The Reciprocal Relationship Between F Distributions

Why Reciprocals Matter for F Distributions

An F random variable is built from a ratio: one scaled chi-square variable divided by another. Taking the reciprocal reverses the ratio, so it also reverses the roles of the numerator and denominator degrees of freedom. This relationship is useful when an F table or software calculation gives a probability or quantile for the degrees of freedom in the opposite order from the one you need.

The key is to track both changes: reciprocating the value and swapping the degrees of freedom. The resulting distribution is not generally the same F distribution as the original one. It is the F distribution with the two degrees of freedom reversed.

Definition: Let $U$ and $V$ be independent chi-square random variables with positive degrees of freedom $d_1$ and $d_2$. Then $$ F=\frac{U/d_1}{V/d_2} $$ has an F distribution with numerator degrees of freedom $d_1$ and denominator degrees of freedom $d_2$, written $F\sim F(d_1,d_2)$. The reciprocal relationship is $$ F\sim F(d_1,d_2)\quad\Longrightarrow\quad \frac{1}{F}\sim F(d_2,d_1). $$

Deriving the Reciprocal Identity

The identity follows directly from the ratio definition. The required model conditions are that $U$ and $V$ are independent chi-square random variables and that both degrees of freedom are positive. A chi-square variable with positive degrees of freedom is positive with probability 1, so $F$ is positive with probability 1 and its reciprocal is defined with probability 1.

Starting with the definition, take the reciprocal and rearrange the ratio:

\frac{1}{F} = \frac{1}{(U/d_1)/(V/d_2)} = \frac{V/d_2}{U/d_1}.

The variables have not changed: $V$ is still chi-square with $d_2$ degrees of freedom, and $U$ is still chi-square with $d_1$ degrees of freedom. They remain independent. In the reciprocal, however, $V/d_2$ is now in the numerator and $U/d_1$ is in the denominator. By the definition of an F distribution, this ratio has an F distribution with numerator degrees of freedom $d_2$ and denominator degrees of freedom $d_1$. Thus $1/F\sim F(d_2,d_1)$.

This is a distributional identity: it describes how the random variable is distributed. For a particular observed value $f$, it also says that its reciprocal $1/f$ can be treated as a value on the reversed-degrees-of-freedom scale. It does not say that $1/F$ has the same distribution as $F$, unless $d_1=d_2$.

Worked Example: Reversing the Ratio and Degrees of Freedom

Suppose $U=12$ and $V=5$ are hypothetical observed values from independent chi-square variables with $d_1=4$ and $d_2=10$ degrees of freedom. Calculate the F ratio and its reciprocal, and identify the distribution associated with the reciprocal.

The original scaled ratio is $$ f=\frac{U/d_1}{V/d_2} =\frac{12/4}{5/10} =\frac{3}{0.5} =6. $$

Its reciprocal is $1/6\approx 0.1667$. Reversing the scaled ratio gives the same result: $$ \frac{V/d_2}{U/d_1} =\frac{5/10}{12/4} =\frac{0.5}{3} =\frac{1}{6} \approx 0.1667. $$

The original ratio has the $F(4,10)$ distribution. Its reciprocal has the $F(10,4)$ distribution. The degrees of freedom switch places because the original denominator becomes the reciprocal’s numerator, and the original numerator becomes its denominator.

Transforming Probabilities

Because $F$ is positive, inequalities involving $F$ reverse direction when taking reciprocals. For a positive cutoff $c$,

F>c \quad\Longleftrightarrow\quad \frac{1}{F}<\frac{1}{c}, \qquad F<c \quad\Longleftrightarrow\quad \frac{1}{F}>\frac{1}{c}.

Consequently, if $F\sim F(d_1,d_2)$, the probability of an upper-tail event for $F$ can be found as a lower-tail probability for a variable with the degrees of freedom reversed:

P(F>c)=P\left(\frac{1}{F}<\frac{1}{c}\right), \qquad \frac{1}{F}\sim F(d_2,d_1).

Likewise, a lower-tail probability becomes an upper-tail probability:

P(F<c)=P\left(\frac{1}{F}>\frac{1}{c}\right).

F distributions are continuous, so the probability of a variable equalling a particular cutoff is zero. As a result, using $<$ or $\leq$ at a cutoff gives the same probability, as does using $>$ or $\geq$. The transformed probability must still use the reciprocal cutoff $1/c$; using $c$ unchanged would describe a different event.

Worked Example: Turning an Upper Tail Into a Lower Tail

Let $X\sim F(4,2)$. Find $P(X>3)$ using the reciprocal identity. For this calculation, use the standard F distribution result that if $Y\sim F(2,4)$, then $P(Y<y)=1-(2/(2+y))^2$ for $y>0$.

Set $Y=1/X$. The reciprocal identity gives $Y\sim F(2,4)$, with the degrees of freedom reversed. Since $X>3$ is equivalent to $Y<1/3$, $$ P(X>3)=P\left(Y<\frac{1}{3}\right). $$

Substitute $y=1/3$ into the given lower-tail formula: $$ P(X>3) =1-\left(\frac{2}{2+1/3}\right)^2 =1-\left(\frac{2}{7/3}\right)^2 =1-\left(\frac{6}{7}\right)^2 =\frac{13}{49} \approx 0.2653. $$

Thus, the probability that an $F(4,2)$ variable exceeds 3 is about 0.2653. The calculation uses the reversed degrees of freedom and the reciprocal cutoff $1/3$, not the original cutoff 3.

