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Symmetry, Tails, and Moments of Student t

Use symmetry and tail behavior to determine Student t probabilities and decide when its mean and variance exist.

Intermediate 10 min read

What You'll Learn

  • Use symmetry to relate left-tail and right-tail probabilities for a Student t variable.
  • Identify the polynomial rate at which Student t tails decay.
  • Determine whether an absolute moment exists from the degrees of freedom.
  • Calculate the mean and variance when those moments exist.
  • Distinguish a finite mean from a finite variance, including at boundary degrees of freedom.

Why Symmetry and Tail Behavior Matter

A Student t distribution is symmetric about zero, but that does not mean all of its moments exist. Its tails decrease at a polynomial rate, rather than the faster rate associated with the normal distribution. The degrees of freedom determine how heavy those tails are, and therefore whether an expected value or variance is mathematically defined.

These properties are useful whenever a standardized statistic is modeled by a t distribution. Symmetry can turn a left-tail probability into a right-tail probability, while the degrees of freedom tell you whether statements about the mean and variance are justified. The results below concern a standard Student t variable, not a shifted or rescaled version.

Definition: For degrees of freedom \(\nu>0\), a standard Student t variable \(T\sim t_\nu\) has density $$ f_T(t)=\frac{\Gamma\left((\nu+1)/2\right)}{\sqrt{\nu\pi}\,\Gamma(\nu/2)} \left(1+\frac{t^2}{\nu}\right)^{-(\nu+1)/2},\qquad -\infty<t<\infty. $$ Here \(\Gamma\) is the gamma function. Equivalently, \(T=Z/\sqrt{U/\nu}\), where \(Z\sim N(0,1)\), \(U\sim\chi^2_\nu\), and \(Z\) and \(U\) are independent.

The ratio representation makes the source of symmetry intuitive: changing \(Z\) to \(-Z\) changes \(T\) to \(-T\), while leaving the distribution of \(Z\) unchanged. The density gives a direct way to establish the same result and to examine the tails.

Symmetry and Probability Identities

The density depends on \(t\) only through \(t^2\), so \(f_T(-t)=f_T(t)\). Thus the distribution is symmetric around zero. For \(x\geq 0\), equal-length intervals on opposite sides of zero have equal probability. In particular, the left and right tails beyond \(x\) have equal probability.

Theorem: If \(T\sim t_\nu\), then \(T\) and \(-T\) have the same distribution. For \(x\geq0\), $$ P(T<-x)=P(T>x). $$ If \(F_\nu(x)=P(T\leq x)\), then \(F_\nu(-x)=1-F_\nu(x). $$ For \(0\leq a<b\), \(P(-b<T<-a)=P(a<T<b)\).

The CDF identity follows by reflecting the part of the density to the left of \(-x\) onto the right of \(x\). Since the density is continuous, including or excluding a single endpoint does not change any of these probabilities.

Worked Example: Reflecting Tail Probabilities

A standardized measurement \(T\) is modeled as \(t_8\). Suppose a calculation provides the right-tail probabilities \(P(T>1)=0.17\) and \(P(T>2)=0.04\), rounded to two decimal places. Find the probabilities that \(T\) lies between \(-2\) and \(-1\), and between \(-2\) and \(1\). Use the supplied probabilities at their stated precision.

Method. Apply symmetry to reflect negative values to the corresponding positive values. The common mistake is to treat the probability between two cutoffs as a tail probability, rather than subtracting the nested tails.

First, reflection gives $$ P(-2<T<-1)=P(1<T<2). $$ The event \(T>1\) consists of the disjoint events \(1<T<2\) and \(T>2\). Therefore, $$ P(1<T<2)=P(T>1)-P(T>2)=0.17-0.04=0.13. $$ So \(P(-2<T<-1)=0.13\), to the precision of the supplied inputs.

For the second interval, split \((-2,1)\) at zero. By symmetry, \(P(-2<T<0)=P(0<T<2)\). Alternatively, the CDF identity gives \(F_8(-2)=0.04\) and \(F_8(1)=1-0.17=0.83\). Hence $$ P(-2<T<1)=F_8(1)-F_8(-2)=0.83-0.04=0.79. $$ The results are consistent: the interval \((-2,1)\) contains \((-2,-1)\), and its probability exceeds \(0.13\).

How Heavy Are the Tails?

