Tutorials › Generalized Linear Models › Specifying Continuous Response Models

Home  /  Tutorials  /  Probability Models for Responses

Probability Models for Responses · Tutorial 7 of 981

Specifying Continuous Response Models

Build continuous response models by matching a valid support to a density that is nonnegative and integrates to one for every allowed parameter value.

Beginner 10 min read

What You'll Learn

  • State a continuous response model as a support together with a parameterized density.
  • Check that a proposed density is measurable, nonnegative, and integrates to one.
  • Derive the valid parameter range for a bounded-response density.
  • Construct models for bounded measurements with either fixed or parameter-dependent support.
  • Specify positive-response models and calculate probabilities by integration.
  • Recognize why support and density must be checked together.

From Possible Values to a Continuous Response Model

A continuous response model is more than a choice of a familiar distribution. It states which values a response can take and gives a density that describes how probability is distributed over those values. Both pieces matter: a density must be valid on its stated support, and its parameter values must preserve that validity.

For example, a laboratory measurement might be constrained to an interval, while a waiting time is positive and has no fixed upper bound. Those different constraints suggest different supports. The density then describes variation within the support. A model is fully specified only when the support, density, and allowed parameter values are clear.

Definition: A parameterized continuous response model consists of a parameter space \(\Theta\), a support \(S_\theta\) for each \(\theta\in\Theta\), and a density \(f(y;\theta)\) such that, for every \(\theta\in\Theta\), \(f(\,\cdot\,;\theta)\) is measurable, nonnegative, zero outside \(S_\theta\), and satisfies
$$ \int_{S_\theta} f(y;\theta)\,dy=1. $$
For a response \(Y\) following this model, probabilities are obtained by integrating the density over the event of interest. The set \(S_\theta\) identifies the values the model permits; endpoints of continuous supports do not affect integrals when they have probability zero.

The parameter \(\theta\) may change the shape of the density, the support, or both. A fixed support is often appropriate when the measurement scale has known limits. A parameter-dependent support can be appropriate when the unknown parameter itself describes a bound. In either case, the normalization and nonnegativity requirements must hold for every allowed parameter value, not just for one example value.

Checking and Constructing Densities

A convenient way to build a density is to start with a nonnegative function, often called a kernel, and divide by its integral. This works only when that integral is finite and strictly positive.

Theorem: Normalization of a nonnegative kernel. Fix \(\theta\). Suppose \(g(y;\theta)\) is measurable and nonnegative on a support \(S_\theta\), and define
$$ Z(\theta)=\int_{S_\theta}g(y;\theta)\,dy. $$
If \(0<Z(\theta)<\infty\), then \(f(y;\theta)=g(y;\theta)/Z(\theta)\) on \(S_\theta\), and zero outside it, is a density on that support.

Proof. Measurability and nonnegativity of \(g\), together with the positive finite constant \(Z(\theta)\), imply measurability and nonnegativity of \(f\). Outside \(S_\theta\), \(f\) is defined to be zero. Finally,

$$ \int_{S_\theta}f(y;\theta)\,dy =\int_{S_\theta}\frac{g(y;\theta)}{Z(\theta)}\,dy =\frac{1}{Z(\theta)}\int_{S_\theta}g(y;\theta)\,dy =1. $$

Thus \(f\) satisfies the density requirements. \(\square\)

The integral \(Z(\theta)\) can depend on the parameter. Dividing by a constant that is correct for only one value of \(\theta\) does not generally define a model for the whole parameter space. Also, if \(Z(\theta)=0\) or is infinite, this normalization argument does not produce a density.

Worked Example: A Bounded Response With a Fixed Support

Let \(Y\) be a continuous measurement scaled to lie between 0 and 1. Consider the proposed density

$$ f(y;\theta)= \begin{cases} 1+\theta(2y-1),&0<y<1,\\ 0,&\text{otherwise}, \end{cases} \qquad -1\leq\theta\leq 1. $$

First check normalization. Since \(\int_0^1 1\,dy=1\) and \(\int_0^1(2y-1)\,dy=[y^2-y]_0^1=0\),

$$ \int_0^1 f(y;\theta)\,dy =1+\theta\cdot 0 =1. $$

Next check nonnegativity. For \(0<y<1\), the quantity \(2y-1\) lies strictly between \(-1\) and \(1\). If \(-1\leq\theta\leq 1\), then \(\theta(2y-1)\geq-1\), so \(f(y;\theta)\geq0\). Thus every stated parameter value gives a valid density.

For instance, with \(\theta=1/2\), the density is \(f(y;1/2)=1/2+y\). The probability that the measurement exceeds \(1/2\) is

$$ P(Y>1/2) =\int_{1/2}^1(1/2+y)\,dy =\left[\frac{y}{2}+\frac{y^2}{2}\right]_{1/2}^1 =1-\left(\frac14+\frac18\right) =\frac58. $$

The support remains \((0,1)\) as \(\theta\) changes, but the density shifts weight toward larger or smaller values. This example specifies a model without assuming that every bounded measurement should have a constant density.

Parameter Restrictions Are Part of the Model

A formula can integrate to one and still fail to be a density if it takes negative values. Therefore, the allowed parameter space cannot be chosen independently of the density formula.

Result: Valid parameter range for the linear bounded model. On the support \(0<y<1\), the function \(f(y;\theta)=1+\theta(2y-1)\) integrates to one for every real \(\theta\). It is nonnegative throughout the support if and only if \(-1\leq\theta\leq1\).

