5.11: The F Distribution (2024)

The trick, once again, is conditioning. The conditional distribution of \( X \) given \( V = v \in (0, \infty) \) is gamma with shape parameter \( n/2 \) and scale parameter \( 2 d / n v \). Hence the conditional PDF is \[ x \mapsto \frac{1}{\Gamma(n/2) \left(2 d / n v\right)^{n/2}} x^{n/2 - 1} e^{-x(nv /2d)} \] By definition, \( V \) has the chi-square distribution with \( d \) degrees of freedom, and so has PDF \[ v \mapsto \frac{1}{\Gamma(d/2) 2^{d/2}} v^{d/2 - 1} e^{-v/2} \] The joint PDF of \( (X, V) \) is the product of these functions: \[g(x, v) = \frac{1}{\Gamma(n/2) \Gamma(d/2) 2^{(n+d)/2}} \left(\frac{n}{d}\right)^{n/2} x^{n/2 - 1} v^{(n+d)/2 - 1} e^{-v( n x / d + 1)/2}; \quad x, \, v \in (0, \infty)\] The PDF of \( X \) is therefore \[ f(x) = \int_0^\infty g(x, v) \, dv = \frac{1}{\Gamma(n/2) \Gamma(d/2) 2^{(n+d)/2}} \left(\frac{n}{d}\right)^{n/2} x^{n/2 - 1} \int_0^\infty v^{(n+d)/2 - 1} e^{-v( n x / d + 1)/2} \, dv \] Except for the normalizing constant, the integrand in the last integral is the gamma PDF with shape parameter \( (n + d)/2 \) and scale parameter \( 2 d \big/ (n x + d) \). Hence the integral evaluates to \[ \Gamma\left(\frac{n + d}{2}\right) \left(\frac{2 d}{n x + d}\right)^{(n + d)/2} \] Simplifying gives the result.

These properties follow from standard calculus. The first derivative of \( f \) is \[ f^\prime(x) = \frac{1}{B(n/2, d/2)} \left(\frac{n}{d}\right)^2 \frac{[(n/d)x]^{n/2-2}}{[1 + (n/2)x]^{n/2 + d/2 + 1}} [(n/2 - 1) - (n/d)(d/2 + 1)x], \quad x \in (0, \infty) \]

These results follow from standard calculus. The second derivative of \( f \) is \[ f^{\prime\prime}(x) = \frac{1}{B(n/2, d/2)} \left(\frac{n}{d}\right)^3 \frac{[(n/d)x]^{n/2-3}}{[1 + (n/d)x]^{n/2 + d/2 + 2}}\left[(n/2 - 1)(n/2 - 2) - 2 (n/2 - 1)(d/2 + 2) (n/d) x + (d/2 + 1)(d/2 + 2)(n/d)^2 x^2\right], \quad x \in (0, \infty) \]

By independence, \( \E(X) = \frac{d}{n} \E(U) \E\left(V^{-1}\right) \). Recall that \( \E(U) = n \). Similarly if \( d \le 2 \), \( \E\left(V^{-1}\right) = \infty \) while if \( d \gt 2 \), \[ \E\left(V^{-1}\right) = \frac{\Gamma(d/2 - 1)}{2 \Gamma(d/2)} = \frac{1}{d - 2} \]

By independence, \( \E\left(X^2\right) = \frac{d^2}{n^2} \E\left(U^2\right) \E\left(V^{-2}\right) \). Recall that \[ E(\left(U^2\right) = 4 \frac{\Gamma(n/2 + 2)}{\Gamma(n/2)} = (n + 2) n \] Similarly if \( d \le 4 \), \( \E\left(V^{-2}\right) = \infty \) while if \( d \gt 4 \), \[ \E\left(V^{-2}\right) = \frac{\Gamma(d/2 - 2)}{4 \Gamma(d/2)} = \frac{1}{(d - 2)(d - 4)} \] Hence \( \E\left(X^2\right) = \infty \) if \( d \le 4 \) while if \( d \gt 4 \), \[ \E\left(X^2\right) = \frac{(n + 2) d^2}{n (d - 2)(d - 4)} \] The results now follow from the previous result on the mean and the computational formula \( \var(X) = \E\left(X^2\right) - \left[\E(X)\right]^2 \).

