How Degrees of Freedom Change the T Curve

Why Degrees of Freedom Change the Curve

Student t distributions form a family of curves, not one fixed curve. The degrees of freedom determine which member of the family is being used. With few degrees of freedom, a t curve has relatively substantial tails: values far from zero are less surprising under the model than they would be under a standard normal model. As the degrees of freedom increase, the t curve becomes more like the standard normal curve.

This difference matters whenever uncertainty is modeled with a t distribution. Degrees of freedom affect how much probability the model assigns to distant values. They are therefore not just a label attached to a curve: they change its shape and the probabilities calculated from it.

Definition: A random variable \(T\) has a Student t distribution with \(df\) degrees of freedom if its density is

f_{df}(t)= \frac{\Gamma\left((df+1)/2\right)} {\sqrt{df\pi}\,\Gamma(df/2)} \left(1+\frac{t^2}{df}\right)^{-(df+1)/2}, \qquad -\infty<t<\infty.

Here \(\Gamma\) is the gamma function, which extends the factorial: for a positive integer \(k\), \(\Gamma(k)=(k-1)!\). Each t curve is symmetric about zero. The standard normal density is \(\phi(t)=\frac{1}{\sqrt{2\pi}}e^{-t^2/2}\).

The t density is centered at zero for every \(df\). What changes is how its probability is distributed around that center. For small \(df\), the curve is comparatively low near zero and has heavier tails than the standard normal curve. As \(df\) rises, the center becomes more concentrated and the tails become less pronounced. The curves do not become identical to the normal curve at any finite \(df\); they approach it as \(df\) grows.

Comparing the Center and the Tails

A curve’s height at a point is its density there, not the probability of getting exactly that value. For a continuous distribution, probabilities are areas under the curve over intervals. Comparing densities is nevertheless useful for seeing how the shapes differ. At zero, the t density is

f_{df}(0)= \frac{\Gamma\left((df+1)/2\right)} {\sqrt{df\pi}\,\Gamma(df/2)}.

For example, the standard normal density at zero is about \(0.3989\). The t density at zero is lower when \(df=5\), about \(0.3796\). Away from zero, however, a t density can be higher than the normal density. This redistribution—less density in the center and more in the tails—is why a low-degree-of-freedom t curve looks broader.

Worked Example: Comparing Densities at the Center and Away From It

Suppose a hypothetical model calls for comparing the standard normal curve with a t curve having \(df=5\). Calculate their densities at zero and at \(t=2\). These are curve heights, not probabilities at individual points.

For \(df=5\), the constant in the t density simplifies as follows:

\frac{\Gamma(3)}{\sqrt{5\pi}\,\Gamma(5/2)} = \frac{2}{\sqrt{5\pi}\,(3\sqrt{\pi}/4)} = \frac{8}{3\pi\sqrt{5}} \approx 0.3796.

At zero, the power term is \(1\), so \(f_5(0)\approx 0.3796\), compared with \(\phi(0)=1/\sqrt{2\pi}\approx 0.3989\). At \(t=2\), substitute into the full density:

f_5(2)=0.3796\left(1+\frac{2^2}{5}\right)^{-3} =0.3796(1.8)^{-3} \approx 0.0651.

The normal density there is

\phi(2)=\frac{1}{\sqrt{2\pi}}e^{-2}\approx 0.0540.

Thus the \(df=5\) curve is lower at zero but higher at \(t=2\). The comparison illustrates the shape difference: relative to the normal curve, density has been redistributed away from the center and into more distant values.

For \(df>2\), the t distribution has a finite variance:

\operatorname{Var}(T)=\frac{df}{df-2}.

This variance exceeds the standard normal variance of \(1\) for every finite \(df>2\), and it approaches \(1\) as \(df\) increases. For \(df=1\) or \(df=2\), the t distribution has no finite variance. This does not mean the curve is undefined; it means its tails are too substantial for the variance to be finite.

Worked Example: Tracking the Variance as Degrees of Freedom Rise

Consider hypothetical t models with \(df=3\), \(df=5\), \(df=10\), and \(df=30\). For each, use the variance formula where it applies.

For \(df=3\), the variance is \(3/(3-2)=3\). For \(df=5\), it is \(5/(5-2)=5/3\approx1.667\). For \(df=10\), it is \(10/(10-2)=1.25\). For \(df=30\), it is \(30/(30-2)=30/28\approx1.071\).

All four variances exceed \(1\), but they move closer to \(1\) as the degrees of freedom rise. The \(df=3\) model has much greater spread than the standard normal model, while the \(df=30\) model has a variance much nearer to the normal variance. This calculation quantifies one aspect of the curves’ convergence; it does not by itself describe their entire shape.

Why the Tails Are Heavier

The t density has a power-law tail. To see this, take a large positive or negative \(t\). Then \(1+t^2/df\) is approximately \(t^2/df\), so the density behaves approximately as a constant times \(|t|^{-(df+1)}\). By contrast, the standard normal density decreases exponentially in \(t^2\). Consequently, at sufficiently large distances, the t density decreases more slowly than the normal density. Its far tails are heavier.

“Heavier tails” is a statement about how the density behaves far from the center. It does not mean that the t density is higher than the normal density at every nonzero point. As the first worked example showed, a t curve can be lower at the center, cross the normal curve, and be higher farther out. A density comparison at one point also does not, on its own, determine the probability over an interval; that probability is an area.

