Deducing F-distribution PDF

Question

Let $V\sim \chi^2(n)$ and $W\sim \chi^2(m)$ indep. r.v.

I want to find the PDF for $X=\frac{V/n}{W/m}$.

For that I define $h(v,w)=(v,v/n\cdot m/w)=(v,x)$. So, $h^{-1}(v,x)=(v,\frac{v \cdot m}{x\cdot n})$

We know that $f_{V,X}=f_{V,W}(h^{-1}(v,x))\cdot |det(Dh^{-1}(v,x))|$.

Because they are independent, $f_{V,W}=f_{V}\cdot f_{W}$, so doing the math I get, $\displaystyle f_{V,X}=\frac{v^{(n+m)/2-1}e^{-\frac{v}{2}(1+m/(x\cdot n))}}{2^{(m+n)/2} \cdot \Gamma(n/2)\cdot \Gamma(m/2)}\cdot \frac{m}{x^2 n} \cdot \left(\frac{m}{xn}\right)^{m/2-1}$.

Now integrating with respect to $v$, we get $\displaystyle f_{X}=\frac{\Gamma((n+m)/2)}{ \Gamma(n/2)\cdot \Gamma(m/2)}\cdot \frac{1}{x} \cdot \left(\frac{m}{xn}\right)^{(m/2)}\cdot \left(1+\frac{m}{xn}\right)^{-(m+n)/2}$.

According to the book I'm using, I should have got instead: $\displaystyle f_{X}=\frac{\Gamma((n+m)/2)}{ \Gamma(n/2)\cdot \Gamma(m/2)}\cdot x^{(n-2)/2} \cdot \left(\frac{n}{m}\right)^{(m/2)}\cdot \left(1+\frac{nx}{m}\right)^{-(m+n)/2}$.

So, my question is where did I go wrong?

Any help would be appreciated.

Answer 1

Your computation is 100% correct. The answer provided by your book is wrong. You can see this by integrating the book's density over $x \in [0,\infty)$: it will give $(n/m)^{(m-n)/2} \ne 1$ if $m \ne n$. Feel free to refer to the Wikipedia page for the $F$-ratio distribution.

Answer 2

You are correct. I can deduce like this:

When V and U are two $\chi^2$ independent random variables: $f_V(v)=\frac{(\frac{1}{2})^{\frac{m}{2}}}{\Gamma(\frac{m}{2})}v^{(\frac{m}{2})-1}e^{-\frac{1}{2}v}$
$f_U(u)=\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}u^{(\frac{n}{2})-1}e^{-\frac{1}{2}u}$
with $m$ and $n$ degrees of freedom, then, the pdf for $W=V/U$ is:
$$\begin{align} f_{V/U}(\omega)&=\int_{0}^{+\infty}|u|f_U(u)f_V(u\omega)du\\ &=\int_{0}^{+\infty}u\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}u^{\frac{n}{2}-1}e^{-\frac{1}{2}u} \frac{(\frac{1}{2})^{\frac{m}{2}}}{\Gamma(\frac{m}{2})}(u\omega)^{\frac{m}{2}-1}e^{-\frac{1}{2}u\omega}du\\ &=\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\frac{(\frac{1}{2})^{\frac{m}{2}}}{\Gamma(\frac{m}{2})} \omega^{\frac{m}{2}-1} \int_{0}^{+\infty}u^{\frac{n}{2}}u^{\frac{m}{2}-1} e^{-\frac{1}{2}u(1+\omega)}du\\ &=\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\frac{(\frac{1}{2})^{\frac{m}{2}}}{\Gamma(\frac{m}{2})} \omega^{\frac{m}{2}-1} \int_{0}^{+\infty}u^{\frac{n+m}{2}-1} e^{-\frac{1}{2}u(1+\omega)}du\\ &=\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\frac{(\frac{1}{2})^{\frac{m}{2}}}{\Gamma(\frac{m}{2})} \omega^{\frac{m}{2}-1} (\frac{\Gamma(\frac{n+m}{2})}{(\frac{1}{2}(1+\omega))^{\frac{n+m}{2}}})\\ &=\frac{\Gamma(\frac{n+m}{2})}{\Gamma(\frac{n}{2})\Gamma(\frac{m}{2})}\frac{\omega^{\frac{m}{2}-1}}{(1+\omega)^{\frac{n+m}{2}}} \end{align}$$

Then, the pdf for $W=\frac{V/m}{U/n}$ is: $$\begin{align} f_{\frac{V/m}{U/n}}&=f_{\frac{n}{m}V/U}\\ &=\frac{m}{n}f_{V/U}(\frac{m}{n}\omega)\\ &=\frac{m}{n}\frac{\Gamma(\frac{n+m}{2})}{\Gamma(\frac{n}{2})\Gamma(\frac{m}{2})}\frac{(\frac{m}{n}\omega)^{\frac{m}{2}-1}}{(1+\frac{m}{n}\omega)^{\frac{n+m}{2}}}\\ &=\frac{\Gamma(\frac{n+m}{2})}{\Gamma(\frac{n}{2})\Gamma(\frac{m}{2})}\frac{m}{n}\frac{(\frac{m}{n}\omega)^{\frac{m}{2}-1}}{(n+m\omega)^{\frac{n+m}{2}}}n^{\frac{n+m}{2}}\\ &=\frac{\Gamma(\frac{n+m}{2})}{\Gamma(\frac{n}{2})\Gamma(\frac{m}{2})}\frac{m^{\frac{m}{2}}n^{\frac{n}{2}}\omega^{\frac{m}{2}-1}}{(n+m\omega)^{\frac{n+m}{2}}} \end{align}$$