Moments and Existence Conditions for F Variables

Why the Denominator Degrees of Freedom Matter

An F variable is formed by dividing two scaled chi-square variables. Its moments therefore depend on both degrees of freedom, but the denominator has a special role: values of the denominator close to zero can make the ratio very large. As a result, the denominator degrees of freedom determine whether the mean and variance are finite.

This tutorial derives the mean and variance from the chi-square representation of an F variable. The main practical distinction is that a finite mean requires more denominator degrees of freedom than zero, while a finite variance requires more still.

Definition: Let \(U\sim\chi^2_m\) and \(V\sim\chi^2_n\) be independent chi-square variables with \(m>0\) and \(n>0\) degrees of freedom. The variable \(F=(U/m)/(V/n)\) has an F distribution with numerator degrees of freedom \(m\) and denominator degrees of freedom \(n\), written \(F\sim F_{m,n}\). In sampling applications, \(m\) and \(n\) are usually positive integers.

The numerator degrees of freedom come from \(U\); the denominator degrees of freedom come from \(V\). Keeping this order straight is essential. For the moment formulas below, the condition is imposed on \(n\), the denominator degrees of freedom.

Deriving the Raw Moments

The chi-square density with \(k\) degrees of freedom is \(f(x)=x^{k/2-1}e^{-x/2}/(2^{k/2}\Gamma(k/2))\) for \(x>0\), where \(\Gamma\) is the gamma function. We use this density to find positive moments of \(U\) and negative moments of \(V\).

Write the \(r\)th power of the ratio.
For \(r>0\), \(F^r=(n/m)^r U^r V^{-r}\). Independence gives \(E[F^r]=(n/m)^rE[U^r]E[V^{-r}]\), provided the expectations are finite.

Integrate the numerator moment.
Using the chi-square density and the substitution \(x=2t\), \(E[U^r]=2^r\Gamma(m/2+r)/\Gamma(m/2)\). This is finite for every \(r>0\).

Check the denominator moment.
The integral for \(E[V^{-r}]\) is proportional to \(\int_0^\infty x^{n/2-r-1}e^{-x/2}\,dx\). It is finite near zero exactly when \(n/2-r>0\), or \(n>2r\). When this holds, \(E[V^{-r}]=2^{-r}\Gamma(n/2-r)/\Gamma(n/2)\).

Combining these expressions cancels the powers of 2. This yields the general positive raw-moment formula:

E[F^r]=\left(\frac{n}{m}\right)^r \frac{\Gamma(m/2+r)}{\Gamma(m/2)} \frac{\Gamma(n/2-r)}{\Gamma(n/2)}, \qquad r>0,\quad n>2r.

If \(n\leq 2r\), the integral for the negative moment of \(V\) diverges at zero, so \(E[F^r]\) is infinite. This is not a restriction from the numerator: it is the possibility of a small denominator that controls existence.

Mean and Variance

Set \(r=1\) in the raw-moment formula. The gamma recurrence \(\Gamma(z+1)=z\Gamma(z)\) gives the mean when \(n>2\):

E[F] =\frac{n}{m}\left(\frac{m}{2}\right) \frac{1}{n/2-1} =\frac{n}{n-2}, \qquad n>2.

For the variance, first calculate the second raw moment. It exists only when \(n>4\):

E[F^2] =\left(\frac{n}{m}\right)^2 \left(\frac{m}{2}\right)\left(\frac{m}{2}+1\right) \frac{1}{(n/2-1)(n/2-2)} =\frac{n^2(m+2)}{m(n-2)(n-4)}.

When \(n>4\), the mean also exists, so \(\operatorname{Var}(F)=E[F^2]-(E[F])^2\). Substitution and simplification give:

\begin{aligned} \operatorname{Var}(F) &=\frac{n^2(m+2)}{m(n-2)(n-4)}-\frac{n^2}{(n-2)^2}\\ &=\frac{n^2\big((m+2)(n-2)-m(n-4)\big)} {m(n-2)^2(n-4)}\\ &=\frac{2n^2(m+n-2)}{m(n-2)^2(n-4)},\qquad n>4. \end{aligned}

Theorem: For \(F\sim F_{m,n}\), the mean is finite exactly when \(n>2\), and then \(E[F]=n/(n-2)\). The variance is finite exactly when \(n>4\), and then \(\operatorname{Var}(F)=2n^2(m+n-2)/(m(n-2)^2(n-4))\). If \(2<n\leq4\), the mean is finite but the variance is infinite. If \(n\leq2\), the mean is infinite, and the variance is not finite.

The variance condition follows from the second moment, not merely from the mean condition. In particular, having a finite mean does not guarantee a finite variance. For \(n=3\) or \(n=4\), the first moment exists but the second moment diverges; for \(n=4\), the relevant integral diverges logarithmically at zero.

Worked Applications

Worked Example: Calculating Both Moments

A hypothetical model defines a statistic as \(F=(U/6)/(V/10)\), where \(U\sim\chi^2_6\) and \(V\sim\chi^2_{10}\) are independent. Find its mean and variance. Use the formulas for an F variable, and check the required existence conditions before calculating.

