Population Targets and Fixed-Design Targets

Why the Target Depends on What Is Being Asked

A learning problem may ask for a response characteristic at a particular covariate value, or for an average response characteristic across a population of covariates. These are different targets. Even if the conditional response mechanism stays exactly the same, changing which covariate values are common in the population can change the population average.

For example, suppose a treatment response tends to be higher for older patients. The conditional mean response at age 40 is a pointwise target. The mean response across a population is an average of those conditional means, weighted by how ages are distributed in that population. A population with more older patients can therefore have a higher average without any change in the response mechanism at a given age.

The distinction is about the quantity to be learned, not the method used to estimate it. A target is specified by the data-generating law and the scientific question; an estimator is a rule for using observed data to learn about that target.

Conditional Targets and Population Averages

Let \(X\) denote the covariates and \(Y\) the response, with joint distribution \(P\) and covariate marginal distribution \(P_X\). Suppose the conditional distribution \(P_{Y\mid X=x}\) is defined for the covariate values under discussion. Choose a response functional \(T\), such as the conditional mean, a conditional quantile, or a class probability. Applying \(T\) to the conditional response distribution gives a target function of the covariates.

Definition: The conditional target function associated with \(T\) is \(\tau_P(x)=T(P_{Y\mid X=x})\). At specified design points \(x_1,\ldots,x_m\), the fixed-design target is the vector \((\tau_P(x_1),\ldots,\tau_P(x_m))\). For a covariate distribution \(Q\), the population-averaged target is \(\tau(P,Q)=\int \tau_P(x)\,dQ(x)\), provided the integrand is measurable and the integral exists. The target for the population generating the data uses \(Q=P_X\).

For the conditional mean, \(T\) is expectation and \(\tau_P(x)=E[Y\mid X=x]\), when this conditional expectation exists. For binary classification, one possible choice is the conditional probability of a specified class. The same definitions apply to other conditional response functionals, as long as the resulting target can be integrated under \(Q\).

A fixed-design target names the covariate values at which the conditional target is requested. It does not average over how often those values occur in a population. A population-averaged target does average: covariate values receive weight according to \(Q\). Thus, a population target is incomplete unless its covariate distribution is understood or explicitly named.

There is a technical qualification when a specified point has probability zero under \(P_X\), as is usual for an exact value of a continuous covariate. A joint distribution determines conditional distributions only up to choices on sets of \(P_X\)-probability zero. To speak of a pointwise target there, one must specify a version of the conditional distribution or a model that defines it at that point. The fixed-design examples below take the conditional mechanism at the named points as specified.

Worked Examples

Worked Example: A Binary Covariate and a Changing Population Mean

Let \(X\) be binary, and suppose the conditional mean response is \(E[Y\mid X=0]=2\) and \(E[Y\mid X=1]=6\). In a population with \(P(X=1)=p\), find the population mean response. Compare populations with \(p=0.25\) and \(p=0.75\), and give the fixed-design targets at \(x=0\) and \(x=1\).

The conditional mean target function is \(\tau_P(0)=2\) and \(\tau_P(1)=6\). The population target weights these two values by their covariate probabilities:

\tau(P,P_X)=E[Y]=2P(X=0)+6P(X=1)=2(1-p)+6p=2+4p.

For \(p=0.25\), the population target is \(2+4(0.25)=3\). For \(p=0.75\), it is \(2+4(0.75)=5\). The fixed-design targets remain \(2\) at \(x=0\) and \(6\) at \(x=1\) in both populations. The average changes because the weights change, not because either conditional mean changes.

Worked Example: Averaging a Nonlinear Conditional Mean

Suppose \(X\) is uniform on \([0,2]\), and the specified conditional mean is \(E[Y\mid X=x]=x^2\). Find the population-averaged conditional mean. Then compare it with the average under a different covariate distribution that puts probability \(1/2\) at \(x=0\) and probability \(1/2\) at \(x=2\), while keeping the same conditional mean function.

Under the uniform distribution, the density is \(1/2\) on \([0,2]\). The population target is

\int_0^2 x^2\frac{1}{2}\,dx =\frac{1}{2}\left[\frac{x^3}{3}\right]_0^2 =\frac{1}{2}\cdot\frac{8}{3} =\frac{4}{3}.

Under the two-point covariate distribution, the target is

\frac{1}{2}(0^2)+\frac{1}{2}(2^2)=0+2=2.

The conditional target at any specified \(x\) is still \(x^2\). In particular, its values at \(x=0\) and \(x=2\) are \(0\) and \(4\). The population average differs because the two covariate distributions assign different weights across values of \(X\).

