitted.
Using one parameter value to represent a composite hypothesis. An event probability calculated at one permitted value is only the probability at that value. It need not match the probability at other covered values.
Confusing a hypothesis with the observed data. Simple or composite describes the set of distributions posited by the hypothesis, not whether a particular sample looks unusual.
Forgetting nuisance parameters. A parameter left unspecified is still part of the model and can change its distribution.

Key takeaway: Classify a hypothesis by the distributions it permits. A simple hypothesis specifies one distribution; a composite hypothesis permits multiple distributions. When the hypothesis is composite, calculations at one parameter value describe only that value, not the entire set.

Check Your Understanding

For each problem, identify the parameter values allowed and use them to classify the hypotheses or compare the requested probabilities.

Let \(X_1,\ldots,X_5\) be independent Bernoulli observations with unknown \(p\in[0,1]\). Classify \(H_0:p=0.60\) and \(H_1:p<0.60\). (Answer: \(H_0\) is simple; \(H_1\) is composite.)
Let \(Y\sim\operatorname{Poisson}(\lambda)\), with \(\lambda>0\). Classify \(H_0:\lambda\geq3\) and \(H_1:\lambda<3\). (Answer: Both hypotheses are composite.)
Suppose \(X_1,\ldots,X_9\) are independent \(N(\mu,16)\) observations, with known variance \(16\). Classify \(H_0:\mu=12\) and \(H_1:\mu\neq12\), and give the standard deviation of \(\bar X\). (Answer: \(H_0\) is simple, \(H_1\) is composite, and \(\operatorname{SD}(\bar X)=4/3\).)
Let \(X\sim N(5,\tau^2)\), where \(\tau>0\) is unknown, and \(H_0:\mu=5\). For \(X>7\), find the probabilities at \(\tau=1\) and \(\tau=2\). (Answer: \(H_0\) is composite; the probabilities are \(1-\Phi(2)\approx0.0228\) and \(1-\Phi(1)\approx0.1587\), respectively.)
For six independent Bernoulli observations with parameter \(p\), let \(H_0:p\leq0.30\). Compare the probability of six zeros at \(p=0.10\) and \(p=0.30\), and classify \(H_0\). (Answer: \(H_0\) is composite; the probabilities are \(0.9^6=0.5314\) and \(0.7^6=0.1176\), respectively, to four decimals.)

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