Supports and Symmetry of the T, Chi-Square, and F Families

Why Supports and Shapes Matter

A probability distribution describes more than how likely values are. It also tells us which values are possible, whether the curve is symmetric, and how probability is spread into the tails. These features help us recognize which probability statements make sense. For example, a negative value is possible for a t variable but not for a chi-square or F variable.

This tutorial compares the Student t, chi-square, and F distribution families through their supports and density curves. The degrees of freedom determine a particular member of each family. Here, \(T\) denotes a t variable with \(\nu\) degrees of freedom, \(Q\) a chi-square variable with \(k\) degrees of freedom, and \(F\) an F variable with numerator and denominator degrees of freedom \(d_1\) and \(d_2\). Degrees of freedom are treated as positive parameters.

Definition: The support of a probability distribution is the set of points for which every open neighborhood has positive probability. Equivalently, it is the smallest closed set that contains the random variable with probability one. Support is not defined merely as the set of points where a particular version of a density is positive.

The Three Families at a Glance

The density curves give a concise comparison. The formulas below are the standard densities for these families. For the chi-square and F formulas, the displayed expressions apply for positive arguments; their behavior at zero is considered through probabilities in neighborhoods of zero.

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

f_Q(q)= \frac{q^{k/2-1}e^{-q/2}} {2^{k/2}\Gamma(k/2)}, \qquad q>0

f_F(x)= \frac{\Gamma\!\left((d_1+d_2)/2\right)} {\Gamma(d_1/2)\Gamma(d_2/2)} \left(\frac{d_1}{d_2}\right)^{d_1/2} x^{d_1/2-1} \left(1+\frac{d_1}{d_2}x\right)^{-(d_1+d_2)/2}, \qquad x>0

Here, \(\Gamma\) is the gamma function, which supplies the constants that make each density integrate to one. The parameterized formulas are useful for recognizing the shapes; the key comparisons are summarized below.

Family	Support	Symmetry and broad shape	Tail behavior
t with \(\nu\) degrees of freedom	\((-\infty,\infty)\)	Symmetric about zero	Both tails decrease at a polynomial rate
Chi-square with \(k\) degrees of freedom	\([0,\infty)\)	Not symmetric; has a right tail	Density decreases exponentially, multiplied by a power of the argument
F with \(d_1,d_2\) degrees of freedom	\([0,\infty)\)	Not symmetric; has a right tail	Density decreases at a polynomial rate

The intervals in the support column include their endpoints. This does not mean that the distributions assign positive probability to every individual point in the support. Each family is continuous, so the probability of any single point is zero. In particular, zero belongs to the support of the chi-square and F distributions, even though each variable is strictly positive with probability one.

Support and the Boundary at Zero

For \(T\), the density is positive at every real number, so every open interval around any real point has positive probability. No value outside the real line is possible. Thus the support is \((-\infty,\infty)\).

For \(Q\) and \(F\), negative values have probability zero. Their densities are positive for positive arguments, however, so every neighborhood around a positive value has positive probability. Every neighborhood of zero also contains positive values, and the density assigns positive probability to a sufficiently small interval of positive values. Therefore zero is in the support, although the event that the variable equals zero has probability zero.

Key distinction: The support of a chi-square or F distribution is \([0,\infty)\), while the variable is strictly positive with probability one. Support describes which points can be approached with positive probability in every neighborhood; it does not require positive probability at the point itself.

Worked Example: Checking Which Events Are Possible

Suppose \(T\) is a t variable with 8 degrees of freedom, \(Q\) is a chi-square variable with 5 degrees of freedom, and \(F\) is an F variable with degrees of freedom 3 and 10. Decide which of the following events have probability zero: \(T<-1\), \(Q<0\), \(Q=0\), and \(F>0\).

The support of \(T\) includes every real number, so \(T<-1\) is possible and does not have probability zero. The support of \(Q\) excludes negative numbers, so \(P(Q<0)=0\). The event \(Q=0\) is a single point in a continuous distribution, so \(P(Q=0)=0\), even though zero is in the support. Finally, \(F>0\) is possible and in fact occurs with probability one for this continuous F variable.