Worked Example: A Population Class Probability

Let \(Y\) indicate whether a person has a condition, and let \(X\) identify one of two groups, \(A\) or \(B\). Suppose \(P(Y=1\mid X=A)=0.2\) and \(P(Y=1\mid X=B)=0.8\). If a population has proportion \(p\) in group \(B\), calculate its overall condition probability when \(p=0.25\) and when \(p=0.75\).

Take \(T\) to be the conditional probability of class \(Y=1\). The population target is the average of the two conditional probabilities using the group proportions:

$$ P(Y=1)=0.2(1-p)+0.8p=0.2+0.6p. $$

When \(p=0.25\), this is \(0.2+0.6(0.25)=0.35\). When \(p=0.75\), it is \(0.2+0.6(0.75)=0.65\). The conditional probabilities remain \(0.2\) in group \(A\) and \(0.8\) in group \(B\); only the population composition has changed.

Two Useful Identities

The first identity explains why averaging conditional means over the data-generating covariate distribution recovers the marginal mean response. It is the law of total expectation, stated here for an integrable response.

Result: If \(E[|Y|]<\infty\), then \(E[Y]=\int E[Y\mid X=x]\,dP_X(x)\), using a version of the conditional expectation.

To verify the identity, write \(m(X)=E[Y\mid X]\). By the defining property of conditional expectation, \(m(X)\) is integrable and has the same expectation as \(Y\). Expressing that expectation using the distribution of \(X\) gives the integral.

E[Y]=E[m(X)] =\int m(x)\,dP_X(x) =\int E[Y\mid X=x]\,dP_X(x).

This identity requires averaging under \(P_X\), the covariate distribution associated with the joint law of \((X,Y)\). Replacing it with another distribution \(Q\) defines a different population-averaged target; it need not equal \(E[Y]\) for the original joint law. \(\square\)

A second identity makes the effect of changing covariate weights explicit. Suppose \(X\) takes finitely many values \(x_1,\ldots,x_k\), and the conditional target values \(t_j=\tau_P(x_j)\) are held fixed. Let \(Q\) and \(R\) assign probabilities \(q_j\) and \(r_j\) to these values. Then the difference between their population targets is exactly the sum of each target value multiplied by its change in weight.

Result: For finite covariate values and fixed conditional target values, \(\tau(P,Q)-\tau(P,R)=\sum_{j=1}^k t_j(q_j-r_j)\).

Indeed, by the definition of a population-averaged target,

\tau(P,Q)-\tau(P,R) =\sum_{j=1}^k t_jq_j-\sum_{j=1}^k t_jr_j =\sum_{j=1}^k t_j(q_j-r_j).

Because both sets of probabilities sum to one, \(\sum_j(q_j-r_j)=0\). Therefore adding the same constant \(c\) to every \(t_j\) does not change this difference: \(\sum_j(t_j+c)(q_j-r_j)=\sum_j t_j(q_j-r_j)\). The change depends on how the weights shift across covariate values with different target values. If all \(t_j\) are equal, the population target does not change at all. \(\square\)

Keeping the Target Straight

When comparing analyses or populations, state both the conditional target and the covariate distribution used for the average. If the conditional response mechanism is held fixed, a target evaluated at the same specified covariate value is unchanged. An average over a population can nevertheless change when that population has a different distribution of covariates.

A common mistake is to treat the population average as an intrinsic summary of the conditional response mechanism alone. For example, someone may see that \(E[Y\mid X=x]\) is unchanged for every \(x\) and conclude that \(E[Y]\) must also be unchanged. That conclusion omits the averaging distribution. The correct calculation is \(E[Y]=\int E[Y\mid X=x]\,dP_X(x)\); if \(P_X\) changes, the integral may change even when the integrand does not.

A related mistake is to treat the observed design points as if they automatically define a population target. In fixed-design sampling, the specified covariates are not themselves sampled from a population distribution. One can still ask for the conditional target at each design point, or choose a distribution \(Q\) and average over it, but that distribution must be part of the target definition.

Key takeaway: A fixed-design target specifies conditional response characteristics at named covariate values. A population-averaged target combines conditional characteristics using a stated covariate distribution. Changing that distribution can change the average even when the conditional mechanism is unchanged.

Check Your Understanding

Use the definitions and identities above to answer the following.

If the conditional mean at two covariate values is 1 and 5, what is the population target when the second value has probability 0.4?
In the first worked example, which quantities changed between the two populations, and which fixed-design targets stayed the same?
Suppose a binary response has conditional class probabilities 0.1 and 0.7 in two groups. Write the overall class probability when the second group has population proportion \(p\).
Why does the law of total expectation use \(P_X\) when it expresses \(E[Y]\) for a joint distribution \(P\)?
Can two populations have the same conditional target function but different population-averaged targets? State the condition under which their averages would instead necessarily agree.