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Confidence and Prediction Regions · Tutorial 269 of 972

Common Errors in Reading Confidence and Prediction Intervals

Learn what confidence and prediction intervals cover, when their nominal coverage is exact, and why leverage and residual degrees of freedom matter.

Intermediate 10 min read

What You'll Learn

  • Distinguish uncertainty about a mean response from variability in a future outcome
  • Interpret confidence and prediction coverage as repeated-sampling properties
  • Recognize why pointwise intervals do not guarantee simultaneous coverage
  • Use residual degrees of freedom to select the correct \(t\) critical value
  • Handle the zero-leverage mean interval and prediction interval correctly

What an Interval Does—and Does Not—Say

A reported “95% interval” can be misread in several ways. A confidence interval for a mean response and a prediction interval for a future observation may be centered at the same fitted value, but they answer different questions. Neither statement means that, after the interval has been calculated, there is a 95% frequentist probability that a fixed parameter lies inside it. Nor does 95% coverage at one design point automatically mean 95% coverage for a whole collection of points.

We make these distinctions precise in the fixed-design normal linear model. The key ingredients are the target being covered, the randomness over which coverage is calculated, and the residual degrees of freedom used to estimate the error variance.

Model and notation: Let \(Y=X\beta+\varepsilon\), where \(X\) is a fixed \(n\times p\) matrix of rank \(p\), \(n>p\), and \(\varepsilon\sim N_n(0,\sigma^2 I_n)\), with \(\sigma^2>0\). Write \(\hat{\beta}=(X^\top X)^{-1}X^\top Y\), \(s^2=\|Y-X\hat{\beta}\|^2/(n-p)\), and \(\nu=n-p\). For a specified design vector \(x_0\in\mathbb{R}^p\), the mean response is \(\mu_0=x_0^\top\beta\), and its fitted value is \(\hat{\mu}_0=x_0^\top\hat{\beta}\). Define \(h_0=x_0^\top(X^\top X)^{-1}x_0\). A future response at this design vector is \(Y_0=x_0^\top\beta+\varepsilon_0\), where \(\varepsilon_0\sim N(0,\sigma^2)\) is independent of the training data.

Two Targets, Two Sources of Uncertainty

A confidence interval for \(\mu_0\) accounts for uncertainty in estimating the model’s mean response. A prediction interval for \(Y_0\) must account for that same estimation uncertainty and the new error \(\varepsilon_0\). Under the stated model, when \(h_0>0\), the standard pointwise intervals are

$$ \hat{\mu}_0 \pm t_{1-\alpha/2,\nu}\,s\sqrt{h_0} \qquad\text{and}\qquad \hat{\mu}_0 \pm t_{1-\alpha/2,\nu}\,s\sqrt{1+h_0}, $$

respectively. Here \(t_{1-\alpha/2,\nu}\) is the \(1-\alpha/2\) quantile of Student’s \(t\) distribution with \(\nu\) degrees of freedom. The additional \(1\) in the prediction standard error represents the variance of the future error, in units of \(\sigma^2\). It is not an adjustment that can be omitted merely because the interval is centered on a fitted mean.

Theorem: Under the model above, conditional on the fixed design matrix \(X\), the confidence interval for \(\mu_0\) has coverage \(1-\alpha\) when \(h_0>0\). The prediction interval for \(Y_0\) has coverage \(1-\alpha\) for every \(x_0\), including \(x_0=0\). If \(h_0=0\), then \(x_0=0\), the mean is known to be zero, and its singleton interval \([0,0]\) has coverage \(1\), not \(1-\alpha\).

Proof. Let \(P_X=X(X^\top X)^{-1}X^\top\) and \(M_X=I_n-P_X\). The fitted error \(\hat{\beta}-\beta=(X^\top X)^{-1}X^\top\varepsilon\) depends on \(P_X\varepsilon\), whereas the residual vector is \(M_X\varepsilon\). These are jointly normal and have zero covariance because \(P_XM_X=0\); hence they are independent. Also, \(\|M_X\varepsilon\|^2/\sigma^2\sim\chi^2_\nu\), since \(M_X\) is a projection of rank \(\nu\). Thus \(s\) is independent of \(\hat{\beta}\), with \(\nu s^2/\sigma^2\sim\chi^2_\nu\).