Proof. The integral calculation in the worked example applies to every real \(\theta\), so normalization imposes no restriction. For nonnegativity, if \(-1\leq\theta\leq1\), then \(|\theta(2y-1)|\leq1\), and hence \(1+\theta(2y-1)\geq0\). Conversely, suppose \(\theta>1\). For \(y\) sufficiently close to 0, \(2y-1\) is sufficiently close to \(-1\) that \(1+\theta(2y-1)<0\); for example, any \(y\) with \(0<y<(\theta-1)/(2\theta)\) has this property. If \(\theta<-1\), for \(y\) sufficiently close to 1 the function is negative; specifically, \(1-\frac{1-y}{1}\) is not needed, and solving directly gives \(1+\theta(2y-1)<0\) whenever \(y>(1-\theta)/( -2\theta)\), a threshold strictly less than 1. Thus values outside \([-1,1]\) violate nonnegativity on a set of positive length. The stated range is necessary and sufficient. \(\square\)

This check illustrates an important modeling habit: derive the parameter space from the requirements on the density. Checking only the integral would miss invalid parameter values.

Bounded Measurements With a Parameter-Dependent Support

Sometimes the upper bound is not fixed in advance. A model may instead use a positive parameter \(\theta\) as the unknown maximum possible measurement. In that case, changing \(\theta\) changes both the support and the density formula's normalization.

Worked Example: A Bounded Response With an Unknown Upper Limit

Suppose \(Y\) is nonnegative and cannot exceed an unknown bound \(\theta>0\). Consider

$$ f(y;\theta)= \begin{cases} \dfrac{2y}{\theta^2},&0<y<\theta,\\ 0,&\text{otherwise}, \end{cases} \qquad \theta>0. $$

The support is \(S_\theta=(0,\theta)\). The density is nonnegative there because \(y>0\) and \(\theta^2>0\). Its integral is

$$ \int_0^\theta \frac{2y}{\theta^2}\,dy =\frac{1}{\theta^2}[y^2]_0^\theta =\frac{\theta^2}{\theta^2} =1. $$

For a numerical check, take \(\theta=4\). Then \(f(y;4)=y/8\) for \(0<y<4\), and \(\int_0^4 y/8\,dy=16/16=1\). The probability of a response between 1 and 2 is

$$ P(1<Y<2\mid\theta=4) =\int_1^2\frac{y}{8}\,dy =\left[\frac{y^2}{16}\right]_1^2 =\frac{4-1}{16} =\frac{3}{16}. $$

As \(\theta\) changes, the interval of possible values changes too. The support is therefore an essential part of this model, not merely a place where the formula happens to be nonzero.

Positive Measurements Without a Fixed Upper Bound

A positive measurement, such as a duration or concentration, may have support \((0,\infty)\). One simple model assigns a rate parameter \(\lambda>0\) and uses a density that decreases as the response increases.

Worked Example: A Positive Response With an Exponential Density

Let the support be \(S=(0,\infty)\), and define

$$ f(y;\lambda)= \begin{cases} \lambda e^{-\lambda y},&y>0,\\ 0,&\text{otherwise}, \end{cases} \qquad \lambda>0. $$

The expression is nonnegative for \(y>0\). Its total integral is

$$ \int_0^\infty \lambda e^{-\lambda y}\,dy =\left[-e^{-\lambda y}\right]_0^\infty =0-(-1) =1, $$

where the limit \(e^{-\lambda y}\to0\) as \(y\to\infty\) uses \(\lambda>0\). Thus each positive rate gives a density on the positive support.

For \(\lambda=2\), the probability that \(Y\) is between \(1/2\) and 1 is

$$ P(1/2<Y<1) =\int_{1/2}^1 2e^{-2y}\,dy =\left[-e^{-2y}\right]_{1/2}^1 =e^{-1}-e^{-2} \approx0.2325. $$

The parameter restriction is essential: if \(\lambda=0\), the expression is identically zero and does not integrate to one; if \(\lambda<0\), the function is negative on the positive support. Hence \(\lambda>0\) is part of the model specification.

A Practical Specification Checklist

When writing down a continuous response model, make the full specification explicit. The following sequence keeps support, parameter restrictions, and normalization aligned.

1
State the response scale and support.
Identify the possible values, including whether the support is bounded and whether it depends on a parameter.
2
Write the density on the support.
Give the formula where it applies and state that it is zero outside the support.
3
Restrict the parameter space.
Require the formula to be nonnegative and its integral to be finite and equal to one for every allowed parameter value.
4
Check an event probability if useful.
Integrate over a stated interval or set to verify that the specified model gives interpretable probabilities.

A common mistake is to choose a convenient-looking density formula and check only that its integral is one at a single parameter value. That does not establish a model: at other parameter values the function may become negative, fail to normalize, or assign positive density outside the intended response range. The correction is to verify the density conditions for every parameter in the stated parameter space, and to state any parameter-dependent support explicitly.

Key takeaway: A continuous response model is the combination of a support, a parameterized density, and a valid parameter space. The density must be measurable, nonnegative, and integrate to one for every allowed parameter value.

Check Your Understanding

Use the support and density requirements to assess each specification.

  1. For \(f(y;\theta)=1+\theta(2y-1)\) on \(0<y<1\), which requirement determines the allowed range of \(\theta\), and what is that range?
  2. For a proposed density proportional to \(y\) on \(0<y<\theta\), with \(\theta>0\), find the normalizing constant.
  3. Why must a model with support \((0,\theta)\) state that \(\theta>0\)?
  4. For \(f(y;\lambda)=\lambda e^{-\lambda y}\) on \(y>0\), explain what fails if \(\lambda=0\) and if \(\lambda<0\).
  5. Give one reason a model for a bounded measurement might use a parameter-dependent support rather than a fixed support.

Previous

Next

Back to Tutorials
Clinical Biostats