The distinction between the last two chi-square statements is important: \(P(Q=0)=0\), but zero is still in the support because every interval around zero contains positive values that together have positive probability.

Symmetry Belongs to the t Family

The t density depends on its argument through \(t^2\). Replacing \(t\) by \(-t\) leaves the density unchanged, so the curve is symmetric about zero. Because a continuous variable has probability zero at exactly zero, the probability on each side of zero is one half. More generally, equal-length intervals reflected across zero have equal probabilities.

Theorem: If \(T\) has a t distribution with any positive degrees of freedom, then \(T\) and \(-T\) have the same distribution. Consequently, \(P(T<0)=P(T>0)=0.5\), and for every \(c>0\), \(P(T>c)=P(T<-c)\).

This symmetry also gives a convenient relationship for a two-sided tail probability. For \(c>0\), the events \(T>c\) and \(T<-c\) are disjoint and have equal probability. Therefore:

$$ P(|T|>c)=P(T>c)+P(T<-c)=2P(T>c) $$

Worked Example: Using Symmetry for a t Variable

Let \(T\) have a t distribution with 7 degrees of freedom. Suppose we want to express the probability that \(T\) lies outside the interval from \(-1.5\) to \(1.5\), without evaluating a t-table or software probability.

The event outside this interval is the union of two non-overlapping tails: \(T<-1.5\) or \(T>1.5\). Symmetry gives \(P(T<-1.5)=P(T>1.5)\). Hence:

P(T<-1.5\text{ or }T>1.5) =2P(T>1.5)

Equivalently, \(P(-1.5<T<1.5)=1-2P(T>1.5)\). This identifies the correct probability relationship; a numerical tail value would require a t probability calculation. Symmetry does not mean the two tails are zero, only that they have equal probability.

Chi-square and F curves do not have this symmetry. Their support is nonnegative and their densities generally extend farther to the right than to the left. A left-tail probability cannot be reflected to a matching right-tail probability as it can for \(T\).

How the Tails Differ

The t density is symmetric but has heavier tails than a normal density. In the density formula, when \(|t|\) becomes large, \(1+t^2/\nu\) grows approximately as \(t^2/\nu\). Thus the density decreases approximately in proportion to \(|t|^{-(\nu+1)}\), and the probability beyond a large positive cutoff decreases approximately in proportion to the cutoff raised to the power \(-\nu\). This is a polynomial rate of decrease, rather than the faster exponential decrease of a normal tail.

The chi-square density contains the factor \(e^{-q/2}\), which decreases exponentially as \(q\) grows. A power of \(q\) also appears, but for large \(q\) the exponential factor determines the rapid decrease. The density’s behavior near zero depends on \(k\): it tends to infinity when \(k<2\), approaches a finite positive value when \(k=2\), and tends to zero when \(k>2\). These are differences in density height near the boundary, not differences in support.

The F density has a polynomial right tail. For large \(x\), the factor involving \(1+(d_1/d_2)x\) grows proportionally to \(x\). Combining its resulting negative power with \(x^{d_1/2-1}\) shows that the density decreases approximately in proportion to \(x^{-(d_2/2+1)}\). The denominator degrees of freedom \(d_2\) therefore govern the power in this tail. Like the chi-square density, the F density’s behavior near zero depends on its exponent \(d_1/2-1\): it diverges if \(d_1<2\), has a finite positive limit if \(d_1=2\), and tends to zero if \(d_1>2\).

Justification: For large positive \(x\), \(1+(d_1/d_2)x\) is approximately \((d_1/d_2)x\). In the F density, its power therefore contributes a factor proportional to \(x^{-(d_1+d_2)/2}\). Multiplying by the other power, \(x^{d_1/2-1}\), gives \(x^{-(d_2/2+1)}\). This comparison identifies the density’s polynomial tail rate without requiring a probability calculation.