Worked Example: A Low-Degrees-of-Freedom Tail Probability

Imagine a hypothetical setting in which \(T\) follows a t distribution with \(df=1\). This special case has density \(f_1(t)=1/[\pi(1+t^2)]\). Calculate the probability that \(T\) lies outside the interval from \(-2\) to \(2\), then compare it with the corresponding standard normal probability.

The density is symmetric, and an antiderivative of \(1/[\pi(1+t^2)]\) is \(\arctan(t)/\pi\). Therefore,

P(|T|>2) =1-\int_{-2}^{2}\frac{1}{\pi(1+t^2)}\,dt =1-\frac{2}{\pi}\arctan(2) \approx 1-0.7048 =0.2952.

For a standard normal random variable \(Z\), the corresponding probability is \(P(|Z|>2)\approx0.0455\). Thus, under the \(df=1\) t model, about \(29.52\%\) of the probability lies outside \([-2,2]\), compared with about \(4.55\%\) under the standard normal model. This is a large difference in the tails for this low-degree-of-freedom case.

Why the T Curve Approaches the Standard Normal

The relationship between the t and normal distributions can be seen from the construction of a t variable. If \(Z\) is standard normal and \(U\) is an independent chi-square random variable with \(df\) degrees of freedom, then

T=\frac{Z}{\sqrt{U/df}}

has a t distribution with \(df\) degrees of freedom. This representation, derived from normal and chi-square variables, explains why \(df\) controls the extra spread. The denominator is random, so \(T\) can be farther from zero than \(Z\). As \(df\) increases, \(U/df\) becomes more concentrated near \(1\), and the denominator becomes more like \(1\).

Justification: For a chi-square variable \(U\) with \(df\) degrees of freedom, \(E(U)=df\) and \(\operatorname{Var}(U)=2df\). Thus \(E(U/df)=1\) and \(\operatorname{Var}(U/df)=2/df\). By Chebyshev’s inequality, for every \(\epsilon>0\),

P\left(\left|\frac{U}{df}-1\right|\geq\epsilon\right) \leq \frac{2}{df\,\epsilon^2} \longrightarrow 0 \qquad\text{as }df\longrightarrow\infty.

Therefore \(U/df\) approaches \(1\) in probability, and its square root does too. In \(T=Z/\sqrt{U/df}\), the random denominator consequently approaches \(1\), leaving a variable whose distribution approaches that of \(Z\), the standard normal.

The same approach appears directly in the density. In its formula, the gamma-function ratio satisfies

\frac{\Gamma((df+1)/2)}{\Gamma(df/2)} \sim \sqrt{\frac{df}{2}} \qquad\text{as }df\longrightarrow\infty,

where \(\sim\) means that the ratio of the two sides approaches \(1\). The density’s leading constant therefore approaches \(1/\sqrt{2\pi}\). For any fixed \(t\), the remaining factor approaches the normal exponential term because

\left(1+\frac{t^2}{df}\right)^{-(df+1)/2} = \exp\left[-\frac{df+1}{2}\log\left(1+\frac{t^2}{df}\right)\right] \longrightarrow e^{-t^2/2}.

This follows from \(\log(1+y)/y\to1\) as \(y\to0\). Combining the limits gives \(f_{df}(t)\to\phi(t)\) at every fixed \(t\). Together, the density limit and the random-variable representation explain why the full t distribution approaches the standard normal distribution, including its probabilities over fixed intervals.

Reading the Family of Curves

When comparing t curves, keep three related ideas separate. The density at zero describes the curve’s height at its center. The variance, when finite, summarizes spread around that center. Tail probabilities describe areas beyond chosen cutoffs. As \(df\) increases, the center and spread move toward their standard normal counterparts, and the tails become less pronounced. But neither a curve’s height at one point nor its variance alone gives every tail probability.

Degrees of freedom are especially influential when they are small. Increasing \(df\) from \(3\) to \(5\), for instance, can make a more noticeable shape difference than increasing it from \(100\) to \(102\). In the large-\(df\) range, the t curve is already close to the standard normal curve. It remains a t curve at finite \(df\), however, so the distinction should not be discarded automatically.

Common Mistake

Wrong reasoning: “The t curve is higher than the normal curve at every point away from zero because it has heavier tails.”
Correction: Heavier tails describe the slower decrease far from the center, not a point-by-point ordering across the whole curve. The t curve may be lower near zero, cross the normal curve, and be higher farther out. Compare areas when the question concerns probabilities.
Wrong reasoning: “At a large finite number of degrees of freedom, the t distribution is exactly normal.”
Correction: The t distribution approaches the standard normal as \(df\) increases, but it is not exactly the same at any finite \(df\).

Key takeaway: Small degrees of freedom produce a t curve with heavier tails than the standard normal curve. As \(df\) increases, the curve’s center, spread, and tail behavior approach those of the standard normal, with convergence rather than exact equality at finite \(df\).

Check Your Understanding

Try these questions without looking back.

For \(df=5\), the t density at zero is about \(0.3796\), while the standard normal density there is about \(0.3989\). What does this comparison say, and what does it not say?
Calculate the variance of a t distribution with \(df=8\). Is it greater than, less than, or equal to the standard normal variance?
Explain in words why the t distribution with \(df=1\) has heavier tails than the standard normal distribution.
In the representation \(T=Z/\sqrt{U/df}\), what happens to the denominator as \(df\) increases, and why does that make \(T\) approach \(Z\)?
Does a t density have to be higher than the standard normal density at every nonzero value in order to have heavier tails? Explain.