For large positive \(t\), the term \(1+t^2/\nu\) is approximately \(t^2/\nu\). Substituting this into the density shows that a t density decreases in proportion to \(t^{-(\nu+1)}\). Integrating that density from a large cutoff to infinity gives a right-tail probability proportional to \(x^{-\nu}\). The corresponding left tail has the same behavior by symmetry.

Theorem: For \(T\sim t_\nu\), as \(x\) tends to infinity, $$ f_T(x)\sim K_\nu x^{-(\nu+1)},\qquad P(T>x)\sim \frac{K_\nu}{\nu}x^{-\nu}, $$ where $$ K_\nu= \frac{\Gamma\left((\nu+1)/2\right)}{\sqrt{\nu\pi}\,\Gamma(\nu/2)} \,\nu^{(\nu+1)/2}. $$ The symbol \(\sim\) means that the ratio of the two sides tends to one. These are large-\(x\) approximations, not exact tail formulas.

To see why the tail exponent follows, write the density for large positive \(t\) as $$ f_T(t)\sim K_\nu t^{-(\nu+1)}. $$ The integral of the comparison function from \(x\) to infinity is $$ \int_x^\infty K_\nu t^{-(\nu+1)}\,dt =\frac{K_\nu}{\nu}x^{-\nu}. $$ Because the ratio of the actual density to that comparison function tends to one, its tail integral has the same leading asymptotic form. This argument also shows why a smaller \(\nu\) means a more slowly decreasing tail.

When Moments Exist

Tail decay determines whether moments are finite. For any real \(r>0\), the integral for the absolute moment \(E[|T|^r]\) has, in its tails, the same convergence behavior as $$ \int_1^\infty t^r t^{-(\nu+1)}\,dt =\int_1^\infty t^{r-\nu-1}\,dt. $$ This integral converges exactly when \(r<\nu\). Consequently, \(E[|T|^r]\) is finite if and only if \(r<\nu\). In particular, an ordinary finite mean requires \(\nu>1\), and a finite second moment requires \(\nu>2\).

Symmetry alone is not enough to assert that the mean is zero. When \(\nu\leq1\), the positive and negative parts both have infinite expectation: cancellation of the two tails is not a valid definition of an expectation. When \(\nu>1\), the absolute first moment is finite, so symmetry does imply that the mean is zero.

Theorem: If \(T\sim t_\nu\), then:
  • If \(\nu>1\), \(E[T]=0\). If \(0<\nu\leq1\), the finite mean does not exist.
  • If \(\nu>2\), \(E[T^2]=\operatorname{Var}(T)=\nu/(\nu-2)\).
  • If \(0<\nu\leq2\), the second moment is infinite. For \(1<\nu\leq2\), the mean exists and is zero, but the variance is infinite.

The exact variance can also be obtained from the normal–chi-square representation. When \(\nu>2\), independence gives $$ E[T^2] =E\left[\frac{Z^2}{U/\nu}\right] =\nu E[Z^2]E[U^{-1}]. $$ Here \(E[Z^2]=1\). If \(U\sim\chi^2_\nu\), its density is proportional to \(u^{\nu/2-1}e^{-u/2}\) for \(u>0\). Integrating \(u^{-1}\) against this density, or using the gamma integral, gives \(E[U^{-1}]=1/(\nu-2)\) when \(\nu>2\). Thus \(E[T^2]=\nu/(\nu-2)\). Since the mean is zero for \(\nu>2\), this second moment is also the variance. At \(\nu\leq2\), the second-moment integral diverges, so the finite-variance formula cannot be used.

Worked Example: Mean and Variance at Two Degrees of Freedom

Suppose a hypothetical standardized process error \(T\) has a \(t_6\) distribution. Determine its mean, second moment, and variance. Then state whether a \(t_{2.5}\) model has a finite mean and variance.

Method. Compare the degrees of freedom with the thresholds for the first and second moments. A common error is to use \(\nu/(\nu-2)\) without checking that \(\nu>2\).

For \(T\sim t_6\), \(6>1\), so the mean exists and $$ E[T]=0. $$ Also \(6>2\), so the variance formula applies: $$ E[T^2]=\operatorname{Var}(T)=\frac{6}{6-2}=\frac{6}{4}=\frac{3}{2}=1.5. $$

For a \(t_{2.5}\) variable, \(2.5>1\), so its mean exists and equals zero. Since \(2.5>2\), its variance exists as well: $$ \operatorname{Var}(T)=\frac{2.5}{2.5-2}=\frac{2.5}{0.5}=5. $$ Thus both examples have finite means and variances, but the \(t_{2.5}\) variance is much larger. As a check, both calculated variances are positive, and the lower degrees of freedom give the heavier tail.

