Deriving Student T from Normal and Chi-Square Variables

Why a Student t Variable Has This Form

The Student t distribution can be described by its density, but its defining construction gives a useful explanation for that density. Begin with a standard normal variable \(Z\), then divide it by a random scale factor formed from an independent chi-square variable \(U\). Because the scale is random, the resulting ratio has more probability in its tails than a standard normal variable. The derivation below makes that construction precise.

The key condition is that the normal variable and the chi-square variable are independent. A chi-square variable in the denominator is not enough by itself: without independence, the ratio need not have a Student t distribution.

Definition: Let \(Z\sim N(0,1)\), and let \(U\) have a chi-square distribution with \(\nu\) degrees of freedom, where \(\nu>0\). If \(Z\) and \(U\) are independent, then \(T=Z/\sqrt{U/\nu}\) has a Student t distribution with \(\nu\) degrees of freedom, written \(T\sim t_\nu\).

For the derivation, we use the standard densities of the two variables. The standard normal density is \(\phi(z)=(1/\sqrt{2\pi})e^{-z^2/2}\). The chi-square density, for \(u>0\), is

f_U(u)=\frac{1}{2^{\nu/2}\Gamma(\nu/2)}u^{\nu/2-1}e^{-u/2}.

Here \(\Gamma(a)=\int_0^\infty x^{a-1}e^{-x}\,dx\), for \(a>0\), is the Gamma function. We will also use the corresponding integral identity \(\int_0^\infty x^{a-1}e^{-bx}\,dx=\Gamma(a)/b^a\) when \(a>0\) and \(b>0\). It follows from the definition by substituting \(y=bx\).

Deriving the Density by Conditioning

First hold the value of \(U\) fixed at some \(u>0\). Under that condition, \(T=Z/\sqrt{u/\nu}=Z\sqrt{\nu/u}\). Multiplying a standard normal variable by a positive constant changes its standard deviation by that factor. Therefore, conditional on \(U=u\), \(T\) is normal with mean \(0\) and variance \(\nu/u\). Its conditional density at \(t\) is

f_{T\mid U}(t\mid u) =\sqrt{\frac{u}{\nu}}\, \phi\left(t\sqrt{\frac{u}{\nu}}\right) =\sqrt{\frac{u}{2\pi\nu}}\exp\left(-\frac{t^2u}{2\nu}\right).

The first equality includes the density scaling factor \(\sqrt{u/\nu}\). This factor is required when a random variable is multiplied by a constant: the density must change scale so that its total area remains one.

Since \(Z\) and \(U\) are independent, the conditional distribution of \(Z\) given \(U=u\) is still standard normal. Thus the conditional density just written applies. To get the unconditional density of \(T\), average that conditional density over all possible values of \(U\):

f_T(t)=\int_0^\infty f_{T\mid U}(t\mid u)f_U(u)\,du.

Substitute both densities. Combining their powers of \(u\) and their exponential terms gives

\begin{aligned} f_T(t) &=\int_0^\infty \sqrt{\frac{u}{2\pi\nu}}\exp\left(-\frac{t^2u}{2\nu}\right) \frac{u^{\nu/2-1}e^{-u/2}}{2^{\nu/2}\Gamma(\nu/2)}\,du\\ &=\frac{1}{\sqrt{2\pi\nu}\,2^{\nu/2}\Gamma(\nu/2)} \int_0^\infty u^{(\nu+1)/2-1} \exp\left[-\frac{u}{2}\left(1+\frac{t^2}{\nu}\right)\right]\,du. \end{aligned}

The power of \(u\) in the second line is obtained by adding the exponents \(1/2\) and \(\nu/2-1\). Set \(a=(\nu+1)/2\) and \(b=(1+t^2/\nu)/2\). Both are positive, so the Gamma integral identity applies. It evaluates the remaining integral as \(\Gamma(a)/b^a\). Substituting and simplifying the powers of 2 gives

\begin{aligned} f_T(t) &=\frac{\Gamma((\nu+1)/2)} {\sqrt{2\pi\nu}\,2^{\nu/2}\Gamma(\nu/2)} \left(\frac{1+t^2/\nu}{2}\right)^{-(\nu+1)/2}\\ &=\frac{\Gamma((\nu+1)/2)} {\sqrt{\nu\pi}\,\Gamma(\nu/2)} \left(1+\frac{t^2}{\nu}\right)^{-(\nu+1)/2}, \qquad -\infty<t<\infty. \end{aligned}

Theorem: If \(Z\sim N(0,1)\) and \(U\sim\chi^2_\nu\) are independent, then \(Z/\sqrt{U/\nu}\) has density \(f_T(t)=\dfrac{\Gamma((\nu+1)/2)}{\sqrt{\nu\pi}\Gamma(\nu/2)} (1+t^2/\nu)^{-(\nu+1)/2}\), for every real \(t\). This is the Student t density with \(\nu\) degrees of freedom.

Worked Examples

Worked Example: Calculating a Realized Ratio

Suppose independent variables \(Z\sim N(0,1)\) and \(U\sim\chi^2_8\) take the hypothetical realized values \(z=1.20\) and \(u=8.00\). Calculate the corresponding value of \(T\).

