Why a Normal–Chi-Square Ratio Defines Student’s \(t\)
A standard normal variable has a fixed scale, but dividing it by a random positive scale changes the distribution. Student’s \(t\) distribution is defined by one particular construction: the numerator is standard normal, and the denominator is the square root of an independent chi-square variable divided by its degrees of freedom. The independence and the scaling by degrees of freedom are both essential.
This ratio is the basic definition of a Student \(t\) variable. It provides a precise way to identify the distribution when the numerator and denominator are specified, and it explains why the degrees of freedom are part of the notation. The later application to a studentized sample mean uses this same definition, but the ratio itself can be studied without starting from a sample.
A chi-square variable with a positive integer number \(\nu\) of degrees of freedom can be represented as the sum of squares of \(\nu\) independent standard normal variables. In the definition above, \(Q\) is independent of the numerator \(Z\). The denominator is positive with probability one, so the ratio is defined with probability one.
The degrees of freedom determine how the chi-square variable contributes to the scale. Increasing \(\nu\) changes the distribution of \(Q/\nu\); it does not change the standard normal distribution of \(Z\). For any particular realization, calculate \(Q/\nu\) first, take its square root, and then divide \(Z\) by that result.
Calculating and Interpreting the Ratio
A ratio question may ask for a realized value of \(T\), or it may ask how a condition on \(T\) translates into a condition on its components. In either case, keep the degrees of freedom attached to the chi-square component and take the square root only after dividing by \(\nu\).
Worked Example: Calculating a Realized t Value
Suppose \(Z=-1.2\) and \(Q=6\), where \(Z\sim N(0,1)\), \(Q\sim\chi^2_3\), and the two variables are independent. Calculate the corresponding value of \(T\), and check whether it is less than \(-0.8\).
Method. Use the defining ratio with \(\nu=3\). The first denominator calculation is $$ \frac{Q}{\nu}=\frac{6}{3}=2. $$
Therefore, $$ T=\frac{-1.2}{\sqrt{2}}=\frac{-1.2}{1.4142\ldots}\approx -0.8485. $$
Since \(-0.8485<-0.8\), the calculated value is less than \(-0.8\). The result is negative because the denominator is positive and the numerator is negative.
Check. Squaring the denominator gives \(Q/\nu=2\), as specified, and dividing \(-1.2\) by the positive square root of \(2\) preserves its negative sign. The common mistake to avoid is using \(\sqrt{Q}=\sqrt{6}\) as the denominator; the definition requires \(\sqrt{Q/\nu}=\sqrt{2}\).
For a positive threshold \(c\), the inequality \(T>c\) can be rearranged while accounting for the positive random denominator. For fixed values of \(Z\) and \(Q\), $$ \frac{Z}{\sqrt{Q/\nu}}>c \quad\Longleftrightarrow\quad Z>c\sqrt{Q/\nu}. $$ Because \(c>0\), this also requires \(Z>0\). If \(Z>0\), squaring both positive sides gives the equivalent condition $$ Q<\nu\left(\frac{Z}{c}\right)^2. $$ This algebra can help check a realized ratio or describe an event. It does not remove the need to use the joint distribution—and, in particular, the independence assumption—when calculating a probability.
Worked Example: Checking a Threshold by Rearranging the Ratio
Let \(Z\sim N(0,1)\) and \(Q\sim\chi^2_5\) be independent. For one realization, suppose \(Z=2\) and \(Q=8\). Calculate \(T\), then determine whether \(T>1.5\) by both direct calculation and the equivalent condition on \(Q\).
Method. First evaluate the scale: $$ \sqrt{\frac{Q}{\nu}}=\sqrt{\frac{8}{5}}=\sqrt{1.6}\approx 1.2649. $$ Hence $$ T=\frac{2}{\sqrt{1.6}}\approx 1.5811, $$ so the realized value exceeds \(1.5\).
For the second check, since \(Z=2>0\), the equivalent condition is $$ Q<5\left(\frac{2}{1.5}\right)^2 =5\left(\frac{4}{3}\right)^2 =5\cdot\frac{16}{9} =\frac{80}{9} \approx 8.8889. $$ The realized \(Q\) is \(8\), and \(8<8.8889\), confirming that \(T>1.5\).
Check. Both methods give the same conclusion. The common mistake is to compare \(Q\) with \((Z/c)^2\) without multiplying by \(\nu\); that would omit the degrees-of-freedom scaling in \(Q/\nu\).
Why This Ratio Has a Student t Density
The ratio’s density follows from the joint density of its independent components. The calculation also shows that the degrees of freedom enter the shape through the exponent and the scale factor. Use the standard normal density for \(Z\) and the chi-square density for \(Q\): $$ f_Z(z)=\frac{1}{\sqrt{2\pi}}e^{-z^2/2}, \qquad f_Q(q)=\frac{q^{\nu/2-1}e^{-q/2}}{2^{\nu/2}\Gamma(\nu/2)},\quad q>0. $$ Here \(\Gamma\) is the gamma function, with \(\Gamma(a)=\int_0^\infty x^{a-1}e^{-x}\,dx\) for \(a>0\). Independence means the joint density is the product \(f_Z(z)f_Q(q)\).
To transform from \((Z,Q)\) to \((T,Q)\), write \(z=t\sqrt{q/\nu}\). The absolute Jacobian determinant for this change of variables is \(\sqrt{q/\nu}\). Integrating the resulting joint density over all \(q>0\) gives the marginal density of \(T\):
Since \(z=t\sqrt{q/\nu}\), its square is \(z^2=t^2q/\nu\). The exponential terms in the product density combine to \(e^{-q(1+t^2/\nu)/2}\).