Worked Example: A Chi-Square Right-Tail Probability

Suppose \(Q\) has a chi-square distribution with 2 degrees of freedom. Find the probability that \(Q\) is greater than 4. For \(k=2\), the chi-square density simplifies to \(f_Q(q)=\frac{1}{2}e^{-q/2}\) for \(q>0\).

The probability to the right of 4 is the integral of this density from 4 to infinity:

P(Q>4) =\int_4^\infty \frac{1}{2}e^{-q/2}\,dq =\left[-e^{-q/2}\right]_4^\infty =e^{-2} \approx 0.1353

Thus, in this hypothetical distributional example, about 13.53% of the probability lies above 4. The integral is over positive values, consistent with the support \([0,\infty)\). The event \(Q>4\) is a right-tail event; there is no corresponding negative tail.

Worked Example: An F Right-Tail Probability

Let \(F\) have an F distribution with \(d_1=2\) and \(d_2=2\). Substituting these degrees of freedom in the F density gives \(f_F(x)=(1+x)^{-2}\) for \(x>0\). Find \(P(F>3)\).

Integrate the density from 3 to infinity:

P(F>3) =\int_3^\infty (1+x)^{-2}\,dx =\left[-(1+x)^{-1}\right]_3^\infty =0-\left(-\frac{1}{4}\right) =0.25

So 25% of the probability lies above 3 for this F distribution. The density is defined for positive values, and its right tail continues indefinitely. Unlike a t variable, an F variable has no negative tail and no symmetry around zero.

Reading Probability Statements in Context

A density curve helps identify the region whose area represents a probability. Because these distributions are continuous, a probability such as \(P(T>c)\) is an area under the curve, and \(P(T=c)=0\) for a single specified value. The value of the density at a point is not itself the probability of that point.

The family determines which statements are meaningful and which relationships follow from symmetry. For example, \(P(T<-c)=P(T>c)\) for \(c>0\). But for a chi-square variable \(Q\), \(P(Q<-c)=0\), and there is no symmetry rule equating it with \(P(Q>c)\). Similarly, an F event such as \(F>c\) is an ordinary right-tail event when \(c>0\); an event \(F<-c\) is impossible.

Common Mistakes

Treating support as the set where a density is positive. A density’s value at an individual point can be changed without changing the distribution. Define support using neighborhoods. For chi-square and F variables, zero is in the support because every neighborhood of zero contains positive values with positive probability.
Concluding that a support endpoint must have positive probability. Zero belongs to the support of \(Q\) and \(F\), but \(P(Q=0)=P(F=0)=0\). Each variable is strictly positive with probability one.
Assuming all three families are symmetric. Only the t family is symmetric about zero. Chi-square and F variables are nonnegative and have right tails.
Confusing density height with probability. A density value such as \(f_Q(4)\) is not \(P(Q=4)\). For a continuous variable, the probability at a single point is zero; probabilities are areas over intervals.
Using a t symmetry identity for an F or chi-square variable. The identity \(P(T>c)=P(T<-c)\) depends on t symmetry and does not apply to the other two families.

Key takeaway: The t family has real-line support and symmetry about zero. Chi-square and F families have support \([0,\infty)\), are strictly positive with probability one, and have right tails rather than symmetric pairs of tails. Their tail rates differ: chi-square densities decrease exponentially, while t and F densities decrease at polynomial rates.

Check Your Understanding

Use the supports and density shapes to answer the following questions.

What is the support of a chi-square variable, and why can zero be in its support even though the variable equals zero with probability zero?
If \(T\) has a t distribution and \(c>0\), express \(P(T<-c)\) using a right-tail probability.
For a chi-square variable \(Q\), what is \(P(Q<0)\)? Does this imply that the support excludes zero? Explain.
Which family is symmetric about zero: t, chi-square, or F?
Which density has an exponentially decreasing right tail, chi-square or F?
Explain why \(f_F(3)\) is not the same quantity as \(P(F=3)\).