6

If $U\sim\chi_{m}^2,V\sim\chi_n^2$ and $U,V$ are independent then prove that $\frac{V}{U+V}\sim\beta\left(\frac n2,\frac m2\right)$

The joint pdf of $U$ and $V$ is, \begin{align} f_{UV}(u,v)&=\frac{1}{2^{\frac m2}\Gamma\left(\frac m2\right)}u^{\frac m2-1}e^{-\frac u2}\frac{1}{2^{\frac n2}\Gamma\left(\frac n2\right)}v^{\frac n2-1}e^{-\frac u2}\\ &=\frac{1}{2^{\frac{m+n}{2}}\Gamma\left(\frac m2\right)\Gamma\left(\frac n2\right)}u^{\frac m2-1}v^{\frac n2-1}e^{-\frac12(u+v)} \end{align} Now let $Y=\frac{V}{U+V}$ then CDF of $Y$ is, \begin{align} F_Y(y)&=\mathbb P(Y\le y)\\ &=\mathbb P\left(\frac{V}{U+V}\le y\right)\\ &=\mathbb P\left(\frac VU\le \left(\frac{y}{1-y}\right)\right)\\ &=\mathbb P\left(V\le \left(\frac{y}{1-y}\right)U\right)\\ &=\int_{u=0}^{\infty}\int_{v=0}^{\left(\frac{y}{1-y}\right)u}\frac{1}{2^{\frac{m+n}{2}}\Gamma\left(\frac m2\right)\Gamma\left(\frac n2\right)}u^{\frac m2-1}v^{\frac n2-1}e^{-\frac12(u+v)}\:dv\:du \end{align} Now we can get $f(y)$ using Leibniz integral rule, \begin{align} f_y(y)&=\frac{1}{2^{\frac{m+n}{2}}\Gamma\left(\frac m2\right)\Gamma\left(\frac n2\right)}\underbrace{\int_{u=0}^{\infty}\frac{u}{(1-y)^2}u^{\frac m2-1}{\left(\frac{yu}{1-y}\right)}^{\frac n2-1}e^{-\frac12\left(u+\frac{yu}{1-y}\right)}\:du}_{I} \end{align} But it seems I am far away to $\beta\left(\frac n2,\frac m2\right)$. Is there other way to proof it$?$ Any hint or solution will be appreciated.

Update:
[For seek of completeness]Using @NCh answer,
Replace $t=u\left(\frac{1}{2(1-y)}\right)$, $u=2(1-y)t$, $du=2(1-y)\,dt$: $$ I=\frac{y^{\frac{n}2-1}}{(1-y)^{\frac{n}{2}+1}}\cdot 2^{\frac{n+m}{2}}(1-y)^\frac{n+m}{2}\underbrace{\int_{t=0}^{\infty}t^{\frac{n+m}{2}-1}e^{-t}\:dt}_{\Gamma\left(\frac{n+m}{2}\right)}$$ $$f_Y(y)=\frac{\Gamma\left(\frac{n+m}{2}\right)}{\Gamma\left(\frac m2\right)\Gamma\left(\frac n2\right)}y^{\frac{n}2-1}(1-y)^{\frac m2-1}$$ Hence $f_Y(y)\sim \beta\left(\frac n2,\frac m2\right)$
StubbornAtom
  • 17,052
NJN
  • 527
  • The crucial assumption independence of $U$ and $V$ is missing in the post. See the answer at https://math.stackexchange.com/a/190695/321264 and several other linked threads asking effectively the same question. – StubbornAtom Jan 14 '20 at 11:42
  • https://math.stackexchange.com/questions/2263496/if-x-y-sim-chi-2-with-m-and-n-degrees-of-freedom-then-fracxxy?rq=1 – StubbornAtom Jan 14 '20 at 15:29

2 Answers2

2

In addition to brilliant comment of StubbornAtom, I’ll only give a partial answer on how to get to the beta distribution in your solution.

First, replace $\Gamma\left(\frac12\right)$ in denominators by $\Gamma\left(\frac{n}2\right)$ and $\Gamma\left(\frac{m}2\right)$ respectively to correct misprints in your solution. Then consider the integral $$ \int_{u=0}^{\infty}\frac{u}{(1-y)^2}u^{\frac m2-1}{\left(\frac{yu}{1-y}\right)}^{\frac n2-1}e^{-\frac12\left(u+\frac{yu}{1-y}\right)}\:du $$ $$ = \frac{y^{\frac{n}2-1}}{(1-y)^{\frac{n}{2}+1}}\int_{u=0}^{\infty}u^{\frac{n+m}{2}-1}e^{-u\left(\frac{1}{2(1-y)}\right)}\:du := I $$ Replace $t=u\left(\frac{1}{2(1-y)}\right)$, $u=2(1-y)t$, $du=2(1-y)\,dt$: $$ I=\frac{y^{\frac{n}2-1}}{(1-y)^{\frac{n}{2}+1}}\cdot 2^{\frac{n+m}{2}}(1-y)^\frac{n+m}{2}\underbrace{\int_{t=0}^{\infty}t^{\frac{n+m}{2}-1}e^{-t}\:dt}_{\Gamma\left(\frac{n+m}{2}\right)}. $$ Finally, substitute this value back into the pdf, you will get the desired pdf.

NCh
  • 13,807
  • 4
  • 17
  • 33
0

\begin{align} \Pr\{\frac{V}{U+V}\leq y\}&=\Pr\{\frac{1}{\frac mn\frac{U/m}{V/n}+1}\leq y\}\\ &=\Pr\{\frac{1}{\frac mnF_{m,n}+1}\leq y\}\\ &=\Pr\{F_{m,n} \geq (y^{-1}-1)\frac nm\}\\ &=1-\Pr\{F_{m,n} \leq (y^{-1}-1)\frac nm\}\\ &=1-\frac{B\left(\left(y^{-1}-1\right)\frac nm,m,n\right)}{B(m,n)},\\ \end{align} where $B\left(x,a,b\right)=\int_0^x t^{a-1}(1-t)^{b-1}dt$. See here for some more details. Therefore the pdf of $\frac{V}{U+V}$ is $$\frac d{dy}\left(1-\frac{B\left(\left(y^{-1}-1\right)\frac nm,m,n\right)}{B(m,n)}\right).$$

The rest shall be manageable.

Math-fun
  • 9,507