The Jacobian contributes \(\sqrt{q/\nu}\), so the power of \(q\) in the integrand is \(q^{(\nu+1)/2-1}\). The factor outside the integral is \(1/(\sqrt{2\pi}\,2^{\nu/2}\Gamma(\nu/2)\sqrt{\nu})\).
For \(a>0\) and \(b>0\), \(\int_0^\infty q^{a-1}e^{-bq}\,dq=\Gamma(a)/b^a\). Apply this with \(a=(\nu+1)/2\) and \(b=(1+t^2/\nu)/2\).
The powers of \(2\) and the normalizing constants combine to yield the Student \(t\) density below.
This derivation depends on both component distributions and on their independence. If \(Z\) and \(Q\) are dependent, their joint density need not be the product used in the calculation, so the stated Student \(t\) density is not guaranteed. Likewise, dividing by \(\sqrt{Q}\) instead of \(\sqrt{Q/\nu}\) gives a different scaled ratio, not the \(t_\nu\) variable as defined here.
Worked Example: Evaluating a t Density at Two Points
Let \(T\sim t_2\). Use its density to find \(f_T(0)\) and \(f_T(\sqrt{2})\). State whether either density value is a probability.
Method. Set \(\nu=2\) in the density formula. The constants are \(\Gamma(3/2)=\sqrt{\pi}/2\) and \(\Gamma(1)=1\), so the leading factor is $$ \frac{\Gamma(3/2)}{\sqrt{2\pi}\Gamma(1)} =\frac{\sqrt{\pi}/2}{\sqrt{2\pi}} =\frac{1}{2\sqrt{2}}. $$
At \(t=0\), the factor raised to the power \(-3/2\) is \((1+0)^{-3/2}=1\). Thus $$ f_T(0)=\frac{1}{2\sqrt{2}}\approx 0.3536. $$ At \(t=\sqrt{2}\), the corresponding factor is $$ \left(1+\frac{(\sqrt{2})^2}{2}\right)^{-3/2} =(1+1)^{-3/2} =2^{-3/2} =\frac{1}{2\sqrt{2}}, $$ and therefore $$ f_T(\sqrt{2})=\frac{1}{2\sqrt{2}}\cdot\frac{1}{2\sqrt{2}} =\frac{1}{8}=0.1250. $$
Check. Both density values are nonnegative, as required. Neither is a probability at a single point: for a continuous variable, \(P(T=t)=0\) at every fixed \(t\). The common mistake is to report \(f_T(t)\) as \(P(T=t)\); a probability requires an interval and is found by integrating the density over that interval.
Conditions and Common Exam Traps
To identify a Student \(t\) variable from a ratio, verify each part of the definition rather than relying on the visual appearance of a normal variable divided by something random. The numerator must be standard normal, the denominator must be the square root of a chi-square variable divided by its degrees of freedom, and the two variables must be independent.
| Feature | Required for this definition | Common error |
|---|---|---|
| Numerator | \(Z\sim N(0,1)\) | Using a normal variable with a different mean or variance without first accounting for that change |
| Chi-square component | \(Q\sim\chi^2_\nu\) | Using the wrong degrees of freedom in the ratio |
| Independence | \(Z\) and \(Q\) are independent | Assuming the same density follows when the components are dependent |
| Denominator | \(\sqrt{Q/\nu}\) | Using \(Q\), \(\sqrt{Q}\), or \(Q/\nu\) without the required square root |
A quick numerical check can expose a scaling error. If \(\nu=4\) and \(Q=9\), the denominator in the definition is \(\sqrt{9/4}=1.5\), not \(3\), \(9\), or \(2.25\). The order matters: divide by the degrees of freedom, take the square root, then divide the numerator by that scale.
Check Your Understanding
For each problem, first identify the degrees of freedom and confirm that the ratio is in the defining form.
- Independent variables \(Z\sim N(0,1)\) and \(Q\sim\chi^2_4\) have realized values \(Z=1.5\) and \(Q=9\). Calculate the corresponding \(T\). (Answer: \(T=1.0000\))
- For independent \(Z\sim N(0,1)\) and \(Q\sim\chi^2_6\), a realization has \(Z=-3\) and \(Q=6\). Calculate \(T\) and determine whether \(T<-1\). (Answer: \(T=-3/\sqrt{1}=-3\), so yes.)
- Let \(Z\sim N(0,1)\) and independent \(Q\sim\chi^2_3\), with realized values \(Z=1.2\) and \(Q=3\). For what condition on \(Q\), with this \(Z\) fixed, is \(T>0.8\)? Does the stated realization satisfy it? (Answer: \(Q<3(1.2/0.8)^2=6.75\); yes, since \(3<6.75\).)
- Suppose \(Z\sim N(0,1)\) and \(Q\sim\chi^2_8\) are independent, and a realization gives \(Z=2.4\), \(Q=18\). Calculate \(T\), then check \(T>1\) by rearranging the inequality. (Answer: \(T=2.4/\sqrt{18/8}=1.6000\); the equivalent condition is \(Q<8(2.4/1)^2=46.08\), which is satisfied.)
- A variable is defined as \(R=Z/\sqrt{Q}\), where \(Z\sim N(0,1)\) and \(Q\sim\chi^2_5\) are independent. Express \(R\) as a constant multiple of a standard \(t\) variable, and state the multiplier. (Answer: \(R=T_5/\sqrt{5}\), so the multiplier is \(1/\sqrt{5}\).)