Given that u="\\chi"2m 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
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.
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