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The pdf for the *F* distribution is

$$y=f\left(x|{\nu}_{1},{\nu}_{2}\right)=\frac{\Gamma \left[\frac{\left({\nu}_{1}+{\nu}_{2}\right)}{2}\right]}{\Gamma \left(\frac{{\nu}_{1}}{2}\right)\Gamma \left(\frac{{\nu}_{2}}{2}\right)}{\left(\frac{{\nu}_{1}}{{\nu}_{2}}\right)}^{\frac{{\nu}_{1}}{2}}\frac{{x}^{\frac{{\nu}_{1}-2}{2}}}{{\left[1+\left(\frac{{\nu}_{1}}{{\nu}_{2}}\right)x\right]}^{\frac{{\nu}_{1}+{\nu}_{2}}{2}}}$$

where Γ( · ) is the Gamma function.

The *F* distribution has a natural relationship
with the chi-square distribution. If *χ*_{1 }and *χ*_{2} are
both chi-square with *ν*_{1} and *ν*_{2} degrees
of freedom respectively, then the statistic *F* below
is *F*-distributed.

$$F\left({\nu}_{1},{\nu}_{2}\right)=\frac{\raisebox{1ex}{${\chi}_{1}$}\!\left/ \!\raisebox{-1ex}{${\nu}_{1}$}\right.}{\raisebox{1ex}{${\chi}_{2}$}\!\left/ \!\raisebox{-1ex}{${\nu}_{2}$}\right.}$$

The two parameters, *ν*_{1} and *ν*_{2},
are the numerator and denominator degrees of freedom. That is, *ν*_{1} and *ν*_{2} are
the number of independent pieces of information used to calculate *χ*_{1 }and *χ*_{2},_{ }respectively.

Compute the pdf of an *F* distibution with 5 numerator degrees of freedom and 3 denominator degrees of freedom.

x = 0:0.01:10; y = fpdf(x,5,3);

Plot the pdf.

figure; plot(x,y)

The plot shows that the *F* distribution exists on positive real numbers and is skewed to the right.

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