By independence, \( \E\left(X^k\right) = \left(\frac{d}{n}\right)^k \E\left(U^k\right) \E\left(V^{-k}\right) \). Recall that \[ \E\left(U^k\right) = \frac{2^k \Gamma(n/2 + k)}{\Gamma(n/2)} \] On the other hand, \( \E\left(V^{-k}\right) = \infty \) if \( d/2 \le k \) while if \( d/2 \gt k \), \[ \E\left(V^{-k}\right) = \frac{2^{-k} \Gamma(d/2 - k)}{\Gamma(d/2)} \]

These results follow from the formulas for \( \E\left(X^k\right) \) for \( k \in \{1, 2, 3, 4\} \) and the standard computational formulas for skewness and kurtosis.

This follows easily from the random variable interpretation in the definition. We can write \[ X = \frac{U/n}{V/d} \] where \( U \) and \( V \) are independent and have chi-square distributions with \( n \) and \( d \) degrees of freedom, respectively. Hence \[ \frac{1}{X} = \frac{V/d}{U/n} \]

This follows easily from the random variable representations of the \( t \) and \( F \) distributions. We can write \[ T = \frac{Z}{\sqrt{V/n}} \] where \( Z \) has the standard normal distribution, \( V \) has the chi-square distribution with \( n \) degrees of freedom, and \( Z \) and \( V \) are independent. Hence \[ T^2 = \frac{Z^2}{V/n} \] Recall that \( Z^2 \) has the chi-square distribution with 1 degree of freedom.

We first find the distribution function \( F \) of \( Z \) by conditioning on \( X \): \[ F(z) = \P(Z \le z) = \P(Y \ge X / z) = \E\left[\P(Y \ge X / z \mid X)\right] \] But \( \P(Y \ge y) = e^{-r y} \) for \( y \ge 0 \) so \( F(z) = \E\left(e^{-r X / z}\right) \). Also, \( X \) has PDF \( g(x) = r e^{-r x} \) for \( x \ge 0 \) so \[ F(z) = \int_0^\infty e^{- r x / z} r e^{-r x} \, dx = \int_0^\infty r e^{-r x (1 + 1/z)} \, dx = \frac{1}{1 + 1/z} = \frac{z}{1 + z}, \quad z \in (0, \infty) \] Differentiating gives the PDF of \( Z \) \[ f(z) = \frac{1}{(1 + z)^2}, \quad z \in (0, \infty) \] which we recognize as the PDF of the \( F \) distribution with 2 degrees of freedom in the numerator and the denominator.

The two statements are equivalent and follow from the standard change of variables formula. The function \[ y = \frac{(n/d) x}{1 + (n/d) x} \] maps \( (0, \infty) \) one-to-one onto (0, 1), with inverse \[ x = \frac{d}{n}\frac{y}{1 - y} \] Let \( f \) denote the PDF of the \( F \) distribution with \( n \) degrees of freedom in the numerator and \( d \) degrees of freedom in the denominator, and let \( g \) denote the PDF of the beta distribution with left parameter \( n/2 \) and right parameter \( d/2 \). Then \( f \) and \( g \) are related by

1. \( g(y) = f(x) \frac{dx}{dy} \)
2. \( f(x) = g(y) \frac{dy}{dx} \)

Let \( f \) denote the PDF of \( X \) and \( g \) the PDF of \( Y \).

1. By the change of variables formula, \[ g(y) = \frac{d}{n} f\left(\frac{d}{n} y\right), \quad y \in (0, \infty) \] Substituting into the beta \( F \) PDF shows that \( Y \) has the appropriate beta prime distribution.
2. Again using the change of variables formula, \[ f(x) = \frac{a}{b} g\left(\frac{a}{b} x\right), \quad x \in (0, \infty) \] Substituting into the beta prime PDF shows that \( X \) has the appropriate \( F \) PDF.