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Let $t_{\nu ,p}$ denote the $p$-quantile of Student's t-distribution. Part (b) gives the Student t confidence interval

A $(1-\alpha )$ prediction interval is a pair of statistics $L(Y),U(Y)$ such that, for every $\beta$ and ${\sigma }^2>0$, an independent future response $Y^{\ast }$ at the specified input satisfies

In the normal linear model, assume $X$ has full column rank and writeFor a new observationindependent of the training sample, the prediction error isThe two terms on the right are independent centered normal random variables, whileConsequently, with the new-point regression leveragewe have

The assumption in the question givesThis variable is independent of $\hat{\beta }$; it is also independent of the new error ${\epsilon }^{\ast }$. Therefore it is independent of the standardized prediction error, and the defining normal-over-chi-square representation of Student's t-distribution givesIf $t_{n-p,1-\alpha /2}$ is the corresponding upper quantile, symmetry yieldsSolving these inequalities for $Y^{\ast }$ proves that the prediction interval in a normal linear model isIts center and endpoints depend only on the observed data and $x^{\ast }$, and the pivotal calculation proves the required coverage uniformly in the unknown $\beta$ and ${\sigma }^2$.

Graph. The original plot below places distance on the horizontal axis and recession speed on the vertical axis. Both axes include the origin, so the proportional model and departures from it can be judged directly. The displayed line is the fitted through-origin line, not a sequence joining adjacent observations.

Minimization. Write $S_{dd}={\sum }_i{d}_i^2$ and $S_{dV}={\sum }_i{d}_i{V}_i$. Differentiating the specified chi-squared objective givesThus the unique least-squares minimum isConsequentlyAn unweighted average of $V_i/d_i$ would not minimize this chi-squared objective: equal speed errors give greater slope information to the more distant points. No intercept is fitted because the supplied model constrains it to zero.

Estimated error. For the residual estimate of slope uncertainty through the origin, independent equal-variance speed errors implyEquivalently, completing the square gives ${\chi }^2(H)-{\chi }^2(\hat{H})=S_{dd}(H-\hat{H}{)}^2/{\sigma }^2$. The common $\sigma$ is not supplied. Estimate it from the residuals, using eight residual degrees of freedom after fitting one parameter:The estimated standard error is $s/\sqrt{S_{dd}}=0.0281283$ km s${}^{-1}$ kpc${}^{-1}$, givingwith the uncertainty interpreted as one estimated standard error. If the common errors are Gaussian, $(\hat{H}-H)/(s/\sqrt{S_{dd}})$ has a Student's t-distribution with eight degrees of freedom; an exact confidence interval uses the corresponding quantile rather than treating the estimated $s$ as known. The quoted error is conditional on exact distances, independent constant-variance speed scatter and the proportional model. A common distance-calibration bias or correlated velocities would not be captured by this residual calculation.

For independent errors of common unknown variance in $Y_i=\beta x_i+{\epsilon }_i$, the ordinary least squares estimate is $\hat{\beta }={\sum }_i{x}_i{Y}_i/{\sum }_i{x}_i^2$. Fitting one parameter leaves $n-1$ residual degrees of freedom. Therefore $s^2={\sum }_i(Y_i-\hat{\beta }{x}_i{)}^2/(n-1)$ estimates the common error variance, and $s^2/{\sum }_i{x}_i^2$ estimates the slope variance. For errors with a Gaussian distribution, the slope's standardized deviation follows a Student's t-distribution with $n-1$ degrees of freedom. A common calibration bias does not appear as independent residual scatter.

Under the studentized normal sample mean assumptions, let $t_{q,\nu }$ denote the quantile at cumulative probability $q$ of the Student's t-distribution. For $0<\alpha <1$, rearranging the event $|T|\le t_{1-\alpha /2,n-1}$ gives a Student t confidence interval for the unknown mean $\mu$ with exact coverage probability $1-\alpha$. ^[ref]

Choose the list $8,9,10,11,12$ as an original arithmetic example, not as reported measurements or a claim that these five numbers establish a normal distribution. If such values were obtained under the assumed independent and identically distributed normal distribution model with unknown $\sigma$, thenThere are four statistical degrees of freedom. For a 95% confidence interval, $t_{0.975,4}=2.776445105\dots$, soUsing the standard normal quantile $z_{0.975}=1.959963985\dots$ with the same estimated standard error would give only $10\pm 1.385903824\dots$. That underestimates the correct margin by about 29.4%. Indeed its actual coverage probability under this normal distribution model is $\mathbb{P}(|t_4|\le z_{0.975})\simeq 0.8784$, rather than 0.95.

The observations, the sample mean and the confidence interval all have different meanings. The bar estimates the population mean; it is not an interval intended to contain 95% of individual observations. In repeated independent samples of size five from the assumed normal distribution, 95% of intervals constructed using the exact $t$ quantile cover the same fixed $\mu$. This is not a 95% posterior probability that the fixed statistical parameter lies in this particular realized interval. Such a probability requires a separate Bayesian inference model.

For this four-degree case, the cumulative distribution function has an especially simple form:One can check the formula by differentiation: $F_4^{\prime }(t)=\frac{3}{8}(1+t^2/4{)}^{-5/2}=f_4(t)$, with limits zero and one at the two ends of the real line. Solving the increasing equation $F_4(t)=0.975$ by the bisection method gives the quantile used above.