Worked Example: A \(t_3\) Tail and Its Moments

A hypothetical model assigns a standard \(t_3\) distribution to a standardized measurement. Find the leading large-\(x\) approximation to \(P(T>x)\), use it to estimate the right-tail probability at \(x=10\), and determine the mean and variance.

Method. Substitute \(\nu=3\) into the asymptotic tail formula, then check moment existence separately. The approximation is not an exact probability, and it should not be used as one.

For \(\nu=3\), the density constant is $$ \frac{\Gamma(2)}{\sqrt{3\pi}\,\Gamma(3/2)} =\frac{1}{\sqrt{3\pi}\,(\sqrt{\pi}/2)} =\frac{2}{\pi\sqrt{3}}. $$ Therefore, $$ K_3=\frac{2}{\pi\sqrt{3}}\,3^2 =\frac{18}{\pi\sqrt{3}} =\frac{6\sqrt{3}}{\pi}. $$ The right-tail approximation is consequently $$ P(T>x)\sim\frac{K_3}{3}x^{-3} =\frac{2\sqrt{3}}{\pi}x^{-3}. $$

At \(x=10\), use \(10^{-3}=0.001\): $$ P(T>10)\approx \frac{2\sqrt{3}}{\pi}(0.001) \approx 1.1027(0.001)=0.0011027. $$ This is a large-\(x\) approximation, reported here to six decimal places as \(0.001103\). It is not a claim that the exact probability equals that rounded value.

Finally, \(3>1\), so the mean exists and is zero. Since \(3>2\), the variance exists and is $$ \operatorname{Var}(T)=\frac{3}{3-2}=3. $$ The finite variance is consistent with the tail test: a \(t_3\) distribution has finite moments of positive order less than three, including its second moment.

Common Mistakes and Exam Traps

  • Claiming symmetry guarantees a mean of zero. Symmetry gives equal tails, but for \(0<\nu\leq1\) those tails are too heavy for a finite mean. The expectation is undefined, not zero.
  • Using the variance formula at the boundary. At \(\nu=2\), the denominator \(\nu-2\) is zero and the second moment diverges. The formula applies only for \(\nu>2\).
  • Confusing a finite mean with a finite variance. For \(1<\nu\leq2\), \(E[T]=0\), but \(\operatorname{Var}(T)\) is infinite.
  • Treating an asymptotic tail as an exact probability. The expression involving \(x^{-\nu}\) describes the leading behavior for large \(x\); it does not replace an exact CDF calculation at an arbitrary cutoff.
  • Reflecting the wrong interval. The interval \((-b,-a)\) reflects to \((a,b)\), not \((b,a)\). For a probability between positive cutoffs, subtract the farther right tail from the nearer one.
Key takeaway: A standard \(t_\nu\) distribution is symmetric for every \(\nu>0\), and its right tail decreases on the order of \(x^{-\nu}\). Its mean is zero only when \(\nu>1\); its variance is \(\nu/(\nu-2)\) only when \(\nu>2\).

Check Your Understanding

For each problem, use symmetry, tail behavior, or the moment conditions as appropriate.

  1. A standard \(t_5\) variable has supplied right-tail probabilities \(P(T>1)=0.15\) and \(P(T>2)=0.04\). Find \(P(-2<T<-1)\) and \(P(-2<T<1)\). (Answer: \(0.11\) and \(0.81\), respectively.)
  2. A standardized quantity follows a \(t_4\) distribution. State its mean and variance, and verify that the variance condition is met. (Answer: mean \(0\); variance \(4/(4-2)=2\), with \(4>2\).)
  3. A model uses \(t_{1.5}\). Does its mean exist? Does its variance exist? (Answer: the mean exists and is \(0\); the variance is infinite because \(1.5\leq2\).)
  4. A standard t density has degrees of freedom \(\nu=5\). State the power of \(x\) in the leading right-tail decay and determine whether its third absolute moment is finite. (Answer: the right tail is of order \(x^{-5}\); the third absolute moment is finite because \(3<5\).)
  5. A standard \(t_1\) variable is symmetric. Is its mean zero, and does its variance exist? (Answer: its finite mean does not exist, and its variance does not exist; symmetry does not make the undefined mean zero.)

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