For \(h_0>0\), \(\hat{\mu}_0-\mu_0\) is normal with mean zero and variance \(\sigma^2h_0\). Therefore

$$ \frac{\hat{\mu}_0-\mu_0}{s\sqrt{h_0}} = \frac{Z}{\sqrt{U/\nu}} \sim t_\nu, $$

where \(Z\sim N(0,1)\), \(U\sim\chi^2_\nu\), and \(Z\) and \(U\) are independent. For prediction, \(Y_0-\hat{\mu}_0=\varepsilon_0-(\hat{\mu}_0-\mu_0)\) is normal with mean zero and variance \(\sigma^2(1+h_0)\). It is independent of \(s\): \(\varepsilon_0\) is independent of the training data, and the fitted error is independent of the residual sum of squares. Consequently,

$$ \frac{Y_0-\hat{\mu}_0}{s\sqrt{1+h_0}}\sim t_\nu. $$

For either pivot, the probability that its absolute value is no larger than \(t_{1-\alpha/2,\nu}\) is \(1-\alpha\). These events are exactly the events that the corresponding target lies in its displayed interval. If \(h_0=0\), positive definiteness of \((X^\top X)^{-1}\) implies \(x_0=0\); hence \(\mu_0=0\) is known and the mean interval is \([0,0]\). The prediction pivot remains valid, with denominator \(s\sqrt{1+h_0}=s\). \(\square\)

Worked Example: Reading Mean and Prediction Intervals at One Point

Suppose \(n=12\), \(p=2\), and a full-rank design has \((X^\top X)^{-1}=\begin{pmatrix}1/12&0\\0&1/20\end{pmatrix}\). At \(x_0=(1,2)^\top\), suppose \(\hat{\beta}=(10,2)^\top\), \(s=3\), and use a 95% interval. Here \(\nu=12-2=10\), so \(t_{0.975,10}=2.228\).

The fitted mean is \(\hat{\mu}_0=1(10)+2(2)=14\). The leverage is \(h_0=1^2(1/12)+2^2(1/20)=1/12+1/5=17/60\approx0.2833\). Thus the mean-response margin is \(2.228(3)\sqrt{17/60}\approx2.228(1.5969)=3.558\), giving the interval \([10.442,17.558]\).

The prediction margin is \(2.228(3)\sqrt{1+17/60}=2.228(3)\sqrt{77/60}\approx2.228(3.3985)=7.572\). The prediction interval is \([6.428,21.572]\). The first interval concerns the fixed mean \(14\); the second concerns a future response that varies around that mean. The wider second interval does not mean the fitted mean itself became less precise.

Coverage Is a Repeated-Sampling Statement

For a frequentist confidence interval, the parameter \(\mu_0\) is fixed under the model. The interval’s endpoints vary with the training data. Saying that its coverage is 95% means that, over repeated samples generated under the model with the same fixed design, 95% of the intervals contain \(\mu_0\). It does not assign a 95% probability to \(\mu_0\) after a particular interval has been observed.

For a prediction interval, the target \(Y_0\) is itself random. The exact coverage statement above is a repeated-sampling statement over the training data and the independent future error, conditional on \(X\). It does not, by itself, assert that the conditional probability of \(Y_0\) lying in the particular interval is exactly 95% given the realized training data. That conditional probability would require a different inferential framework or additional assumptions.

Worked Example: The Zero-Leverage Boundary Case

Let \(x_0=(0,0)^\top\), with \(n=10\), \(p=2\), \(s=2\), and a 95% prediction interval. Then \(h_0=x_0^\top(X^\top X)^{-1}x_0=0\) and \(\hat{\mu}_0=0\), regardless of the fitted coefficient values. The mean is exactly \(x_0^\top\beta=0\), so its confidence interval is the singleton \([0,0]\); it covers with probability \(1\).