Method and condition check. The statistic has \(m=6\) numerator degrees of freedom and \(n=10\) denominator degrees of freedom. Since \(10>2\), its mean is finite; since \(10>4\), its variance is finite.

Mean. \[ E[F]=\frac{n}{n-2}=\frac{10}{10-2}=\frac{10}{8}=1.25. \]

Variance. \[ \operatorname{Var}(F) =\frac{2(10)^2(6+10-2)}{6(10-2)^2(10-4)} =\frac{200(14)}{6(64)(6)} =\frac{2800}{2304} =\frac{175}{144} \approx 1.2153. \] The decimal is rounded to four places; the exact value is \(175/144\).

Check. The second raw moment is \(E[F^2]=100(8)/(6\cdot8\cdot6)=25/9\). Subtracting the squared mean gives \(25/9-(1.25)^2=175/144\), agreeing with the variance formula. This method avoids treating the numerator degrees of freedom as the threshold for moment existence.

Worked Example: A Finite Mean but No Finite Variance

A hypothetical quality-control procedure produces \(F=(U/3)/(V/4)\), where \(U\sim\chi^2_3\) and \(V\sim\chi^2_4\) are independent. Determine whether the mean and variance exist, and calculate any finite moment requested.

Method and condition check. Here \(m=3\) and \(n=4\). The mean condition is \(n>2\), which holds, so \[ E[F]=\frac{4}{4-2}=2. \]

For the variance, the required condition is \(n>4\). It fails because \(n=4\). The second-moment integral contains a factor proportional to \[ \int_0^1 x^{4/2-2-1}\,dx=\int_0^1\frac{1}{x}\,dx, \] which diverges. Therefore \(E[F^2]\) and the variance are infinite; the variance is not a finite number.

Check. The result distinguishes the two thresholds: \(n=4\) is enough for a finite mean, but it is not enough for a finite variance. The common mistake is to report a variance after substituting \(n=4\) into a formula whose condition requires \(n>4\); the factor \(n-4\) in its denominator signals that the formula cannot be used there.

Worked Example: Checking a Borderline Case and Computing the Variance

A hypothetical ratio statistic has \(F\sim F_{4,7}\). Find its mean, second raw moment, and variance. State why the moments exist before evaluating them.

Method and condition check. The denominator degrees of freedom are \(n=7\). Since \(7>2\), the mean exists. Since \(7>4\), the second raw moment and variance also exist.

Mean. \[ E[F]=\frac{7}{7-2}=\frac{7}{5}=1.4. \]

Second raw moment. \[ E[F^2]=\frac{7^2(4+2)}{4(7-2)(7-4)} =\frac{49\cdot6}{4\cdot5\cdot3} =\frac{294}{60}=4.9. \]

Variance. \[ \operatorname{Var}(F)=E[F^2]-(E[F])^2 =4.9-(1.4)^2 =4.9-1.96=2.94. \]

Check. Substituting \(m=4,n=7\) into the variance formula gives \[ \frac{2(7)^2(4+7-2)}{4(7-2)^2(7-4)} =\frac{98\cdot9}{4\cdot25\cdot3} =\frac{882}{300}=2.94. \] The two calculations agree. Checking \(n>4\) first avoids using the second-moment formula outside its range.

Common Mistakes and Practical Interpretation

An F distribution can have a long upper tail because its denominator can be small. Increasing the numerator degrees of freedom does not remove the denominator condition: even if \(m\) is large, the mean still requires \(n>2\), and the variance still requires \(n>4\). When the variance exists, both formulas involve \(m\), but the existence thresholds are determined by \(n\).

Swapping the degrees of freedom. In \(F_{m,n}\), \(m\) is for the numerator and \(n\) is for the denominator. Apply the moment thresholds to \(n\), not \(m\).
Using the mean condition for the variance. \(n>2\) guarantees a finite mean, not a finite variance. The variance requires \(n>4\).
Substituting at a boundary. The variance formula is not valid at \(n=4\); its denominator is zero because the second moment diverges there.
Reporting an undefined or infinite moment as a finite value. If a required moment does not exist, do not use a formula that assumes it does.

Key takeaway: For an F variable, check denominator degrees of freedom before calculating moments. The mean requires \(n>2\); the variance requires \(n>4\). These conditions reflect the denominator’s behavior near zero.

Check Your Understanding

For each problem, identify the numerator and denominator degrees of freedom first. Then check moment existence before calculating.

Let \(F\sim F_{8,12}\). Find its mean and variance. (Answer: mean \(=1.2\); variance \(=0.81\).)
Let \(F\sim F_{2,2}\). Does its mean exist as a finite value? Does its variance exist as a finite value? (Answer: neither is finite; the mean is infinite and the variance is not finite.)
Let \(F\sim F_{9,6}\). Determine whether its mean and variance exist, and calculate both if they do. (Answer: both exist; mean \(=1.5\), variance \(=3.25\).)
For \(F\sim F_{5,5}\), calculate the mean and variance, showing the denominator condition for each. (Answer: both exist; mean \(=5/3\), variance \(=80/9\).)
A statistic has distribution \(F_{10,4}\). Find its mean and state whether its variance is finite. (Answer: mean \(=2\); variance is infinite, so it is not finite.)

What You'll Learn