Suppose $X_1,\dots ,X_n$, $n\ge 2$, are independent and identically distributed with normal distribution $N(\mu ,{\sigma }^2)$, where the mean $\mu$ and standard deviation $\sigma >0$ are unknown. Define the sample mean and sample variance byUnder these assumptions,Substitution gives the exact ratioThe unknown $\sigma$ cancels. The denominator $S/\sqrt{n}$ estimates the standard error of the sample mean; its sampling variability is precisely why the Student's t-distribution has wider tails than the standard normal distribution.

The crucial independence has a geometric explanation. Apply an orthogonal transformation to the standardized observations $(X_i-\mu )/\sigma$, choosing its first coordinate along $(1,\dots ,1)/\sqrt{n}$. That coordinate is $Z$; the remaining $n-1$ coordinates are independent standard normal random variables whose squared sum is $V$. Their joint probability density factorizes because the transformation preserves the squared length. This also explains the loss of one statistical degree of freedom when estimating the mean. Uncorrelatedness alone would not establish the required independence for arbitrary data.

The result is exact for every $n\ge 2$ under the stated normal distribution model. A small sample size by itself does not justify it: a sample from an arbitrary probability distribution need not have the displayed chi-squared distribution or the necessary independence.

The Student's t-distribution describes a standard normal random variable divided by an independent estimate of its scale. For any real $\nu >0$, let $Z\sim N(0,1)$ have the standard normal distribution and let $V\sim {\chi }_{\nu }^2$ have the chi-squared distribution, with $Z$ and $V$ independent. ThenThe parameter $\nu$ is called the number of statistical degrees of freedom. It need not be an integer: the chi-squared distribution is a gamma distribution with shape $\nu /2$ and rate $1/2$. The independence assumption is essential; specifying only the two marginal distributions does not determine the probability distribution of their ratio.

Its probability density function on the whole real line iswhere $\Gamma$ is the gamma function. Every member is a symmetric distribution about zero. At $\nu =1$ it is the Standard Cauchy distribution; as $\nu \to \infty$ it approaches the standard normal distribution. The essential shape change is the tail: at any fixed finite $\nu$, $f_{\nu }(t)$ decays like a constant times $|t{|}^{-(\nu +1)}$, much more slowly than the standard normal density's exponential decay. Thus it is a heavy-tailed distribution. ^[ref]

The left panel shows the centers; the right panel uses a logarithmic scale to reveal the tails that a linear plot hides. All four curves have location zero and scale one. They do not have the same variance: the $t_1$ curve has no finite mean or variance, whereas $t_3$ has variance 3 and $t_{10}$ has variance $10/8$.

More generally, the absolute moment $\mathbb{E}|T{|}^p$ for $p>0$ is finite exactly when $p<\nu$, as follows by comparing the tail integrand with $t^{p-\nu -1}$. ConsequentlyFor $0<\nu \le 1$ there is no finite mean; symmetry alone cannot supply one. For $1<\nu \le 2$ the mean exists but the variance is infinite. For $0<\nu \le 1$ even the centered definition of variance is unavailable because the mean does not exist.

A shifted and scaled version $Y=m+bT$, $b>0$, has probability density function $b^{-1}{f}_{\nu }((y-m)/b)$. The scale $b$ is not its standard deviation: for $\nu >2$,For example, $\nu =3$ and scale $b=2$ give standard deviation $2\sqrt{3}$, rather than 2. The distinction matters when using a Student's t-distribution to model data with occasional extreme values, as well as when interpreting a Student t confidence interval.

Student's t-test divides a normally distributed estimated effect by its estimated standard error. Under its null hypothesis the statistic has a Student's t-distribution, with degrees of freedom determined by the variance estimate.

A Student t random-effect model assigns Student's t-distributions to exchangeable study effects, allowing heavier tails than a normal distribution hierarchy. An equivalent Gaussian scale mixture is ${\lambda }_j\sim {\chi }_{\nu }^2$, ${\beta }_j\mid {\lambda }_j,\mu ,\psi \sim N(\mu ,\nu {\psi }^2/{\lambda }_j)$, with independent latent draws. For $\nu >2$ the variance is $\nu {\psi }^2/(\nu -2)$, so $\psi$ is a scale rather than a standard deviation. Small latent precisions weaken shrinkage for atypical studies while retaining partial pooling for the rest.

For independent $Y_i\sim N(\beta x_i,{\sigma }^2)$ with $n\ge 2$ and $\sum x_i^2>0$, the least-squares estimator is $\hat{\beta }=\sum x_i{Y}_i/\sum x_i^2$. The projection of the isotropic Gaussian vector onto the design direction is independent of its orthogonal residual; hence $\hat{\beta }\sqrt{\sum x_i^2}/\sigma$ is standard normal under $\beta =0$ and $\mathrm{SSE}/{\sigma }^2$ is independent chi-squared with $n-1$ degrees of freedom. The displayed statistic therefore has Student's t-distribution with $n-1$ degrees of freedom. A two-sided test rejects at the corresponding upper/lower quantiles. The zero design has no information about the slope; with one observation there is no residual variance degree of freedom.