Answer to Question #242554 in Statistics and Probability for PRS

Question

Given that u=$\chi$²_m and v=$\chi$ _n² further let u and v be independent variables obtain the probability density of F=(U/M)/(V/N) where m and n are the df for U and V

Answer 1

Let us first determine the marginal distribution of random variables U and V. For convenience, let $c=m/2$ and $d=n/2$. To compute the distribution of $F$, we need to find the joint distribution of $U$ and $V$.

Note that,

$f_U(u)=(1/(\varGamma(c)*2^c))*u^{c-1}e^{-u/2},\space u\gt0$ and

$f_V(v)=(1/(\varGamma(d)*2^d))*v^{d-1}e^{-v/2},\space v\gt0$

Since $U$ and $V$ are independent, their joint distribution is given as,

$f_{U,V}(u,v)=(1/\varGamma(c)\varGamma(d)2^{c+d})*u^{c-1}v^{d-1}e^{-u/2}e^{-v/2},\space u,v\gt0$

We will first find the distribution of the random variable $U/V$ by using the $cdf$ method given as,

$F_{U/V}(f)=P(U/V\leqslant f)=P(U\leqslant fV).$

Now,

$P(U\leqslant fV)=(1/(\varGamma(c)\varGamma(d)2^{c+d}))*\int^\infin_0(\int^{fv}_0u^{c-1}e^{-u/2}du)v^{d-1}e^{-v/2}dv$

$f_{U/V}(f)= F'_{U/V}(f)=(f^{c-1}/\varGamma(c)\varGamma(d))\int^\infin_0u^{c+d-1}e^{-((f+1)/2)v}dv$

Observe that,

$f(v)=(f+1)^{c+d}u^{c+d-1}e^{-((f+1)/2)v}$ is the $pdf$of a Gamma random variable with parameters, $\alpha=(c+d)$ and $\beta=(f+1)/2$.

Since $\int^\infin_0f(v)=1,$

$f_{U/V}(f)=(\varGamma(c+d)/\varGamma(c)\varGamma(d))*(f^{c-1}/(f+1)^{c+d})$

We now have that,

$f_F(f)=f_{(U/m,V/n)}(f)=f_{(n/m)(U/V)}(f)=(m/n)f_{U/V}(m/n*f)$

$=\varGamma((m+n)/2)m^{m/2}n^{n/2}f^{(m/2)-1}/\varGamma(m/2)\varGamma(n/2)(n+mf)^{(m+n)/2},\space f\gt0$

Which is an $F$ distribution with $m$ numerator degrees of freedom and $n$ denominator degrees of freedom.