Definition – F-distribution.

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1. Definition

Definition

Recall the definition of the chi-square distribution,

p_Y(y) = \frac{1}{2^{\frac{n}{2}}\Gamma(\frac{n}{2})}y^{\frac{n}{2}-1}e^{-\frac{1}{2} y}

Let U_m^2 and V_n^2 both be independent and follow the chi-square distribution with m and n degrees of freedom respectively. The random variable

F = \frac{(\frac{U_m^2}{m}) }{ (\frac{V_n^2}{n}) }

follows an F-distribution, with pdf

p(F=f) = \frac{ \Gamma(\frac{m+n}{2}) } {\Gamma(\frac{m}{2}) \Gamma(\frac{n}{2}) } \cdot (\frac{m}{n})^{\frac{m}{2}} \cdot f^{\frac{m}{2} -1} \cdot (\frac{m}{n}f +1)^{-\frac{m+n}{2}}

Derivation

Let U_m^2 and V_n^2 follow chi-square distribution and be independent. Their pdfs are

p(U=u) = \frac{1}{2^{\frac{m}{2}}\Gamma(\frac{m}{2})}u^{\frac{m}{2}-1}e^{-\frac{1}{2} u}

and

p(V=v) = \frac{1}{2^{\frac{n}{2}}\Gamma(\frac{n}{2})}v^{\frac{n}{2}-1}e^{-\frac{1}{2} v}

Applying the bijective transformation g(U) = \frac{U}{m} = y_1 gives g^{-1}(y_1) = my_1, and \frac{dg^{-1}(y_1)}{dy_1} = n. Thus,

p(\frac{U}{m} = y_1) = \frac{m^{ \frac{m}{2} } } {\sqrt{2}^m \Gamma(\frac{m}{2})} y_1^{\frac{m}{2} -1} e^{\frac{my_1}{2}}

Which of course for g(V) = \frac{V}{n} gives the pdf

p(\frac{V}{n} = y_2) = \frac{n^{ \frac{n}{2} } } {\sqrt{2}^n \Gamma(\frac{n}{2})} y_2^{\frac{n}{2} -1} e^{\frac{ny_2}{2}}

We can now apply the formula for the ratio of two random variables,

p(F = \frac{Y_1}{Y_2} = f) = \int_{-\infty}^{\infty} |y_2| p(Y = fy_2)\cdot p(Y_2=y_2) dy_2

From now on, rather than writing y_2, we simply write y. From the above formula, we have,

p(F = \frac{Y_1}{Y_2} = f) = \int\limits_{-\infty}^{\infty} |y| \cdot \frac{m^{ \frac{m}{2} } } {\sqrt{2}^m \Gamma(\frac{m}{2})} (fy)^{\frac{m}{2} -1} e^{\frac{mfy}{2}} \cdot \frac{n^{ \frac{n}{2} } } {\sqrt{2}^n \Gamma(\frac{n}{2})} y^{\frac{n}{2} -1} e^{\frac{ny}{2}} dy

Taking all the terms not depending on the variable y,

\frac{m^{\frac{m}{2}} n^{\frac{n}{2}} f^{\frac{m}{2} - 1}} {(\sqrt{2})^{m+n} \Gamma(\frac{n}{2}) \Gamma(\frac{m}{2})} \int\limits_{-\infty}^{\infty} y^{\frac{m+n}{2} -1} e^{-(\frac{mf+n}{2})y} dy

But the integral is now a gamma distribution without its normalising constant. We can see very easily that \alpha = \frac{n+m}{2} and \beta = \frac{mt+n}{2}, hence

\frac{m^{\frac{m}{2}} n^{\frac{n}{2}} f^{\frac{m}{2} - 1}} {(\sqrt{2})^{m+n} \Gamma(\frac{n}{2}) \Gamma(\frac{m}{2})} \int\limits_{-\infty}^{\infty} y^{\frac{m+n}{2}} e^{-(\frac{mt+n}{2})y}dy \\\\ = \frac{m^{\frac{m}{2}} n^{\frac{n}{2}} f^{\frac{m}{2} - 1}} {(\sqrt{2})^{m+n} \Gamma(\frac{n}{2}) \Gamma(\frac{m}{2})} \times \Gamma(\frac{m+n}{2}) (\frac{mt+n}{2})^{-\frac{m+n}{2}}

Finally, focussing on the final term, we have (\frac{mt+n}{2})^{-\frac{m+n}{2}} = \frac{(\sqrt{2})^{m+n}}{n^{\frac{m+n}{2}}}(\frac{m}{n}f + 1)^{-\frac{m+n}{2}} , which leaves

\frac{m^{\frac{m}{2}} n^{\frac{n}{2}} f^{\frac{m}{2} - 1}} {(\sqrt{2})^{m+n} \Gamma(\frac{n}{2}) \Gamma(\frac{m}{2})} \times \Gamma(\frac{m+n}{2}) (\frac{mt+n}{2})^{-\frac{m+n}{2}} \\\\ = \frac{ \Gamma(\frac{m+n}{2}) } {\Gamma(\frac{m}{2}) \Gamma(\frac{n}{2}) } \cdot (\frac{m}{n})^{\frac{m}{2}} \cdot f^{\frac{m}{2} -1} \cdot (\frac{m}{n}f +1)^{-\frac{m+n}{2}} \quad \square