Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ F-distribution ▷ Probability density function

Theorem: Let $F$ be a random variable following an F-distribution:

$\label{eq:f} F \sim \mathrm{F}(u,v) \; .$

Then, the probability density function of $F$ is

$\label{eq:f-pdf} f_F(f) = \frac{\Gamma\left( \frac{u+v}{2} \right)}{\Gamma\left( \frac{u}{2} \right) \cdot \Gamma\left( \frac{v}{2} \right)} \cdot \left( \frac{u}{v} \right)^{\frac{u}{2}} \cdot f^{\frac{u}{2}-1} \cdot \left( \frac{u}{v}f+1 \right)^{-\frac{u+v}{2}} \; .$

Proof: An F-distributed random variable is defined as the ratio of two chi-squared random variables, divided by their degrees of freedom

$\label{eq:f-def} X \sim \chi^2(u), \; Y \sim \chi^2(v) \quad \Rightarrow \quad F = \frac{X/u}{Y/v} \sim \mathrm{F}(u,v)$

where $X$ and $Y$ are independent of each other.

$\label{eq:chi2-pdf} f_X(x) = \frac{1}{\Gamma\left( \frac{u}{2} \right) \cdot 2^{u/2}} \cdot x^{\frac{u}{2}-1} \cdot e^{-\frac{x}{2}} \; .$

Define the random variables $F$ and $W$ as functions of $X$ and $Y$

$\label{eq:FW-XY} \begin{split} F &= \frac{X/u}{Y/v} \\ W &= Y \; , \end{split}$

such that the inverse functions $X$ and $Y$ in terms of $F$ and $W$ are

$\label{eq:XY-FW} \begin{split} X &= \frac{u}{v} F W \\ Y &= W \; . \end{split}$

This implies the following Jacobian matrix and determinant:

$\label{eq:XY-FW-jac} \begin{split} J &= \left[ \begin{matrix} \frac{\mathrm{d}X}{\mathrm{d}F} & \frac{\mathrm{d}X}{\mathrm{d}W} \\ \frac{\mathrm{d}Y}{\mathrm{d}F} & \frac{\mathrm{d}Y}{\mathrm{d}W} \end{matrix} \right] = \left[ \begin{matrix} \frac{u}{v} W & \frac{u}{v} F \\ 0 & 1 \end{matrix} \right] \\ \lvert J \rvert &= \frac{u}{v} W \; . \end{split}$

Because $X$ and $Y$ are independent, the joint density of $X$ and $Y$ is equal to the product of the marginal densities:

$\label{eq:f-XY} f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y) \; .$

With the probability density function of an invertible function, the joint density of $F$ and $W$ can be derived as:

$\label{eq:f-FW-s1} f_{F,W}(f,w) = f_{X,Y}(x,y) \cdot \lvert J \rvert \; .$

Substituting \eqref{eq:XY-FW} into \eqref{eq:chi2-pdf}, and then with \eqref{eq:XY-FW-jac} into \eqref{eq:f-FW-s1}, we get:

$\label{eq:f-FW-s2} \begin{split} f_{F,W}(f,w) &= f_X\left( \frac{u}{v} f w \right) \cdot f_Y(w) \cdot \lvert J \rvert \\ &= \frac{1}{\Gamma\left( \frac{u}{2} \right) \cdot 2^{u/2}} \cdot \left( \frac{u}{v} f w \right)^{\frac{u}{2}-1} \cdot e^{-\frac{1}{2} \left( \frac{u}{v} f w \right)} \cdot \frac{1}{\Gamma\left( \frac{v}{2} \right) \cdot 2^{v/2}} \cdot w^{\frac{v}{2}-1} \cdot e^{-\frac{w}{2}} \cdot \frac{u}{v} w \\ &= \frac{\left( \frac{u}{v} \right)^{\frac{u}{2}} \cdot f^{\frac{u}{2}-1}}{\Gamma\left( \frac{u}{2} \right) \cdot \Gamma\left( \frac{v}{2} \right) \cdot 2^{(u+v)/2}} \cdot w^{\frac{u+v}{2}-1} \cdot e^{-\frac{w}{2} \left( \frac{u}{v} f + 1 \right)} \; . \end{split}$

The marginal density of $F$ can now be obtained by integrating out $W$:

$\label{eq:f-F-s1} \begin{split} f_F(f) &= \int_{0}^{\infty} f_{F,W}(f,w) \, \mathrm{d}w \\ &= \frac{\left( \frac{u}{v} \right)^{\frac{u}{2}} \cdot f^{\frac{u}{2}-1}}{\Gamma\left( \frac{u}{2} \right) \cdot \Gamma\left( \frac{v}{2} \right) \cdot 2^{(u+v)/2}} \cdot \int_{0}^{\infty} w^{\frac{u+v}{2}-1} \cdot \mathrm{exp}\left[ -\frac{1}{2} \left( \frac{u}{v} f + 1 \right) w \right] \, \mathrm{d}w \\ &= \frac{\left( \frac{u}{v} \right)^{\frac{u}{2}} \cdot f^{\frac{u}{2}-1}}{\Gamma\left( \frac{u}{2} \right) \cdot \Gamma\left( \frac{v}{2} \right) \cdot 2^{(u+v)/2}} \cdot \frac{\Gamma\left( \frac{u+v}{2} \right)}{\left[ \frac{1}{2}\left( \frac{u}{v} f + 1 \right) \right]^{(u+v)/2}} \cdot \int_{0}^{\infty} \frac{\left[ \frac{1}{2}\left( \frac{u}{v} f + 1 \right) \right]^{(u+v)/2}}{\Gamma\left( \frac{u+v}{2} \right)} \cdot w^{\frac{u+v}{2}-1} \cdot \mathrm{exp}\left[ -\frac{1}{2} \left( \frac{u}{v} f + 1 \right) w \right] \, \mathrm{d}w \; . \end{split}$

At this point, we can recognize that the integrand is equal to the probability density function of a gamma distribution with

$\label{eq:f-W-gam-ab} a = \frac{u+v}{2} \quad \text{and} \quad b = \frac{1}{2}\left( \frac{u}{v} f + 1 \right) \; ,$

and because a probability density function integrates to one, we finally have:

$\label{eq:f-F-s2} \begin{split} f_F(f) &= \frac{\left( \frac{u}{v} \right)^{\frac{u}{2}} \cdot f^{\frac{u}{2}-1}}{\Gamma\left( \frac{u}{2} \right) \cdot \Gamma\left( \frac{v}{2} \right) \cdot 2^{(u+v)/2}} \cdot \frac{\Gamma\left( \frac{u+v}{2} \right)}{\left[ \frac{1}{2}\left( \frac{u}{v} f + 1 \right) \right]^{(u+v)/2}} \\ &= \frac{\Gamma\left( \frac{u+v}{2} \right)}{\Gamma\left( \frac{u}{2} \right) \cdot \Gamma\left( \frac{v}{2} \right)} \cdot \left( \frac{u}{v} \right)^{\frac{u}{2}} \cdot f^{\frac{u}{2}-1} \cdot \left( \frac{u}{v}f+1 \right)^{-\frac{u+v}{2}} \; . \end{split}$
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Metadata: ID: P264 | shortcut: f-pdf | author: JoramSoch | date: 2021-10-12, 09:00.