Show 1/X is F(n,m) if X is F(m,n)

Question

How to show this? F is the Fisher–Snedecor distribution, B is the beta function. This is what i have done

$$Y=\frac{1}{X}$$

$$f_{Y}(y)=f_{X}(1/y)\times (1/y^{2})=\frac{1}{B(\frac{m}{2},\frac{n}{2})}(\frac{m}{n})^{\frac{m}{2}}(\frac{1}{y})^{\frac{m}{2}-1}(1+\frac{m}{n}(\frac{1}{y}))^{\frac{-(m+n)}{2}}(\frac{1}{y})^{2}$$

I concluded that

$$B(\frac{m}{2},\frac{n}{2})=B(\frac{n}{2},\frac{m}{2})$$

I rearranged it so that I got

$$f_{Y}(y)=\frac{1}{B(n,m)}m^{\frac{m}{2}}n^{\frac{-n}{2}}y^{\frac{-m}{2}}y(ny+m)^{\frac{-m}{2}}(ny+m)^{\frac{-n}{2}}(ny)^{\frac{m}{2}}(ny)^{\frac{n}{2}}y^{-2}$$

Where to go next?

Answer 1

This is the full answer:

$$X\in F(\frac{m}{2},\frac{n}{2})\qquad Y=\frac{1}{X}$$ We want to show that $$Y\in F(\frac{n}{2},\frac{m}{2}).$$ So, $$F_{Y}(y)=P(\frac{1}{X}\leq y)=P(X\geq\frac{1}{y})=1-P(X\leq\frac{1}{y})=1-F_{X}(\frac{1}{y})\\ f_{Y}(y)=-f_{X}(\frac{1}{y})\times(\frac{-1}{y^{2}})=f_{X}(\frac{1}{y})\times(\frac{1}{y^{2}})=\frac{1}{B(\frac{m}{2},\frac{n}{2})}(\frac{m}{n})^{\frac{m}{2}}(\frac{1}{y})^{\frac{m}{2}-1}\Big( 1+(\frac{m}{n})(\frac{1}{y})\Big)^{\frac{-(m+n)}{2}}(\frac{1}{y^{2}})=\frac{1}{\frac{\Gamma{(\frac{m}{2})}\Gamma{(\frac{n}{2})}}{\Gamma{(\frac{m+n}{2})}}}(\frac{m}{n})^{\frac{m}{2}}(\frac{1}{y})^{\frac{m}{2}-1}\Big( 1+(\frac{m}{n})(\frac{1}{y})\Big)^{\frac{-(m+n)}{2}}(\frac{1}{y^{2}})=\frac{\Gamma{(\frac{m+n}{2})}}{\Gamma{(\frac{m}{2})}\Gamma{(\frac{n}{2})}}(\frac{m}{n})^{\frac{m}{2}}(\frac{1}{y})^{\frac{m}{2}-1}\Big( 1+(\frac{m}{n})(\frac{1}{y})\Big)^{\frac{-(m+n)}{2}}(\frac{1}{y^{2}})=\frac{\Gamma{(\frac{m+n}{2})}}{\Gamma{(\frac{m}{2})}\Gamma{(\frac{n}{2})}}(\frac{m}{n})^{\frac{m}{2}}(\frac{1}{y})^{\frac{m}{2}-1}\Big( \frac{ny+m}{ny}\Big)^{\frac{-(m+n)}{2}}(\frac{1}{y^{2}})=\frac{\Gamma{(\frac{m+n}{2})}}{\Gamma{(\frac{m}{2})}\Gamma{(\frac{n}{2})}}(\frac{m}{n})^{\frac{m}{2}}(\frac{1}{y})^{\frac{m}{2}-1}\Big( ny+m\Big)^{\frac{-(m+n)}{2}}(ny)^{\frac{m+n}{2}}(\frac{1}{y^{2}})=\frac{\Gamma{(\frac{m+n}{2})}}{\Gamma{(\frac{m}{2})}\Gamma{(\frac{n}{2})}}(\frac{m}{n})^{\frac{m}{2}}(\frac{1}{y})^{\frac{m}{2}-1}\Big( \frac{n}{m}y+1\Big)^{\frac{-(m+n)}{2}}(m)^{\frac{-(m+n)}{2}}(ny)^{\frac{m+n}{2}}(\frac{1}{y^{2}})=\frac{\Gamma{(\frac{m+n}{2})}}{\Gamma{(\frac{m}{2})}\Gamma{(\frac{n}{2})}}(\frac{m}{n})^{\frac{m}{2}}(\frac{1}{y})^{\frac{m}{2}-1}\Big( \frac{n}{m}y+1\Big)^{\frac{-(m+n)}{2}}(m)^{\frac{-m}{2}}(m)^{\frac{-n}{2}}(y)^{\frac{m+n}{2}}(n)^{\frac{m}{2}}(n)^{\frac{n}{2}}(\frac{1}{y^{2}})=\frac{\Gamma{(\frac{m+n}{2})}}{\Gamma{(\frac{m}{2})}\Gamma{(\frac{n}{2})}}(\frac{m}{n})^{\frac{m}{2}}(\frac{n}{m})^{\frac{m}{2}}(\frac{n}{m})^{\frac{n}{2}}y^{1-\frac{n}{2}+\frac{n}{2}+\frac{m}{2}-2}\Big( \frac{n}{m}y+1\Big)^{\frac{-(m+n)}{2}}=\frac{\Gamma{(\frac{n+m}{2})}}{\Gamma{(\frac{n}{2})}\Gamma{(\frac{m}{2})}}(\frac{n}{m})^{\frac{n}{2}}y^{\frac{m}{2}-1}\Big(1+\frac{n}{m}y\Big)^{\frac{-(n+m)}{2}} $$