For the future response, \(\nu=10-2=8\) and \(t_{0.975,8}=2.306\). The prediction standard error is \(s\sqrt{1+h_0}=2\), so the prediction interval is \(0\pm2.306(2)=[-4.612,4.612]\). Its coverage is exactly 95% under the model. The mean-interval pivot with denominator \(s\sqrt{h_0}\) is undefined here, so it must not be used to claim 95% coverage for the singleton interval.

Pointwise Coverage Is Not Simultaneous Coverage

The intervals above are pointwise: a particular \(x_0\) is specified, and the coverage statement applies to that point. Even if several intervals each have 95% coverage, that does not establish that all their targets are covered together with probability 95%. For example, if two coverage events are independent and each has probability \(0.95\), the probability both occur is \(0.95^2=0.9025\), not \(0.95\). Without independence, the joint probability still cannot be inferred from the separate marginal coverage statements alone.

Therefore, a collection of separate pointwise intervals should not be described as a 95% simultaneous band or region without a method that controls joint coverage. The distinction concerns the probability statement, not just how the intervals are drawn or displayed.

Worked Example: Why the Residual Degrees of Freedom Matter

Suppose \(n=9\), \(p=3\), the design has full column rank, \(s=4\), and at a specified point \(\hat{\mu}_0=20\) and \(h_0=0.5\). For a 95% interval, the residual degrees of freedom are \(\nu=n-p=6\), so \(t_{0.975,6}=2.447\).

The mean-response margin is \(2.447(4)\sqrt{0.5}\approx2.447(2.8284)=6.921\), giving \([13.079,26.921]\). The prediction margin is \(2.447(4)\sqrt{1.5}\approx2.447(4.8990)=11.988\), giving \([8.012,31.988]\).

Using \(n-1=8\) degrees of freedom would substitute \(t_{0.975,8}=2.306\), yielding incorrect margins of approximately \(6.522\) and \(11.297\). The residual sum of squares has \(\nu=n-p\) degrees of freedom because fitting \(p\) linearly independent coefficients removes \(p\) dimensions from the \(n\)-dimensional error vector. It is those residual degrees of freedom—not the sample size alone—that calibrate \(s^2\) and the \(t\) critical value.

Common Errors and Corrections

  • Wrong: “There is a 95% probability that the fixed mean is in this calculated confidence interval.” Correction: The 95% describes the long-run coverage of the interval procedure under repeated sampling.
  • Wrong: “A prediction interval covers the mean, so it needs no extra uncertainty term.” Correction: Its target is a future outcome, and its standard error includes the independent future error through \(1+h_0\).
  • Wrong: “Every one of several 95% intervals is 95%, so they jointly cover with probability 95%.” Correction: Pointwise coverage does not determine joint coverage.
  • Wrong: “The mean interval always has exact \(1-\alpha\) coverage, including when \(h_0=0\).” Correction: At \(h_0=0\), the mean is known to be zero and the singleton interval has coverage one; the usual pivot is undefined.
  • Wrong: “Use \(n-1\) degrees of freedom because there are \(n\) observations.” Correction: In this model, use the residual degrees of freedom \(\nu=n-p\).
Key takeaway: Before interpreting an interval, identify its target and its coverage statement. A mean-response interval describes uncertainty in a fitted mean; a prediction interval also includes future-outcome variability. The displayed exact \(t\) procedures are pointwise, use residual degrees of freedom \(n-p\), and require special care for the mean when \(h_0=0\).

Check Your Understanding

Use the model assumptions and distinctions developed above to answer these questions.

  1. In repeated sampling, what does 95% coverage mean for a confidence interval for a fixed mean response?
  2. Why does the prediction standard error contain \(1+h_0\), while the mean-response standard error contains only \(h_0\)?
  3. If \(h_0=0\), what are the mean-response interval and its coverage? Why is the usual mean pivot unavailable?
  4. Why does a collection of pointwise 95% intervals not automatically provide 95% simultaneous coverage?
  5. For \(n=15\) observations and a full-rank model with \(p=4\) coefficients, what residual degrees of freedom determine the \(t\) critical value?
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