Normal\(, \chi ^{2}, t\), and F distributions form a connected family. Their constructions explain their roles in statistical inference and the assumptions behind exact critical values.
1. Normal probabilities and standardization
The second argument in \(\mathcal{N}(\mu ,\sigma ^{2})\) is the variance. A normal distribution is symmetric, has skewness zero and kurtosis three. Its central 95% lies within approximately 1.96 standard deviations of its mean. Standardization always gives mean zero and variance one when these moments exist; it gives a standard normal distribution only when normality is justified.
Example. If \(Y\sim \mathcal{N}(1,4), \operatorname{SD}(Y)=2\) and \(\Pr(Y\le 2)=\Phi (.5)\approx .6915. \Pr(|Z|>c)=2[1-\Phi (c)]\) for \(c\ge 0\). Symmetry gives \(\Phi (-c)=1-\Phi (c)\).
| Central probability | Positive cutoff | Area in each tail |
|---|---|---|
| 90% | 1.64485 | 5% |
| 95% | 1.95996 | 2.5% |
| 99% | 2.57583 | .5% |
2. Joint normality
A jointly normal vector has normal linear combinations. Consequently, each marginal is normal; aX+bY is normal with mean \(a\mu _{X}+b\mu _{Y}\) and variance \(a^{2}\sigma _{X}^{2}+b^{2}\sigma _{Y}^{2}+2ab\sigma _{XY}\). For jointly normal variables, zero covariance implies independence. Merely having normal marginal distributions does not establish joint normality.
For a nondegenerate bivariate normal distribution, the conditional mean is linear:
Joint normality implies this form, but a linear conditional mean does not imply joint normality. The displayed slope also anticipates the population linear projection in Module 12.
3. Constructing \(\chi ^{2}\), Student \(t\), and F
| Distribution | Construction | Required independence |
|---|---|---|
| \(\chi ^{2}_{m}\) | \(W=Z_{1}^{2}+\cdots +Z_{m}^{2}, Z_{j}\sim \mathcal{N}(0,1)\) | The \(Z_{j}\) are mutually independent |
| \(t_{m}\) | \(T=Z/\sqrt{W/m}, Z\sim \mathcal{N}(0,1), W\sim \chi ^{2}_{m}\) | \(Z\) and \(W\) are independent |
| \(F_{m,k}\) | \((W/m)/(V/k), W\sim \chi ^{2}_{m}, V\sim \chi ^{2}_{k}\) | \(W\) and V are independent |
The degrees of freedom count independent squared standard normals in the \(\chi ^{2}\) construction\(. \chi ^{2}\) and F are nonnegative; \(t\) is symmetric with heavier tails than the standard normal for finite degrees of freedom. Since \(\mathbb{E}[Z^{2}]=1, \mathbb{E}[W]=m\). The denominator \(W/m\) becomes close to one as m grows, explaining \(t_{m} \to \mathcal{N}(0,1)\). Likewise, as k grows\(, F_{m,k}\) approaches \(\chi ^{2}_{m}/m\).
Useful connection. Squaring \(a t_{k}\) variable gives \(F_{1,k}\). The numerator becomes \(Z^{2}\sim \chi ^{2}_{1}\) and the denominator remains \(W/k\). This later connects a two-sided \(t\) test for one restriction to an F test.
Example. The 95th percentile of \(\chi ^{2}_{3}\) is about 7.815, so the 95th percentile of \(F_{3,\infty }\) is about \(7.815/3=2.605. A 95\)th percentile is an upper-tail cutoff leaving 5%, not the positive endpoint of every two-sided interval.
4. Distribution names are conclusions, not labels
A statistic called a “t-statistic” need not have a Student \(t\) distribution. The exact \(t\) result requires a standard normal numerator and an independent chi-squared variance term. Without these facts, the name of the statistic cannot justify \(t\) critical values. The CLT offers a different, asymptotic justification discussed next.
Check your understanding
Question. If \(Z\sim \mathcal{N}(0,1)\) and \(W=Z^{2}\), does \(Z/\sqrt{W}\) have \(a t_{1}\) distribution?
Solution. No. The ratio equals the sign of \(Z\), taking only \(\pm 1\). Independence in the \(t\) construction is essential.
Sources: SW §2.4, pp. 82–88. The explicit conditional-mean formula and \(t^{2}\)–F identity are supporting derivations from the stated distribution properties.