Use the defining ratio with \(\nu=8\):

t=\frac{z}{\sqrt{u/\nu}} =\frac{1.20}{\sqrt{8.00/8}} =\frac{1.20}{1} =1.20.

This is one possible realized value from a \(t\) distribution with 8 degrees of freedom. The calculation alone does not say how likely that value is; that is a question about the distribution of the ratio over repeated draws of \(Z\) and \(U\).

Worked Example: Interpreting a Fixed Denominator

Let \(\nu=5\). If the chi-square variable happens to take the value \(u=5\), what is the conditional distribution of \(T\) given this value?

With \(U=5\), the denominator becomes \(\sqrt{u/\nu}=\sqrt{5/5}=1\). Thus, conditionally, \(T=Z\), so \(T\mid U=5\) has a standard normal distribution. More generally, for a fixed \(u\), the conditional variance is \(\nu/u\); here it is \(5/5=1\).

This conditional result is not the unconditional distribution of \(T\). In the full model, \(U\) varies, so the conditional normal density also varies with \(u\). Integrating across those different conditional densities produces the \(t_5\) density derived above.

Worked Example: Evaluating the Density at a Point

For a \(t\) distribution with \(\nu=4\), find the density at \(t=1\). Use \(\Gamma(2)=1\) and \(\Gamma(5/2)=3\sqrt{\pi}/4\).

Substitute \(\nu=4\) and \(t=1\) into the derived density:

\begin{aligned} f_T(1) &=\frac{\Gamma(5/2)}{\sqrt{4\pi}\,\Gamma(2)} \left(1+\frac{1^2}{4}\right)^{-5/2}\\ &=\frac{3\sqrt{\pi}/4}{2\sqrt{\pi}} (1.25)^{-2.5}\\ &=0.375(0.5724)\approx 0.2147. \end{aligned}

The result \(0.2147\) is a density value, not the probability that \(T=1\). For a continuous distribution, probabilities come from areas over intervals; a single point has probability zero.

Why Independence Is Essential

Independence allows the conditional distribution of \(Z\) given \(U=u\) to remain standard normal. It is what justifies using the same normal density for every value of \(u\) before mixing over the chi-square density. If \(Z\) and \(U\) are dependent, that conditional normal distribution may change, and the derivation no longer follows.

Worked Example: A Dependent Pair Does Not Give a t Variable

Let \(Z\sim N(0,1)\), set \(U=Z^2\), and take \(\nu=1\). Then \(U\) has a chi-square distribution with 1 degree of freedom, but \(U\) is determined by \(Z\), so the two variables are not independent. Examine the ratio.

For \(Z\ne 0\), which occurs with probability 1,

\frac{Z}{\sqrt{U/1}} =\frac{Z}{\sqrt{Z^2}} =\frac{Z}{|Z|} = \begin{cases} 1,&Z>0,\\ -1,&Z<0. \end{cases}

This ratio takes only the values \(-1\) and \(1\). It therefore cannot have the continuous \(t_1\) density. Both marginal distribution statements are true, but the independence condition required by the theorem is false.

What the Construction Tells Us

The ratio separates two sources of randomness. The numerator \(Z\) supplies a symmetric normal fluctuation. The denominator \(\sqrt{U/\nu}\) is a positive random scale. When that scale is smaller than 1, the ratio’s magnitude is enlarged; when it is larger than 1, the magnitude is reduced. Since the denominator varies from draw to draw, the result is not simply a standard normal variable.

The derivation also explains why the degrees of freedom label the resulting family: they specify the chi-square distribution used in the random denominator, and hence the density obtained by the mixture. The formula is symmetric in \(t\), because it depends on \(t^2\), and it is defined over the entire real line. The detailed ways that changing \(\nu\) alters the curve are a separate question.

Common Mistakes

Leaving out independence. Knowing only the marginal distributions of \(Z\) and \(U\) does not establish the result. Independence is needed to use the standard normal conditional density for every \(u\).
Using \(U\) instead of \(\sqrt{U/\nu}\). The defining ratio divides by the square root of the chi-square variable after scaling it by its degrees of freedom.
Confusing the conditional and unconditional distributions. For fixed \(u\), \(T\mid U=u\) is normal with variance \(\nu/u\). The unconditional distribution averages across the random values of \(U\) and is Student t.
Calling a density a point probability. A density value such as \(f_T(1)=0.2147\) is not \(P(T=1)\). Probabilities are areas under the density over intervals.

Key takeaway: Under the stated independence condition, a standard normal variable divided by the square root of an independent chi-square variable per degree of freedom has a Student t distribution. Conditioning on the denominator and integrating its density yields the t density.

Check Your Understanding

Try these questions using the definition and derivation above.

For independent \(Z\sim N(0,1)\) and \(U\sim\chi^2_9\), write the expression for \(T\).
If \(\nu=6\) and a hypothetical draw gives \(z=-2\) and \(u=6\), calculate the corresponding value of \(T\).
For fixed \(U=u\), what are the conditional mean and variance of \(T\)?
Which step in the derivation uses independence, and why is it needed?
Explain why a density value at one point is not the probability of observing exactly that value.