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The Student T Distribution from Its Density

Use the Student t density to identify its degrees of freedom and understand the shape and tail behavior of the distribution.

Intermediate 9 min read

What You'll Learn

  • Read the Student t density and identify its degrees-of-freedom parameter
  • Explain why the density is symmetric about zero and has its mode there
  • Describe the support and distinguish density values from probabilities
  • Verify that the density integrates to one
  • Explain how polynomial tails differ from normal tails
  • Avoid assuming that symmetry guarantees a finite mean or variance

Why Study the Student t Density?

The Student t family is a set of probability distributions used throughout statistical inference. Each member is symmetric and centered at zero, but its tails extend farther than those of the standard normal distribution. The parameter that identifies a particular member is its degrees of freedom. Understanding the density makes these features precise: it shows where the distribution can take values, where it is most concentrated, and how quickly its density falls in the tails.

The degrees-of-freedom idea was introduced earlier in the course. Here, the parameter is written as \(\nu\), and a random variable with the corresponding Student t distribution is written \(X\sim t_\nu\). In many statistical models \(\nu\) is a positive integer, but the density is defined for any real \(\nu>0\).

The Density and Its Parameter

Definition: For degrees of freedom \(\nu>0\), the Student t distribution has density $$ f_\nu(x)= \frac{\Gamma\left(\frac{\nu+1}{2}\right)} {\sqrt{\nu\pi}\,\Gamma\left(\frac{\nu}{2}\right)} \left(1+\frac{x^2}{\nu}\right)^{-\frac{\nu+1}{2}}, \qquad -\infty<x<\infty. $$ Here \(\Gamma\) is the gamma function, defined for \(a>0\) by $$ \Gamma(a)=\int_0^\infty u^{a-1}e^{-u}\,du. $$ In particular, \(\Gamma(m)=(m-1)!\) for a positive integer \(m\), and \(\Gamma(1/2)=\sqrt{\pi}\).

The formula has a normalizing constant—the factor before the power—that depends on \(\nu\). It ensures that the total area under the density is one. The degrees of freedom also appear inside the power term, so changing \(\nu\) changes the curve's shape as well as its constant.

A density is not itself a probability at a single point. For a continuous random variable, \(P(X=c)=0\) for any one value \(c\). Probabilities are areas over intervals: for \(a<b\), \(P(a<X<b)\) is the integral of \(f_\nu(x)\) from \(a\) to \(b\). The height \(f_\nu(c)\) describes density near \(c\), not the probability that \(X=c\).

Worked Example: Writing the Density When \(\nu=4\)

Suppose \(X\sim t_4\). Substitute \(\nu=4\) into the density and simplify its normalizing constant.

The constant is $$ \frac{\Gamma(5/2)}{\sqrt{4\pi}\,\Gamma(2)} = \frac{(3/4)\sqrt{\pi}}{2\sqrt{\pi}\,(1)} = \frac{3}{8}, $$ because \(\Gamma(5/2)=\tfrac{3}{4}\sqrt{\pi}\) and \(\Gamma(2)=1\). Therefore, $$ f_4(x)=\frac{3}{8}\left(1+\frac{x^2}{4}\right)^{-5/2}. $$ At \(x=2\), the density is $$ f_4(2)=\frac{3}{8}(1+1)^{-5/2} =\frac{3}{8\cdot 2^{5/2}} \approx 0.06629. $$ This is a density height, not a probability of \(0.06629\) at \(x=2\). The density is defined for every real \(x\), and its value at \(-2\) is the same because the formula depends on \(x^2\).

Support, Symmetry, and Center

The support of a continuous distribution is the set of values where its density is positive. In the Student t formula, \(1+x^2/\nu\) is positive for every real \(x\), as is the normalizing constant. Thus the density is positive for all real \(x\), and the support is the entire real line. A Student t variable can take negative or positive values, with no finite lower or upper endpoint.

The formula depends on \(x\) only through \(x^2\), so replacing \(x\) by \(-x\) leaves the density unchanged:

$$ f_\nu(-x)=f_\nu(x). $$

This equality means the distribution is symmetric about zero. The density also reaches its maximum at zero. To see why, differentiate its logarithm:

$$ \frac{d}{dx}\log f_\nu(x) = -\frac{(\nu+1)x}{\nu+x^2}. $$

For \(x<0\), this derivative is positive, so the density increases as \(x\) approaches zero. For \(x>0\), it is negative, so the density decreases as \(x\) moves away from zero. Therefore zero is the mode, the point of maximum density, as well as the center of symmetry.

Symmetry and mode should not be confused with a guarantee that every moment exists. The mean of a Student t variable exists when \(\nu>1\), and in that case it equals zero. For \(0<\nu\leq 1\), the distribution is symmetric but its mean is not defined. This distinction matters: a symmetric density can have tails so heavy that its mean does not exist.

Worked Example: Finding the Center and Density at Zero

Suppose \(X\sim t_3\). Find the density at zero and state what the density implies about the curve's center and mode.

For \(\nu=3\), the normalizing constant is $$ \frac{\Gamma(2)}{\sqrt{3\pi}\,\Gamma(3/2)} = \frac{1}{\sqrt{3\pi}\,(\sqrt{\pi}/2)} = \frac{2}{\pi\sqrt{3}} \approx 0.3676. $$ Since the power term equals one at zero, $$ f_3(0)=\frac{2}{\pi\sqrt{3}}\approx 0.3676. $$ The density is symmetric about zero and has its maximum there, so zero is both its symmetry center and its mode. Because \(\nu=3>1\), its mean also exists and equals zero. The value \(f_3(0)\) is the curve's height at zero, not the probability \(P(X=0)\), which is zero.