Transforming Quantiles

A $p$-quantile is a value $q_p$ for which the probability of being at or below it is $p$. Since F distributions are continuous, the reciprocal identity gives a precise relationship between the quantiles of the two distributions. If $q_p(d_1,d_2)$ is the $p$-quantile of $F(d_1,d_2)$, then

q_p(d_1,d_2)=\frac{1}{q_{1-p}(d_2,d_1)},\qquad 0<p<1.

To see why, let $F\sim F(d_1,d_2)$, and write $G=1/F$, so $G\sim F(d_2,d_1)$. For a positive value $x$,

P(F\leq x) =P\left(G\geq \frac{1}{x}\right) =1-P\left(G<\frac{1}{x}\right).

If $x=q_p(d_1,d_2)$, the left side is $p$. Therefore $P(G<1/x)=1-p$, so $1/x$ is the $(1-p)$-quantile of $F(d_2,d_1)$. Rearranging gives the stated quantile relationship.

In upper-tail language, let $c_\alpha(d_1,d_2)$ denote the cutoff with upper-tail probability $\alpha$, so $P(F>c_\alpha(d_1,d_2))=\alpha$. Then

c_\alpha(d_1,d_2)=\frac{1}{c_{1-\alpha}(d_2,d_1)}.

Here $c_{1-\alpha}(d_2,d_1)$ is also an upper-tail cutoff: its upper-tail probability is $1-\alpha$. Keeping track of whether a quantile is specified by a lower-tail or an upper-tail probability prevents a common reversal error.

Worked Example: Finding a Quantile From Reversed Degrees of Freedom

Suppose $X\sim F(2,4)$. Find its 90th percentile using the reciprocal relationship. For $F(2,4)$, use the lower-tail formula $P(X<x)=1-(2/(2+x))^2$ for $x>0$.

First find the value $x$ for which $P(X<x)=0.90$: $$ 0.90=1-\left(\frac{2}{2+x}\right)^2. $$

Rearrange and solve, noting that $2/(2+x)$ is positive: $$ \left(\frac{2}{2+x}\right)^2=0.10, \qquad \frac{2}{2+x}=\sqrt{0.10}, \qquad x=2\left(\frac{1}{\sqrt{0.10}}-1\right) \approx 4.325. $$

Thus $q_{0.90}(2,4)\approx 4.325$. By the reciprocal quantile relationship, this also means $$ q_{0.10}(4,2)=\frac{1}{q_{0.90}(2,4)} \approx\frac{1}{4.325} \approx 0.2312. $$

The 10th percentile of $F(4,2)$ is about 0.2312. A low quantile for the reversed degrees of freedom corresponds to the reciprocal of a high quantile for the original degrees of freedom.

When the Identity Is Useful

The reciprocal identity provides a way to translate a calculation when a table, calculator, or software output is organized around the opposite ordering of degrees of freedom. It is especially useful for upper-tail probabilities and critical values: the degrees of freedom swap, the cutoff is reciprocated, and the tail direction changes. These are separate steps and all matter.

It also provides a useful check on calculations. If a reported quantile for one ordering is $q$, the corresponding quantile at the complementary probability for the reversed ordering should be $1/q$. Small differences can arise from rounding, but a result with the same degrees of freedom or the same tail probability is not the reciprocal relationship.

Key takeaway: If $F\sim F(d_1,d_2)$, then $1/F\sim F(d_2,d_1)$. For probabilities, reciprocate the cutoff and reverse the inequality. For quantiles, swap the degrees of freedom, take the reciprocal, and use the complementary probability.

Common Mistakes

Keeping the degrees of freedom in the original order. The reciprocal puts the original denominator in the numerator, so $F(d_1,d_2)$ becomes $F(d_2,d_1)$.
Reciprocating the variable but not the cutoff. The event $F>c$ becomes $1/F<1/c$, not $1/F<c$.
Forgetting that the inequality reverses. Taking reciprocals of positive quantities reverses their order, so an upper tail becomes a lower tail and vice versa.
Using the same quantile probability after swapping degrees of freedom. A $p$-quantile corresponds to the reciprocal of the $(1-p)$-quantile for the reversed degrees of freedom.
Assuming the reciprocal has the same F distribution. That is generally false; the degrees of freedom must be swapped. The distribution is unchanged only when the two degrees of freedom are equal.

Check Your Understanding

Use the reciprocal identity and show enough steps to justify your answers.

If $F\sim F(7,12)$, what is the distribution of $1/F$?
Rewrite $P(F>c)$, for $F\sim F(d_1,d_2)$ and $c>0$, as a probability involving $1/F$. State its distribution and the transformed cutoff.
What is the relationship between $q_{0.95}(d_1,d_2)$ and a quantile of the F distribution with the degrees of freedom reversed?
Explain why the inequality reverses when an event involving a positive F variable is expressed using its reciprocal.
A student writes $q_{0.20}(d_1,d_2)=1/q_{0.20}(d_2,d_1)$. Identify the error and give the corrected relationship.