Why the Tails Are Heavier Than Normal Tails

A useful way to understand the Student t distribution is to examine its density far from zero. Let \(C_\nu\) denote its positive normalizing constant. For large \(|x|\), factor \(x^2/\nu\) out of the term inside the power:

$$ f_\nu(x) = C_\nu \left(\frac{x^2}{\nu}\right)^{-(\nu+1)/2} \left(1+\frac{\nu}{x^2}\right)^{-(\nu+1)/2}. $$

As \(|x|\) grows, the last factor approaches one. Thus the density has the tail behavior

$$ f_\nu(x)\sim C_\nu\,\nu^{(\nu+1)/2}|x|^{-(\nu+1)} \qquad\text{as } |x|\to\infty. $$

In other words, the Student t density decreases at a polynomial rate. By contrast, the standard normal density is proportional to \(e^{-x^2/2}\), which decreases exponentially in \(x^2\). The Student t density therefore eventually dominates the standard normal density far enough into either tail. This is the precise sense in which its tails are heavier: extreme values retain more density far from the center.

“Heavier tails” does not mean that the t density is higher than the normal density at every nonzero value. The curves can cross; the statement concerns sufficiently distant tail values and the tail probabilities beyond them. Nor does every member of the t family have the same tail behavior: the exponent \(\nu+1\) depends on the degrees of freedom.

Worked Example: Comparing Tail Density Values

For a numerical illustration, suppose \(X\sim t_2\). Compare its density at \(x=4\) with the standard normal density at the same value. These are density heights, not probabilities at exactly \(4\).

For \(\nu=2\), the Student t normalizing constant is $$ \frac{\Gamma(3/2)}{\sqrt{2\pi}\,\Gamma(1)} = \frac{\sqrt{\pi}/2}{\sqrt{2\pi}} = \frac{1}{2\sqrt{2}}. $$ Consequently, $$ f_2(4) = \frac{1}{2\sqrt{2}} \left(1+\frac{16}{2}\right)^{-3/2} = \frac{1}{2\sqrt{2}}\,(9)^{-3/2} = \frac{1}{54\sqrt{2}} \approx 0.01309. $$ The standard normal density at \(4\) is $$ \phi(4)=\frac{1}{\sqrt{2\pi}}e^{-4^2/2} =\frac{e^{-8}}{\sqrt{2\pi}} \approx 0.0001338. $$ At this particular point, the \(t_2\) density is much higher. This comparison illustrates the tail difference; it does not mean the probability of either distribution taking exactly the value \(4\) is positive.

Checking That the Formula Is a Density

A valid density must be nonnegative and have total area one. Nonnegativity follows directly from the formula. The normalizing constant can be checked using the beta integral \(B(a,b)=\int_0^\infty u^{a-1}(1+u)^{-(a+b)}\,du\), for \(a,b>0\), and the identity \(B(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)\). The integral of the unnormalized part is

$$ \begin{aligned} \int_{-\infty}^{\infty} \left(1+\frac{x^2}{\nu}\right)^{-(\nu+1)/2}\,dx &=2\int_0^\infty \left(1+\frac{x^2}{\nu}\right)^{-(\nu+1)/2}\,dx\\ &=\sqrt{\nu}\int_0^\infty u^{-1/2}(1+u)^{-(\nu+1)/2}\,du\\ &=\sqrt{\nu}\,B\left(\frac12,\frac{\nu}{2}\right)\\ &=\frac{\sqrt{\nu\pi}\,\Gamma(\nu/2)} {\Gamma((\nu+1)/2)}. \end{aligned} $$

The substitution in the second line is \(u=x^2/\nu\), so \(dx=\sqrt{\nu}\,u^{-1/2}du/2\); the factor of two from symmetry cancels the \(1/2\). Multiplying this integral by the density's normalizing constant gives one. This verifies the total area and explains why that particular constant appears in the formula.

Common Mistakes and Why the Density Matters

  • Calling a density value a probability. A density of \(0.1\) at a point does not mean a 10% chance of that exact value. Probabilities for continuous variables are areas over intervals.
  • Assuming symmetry guarantees a mean. Symmetry gives a center at zero, but the mean exists only when \(\nu>1\).
  • Assuming heavier means higher everywhere. The t density has polynomial tails and eventually exceeds the standard normal density far out, but that does not imply it is higher at every point.
  • Ignoring the degrees of freedom. The parameter \(\nu\) appears in both the constant and the power term; it identifies the particular density in the Student t family.

The density is useful because it gives an exact description of a t distribution, not just a sketch of its shape. Its support and symmetry identify the range and center, while its degrees of freedom and polynomial tail behavior explain why t-based procedures account for more extreme values than a standard normal model.

Check Your Understanding

Use the density and its properties to answer these questions.

  1. For \(X\sim t_5\), what is the support, and what does symmetry imply about \(f_5(-x)\) and \(f_5(x)\)?
  2. Why is the density value \(f_\nu(1)\) not the probability \(P(X=1)\)?
  3. Where is the mode of a Student t density, and how can the sign of its log-derivative justify your answer?
  4. What feature of the Student t density makes its tails heavier than those of a standard normal density?
  5. For which values of \(\nu\) does the Student t distribution have a defined mean, and what is that mean?

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