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Random variables $X, Y$ are independent with densities $f_{X}$ and $f_{Y}$ where $f_{X}(x) = \begin{cases} 4x^{2}e^{-2x} & \text{if $x>0$}\\ 0 & \text{otherwise} \end{cases}$

$f_{Y}(x) = \begin{cases} \frac{8}{3}x^{3}e^{-2x} & \text{if $x>0$}\\ 0 & \text{otherwise} \end{cases}$

Let $V=\frac{X}{X+Y}$. Find $Var(V)$.

My problem is that this might be done right from the definition of $Var$ (and the answer is $\frac{3}{98}$) but it takes a lot of time. I am looking for some smarter solution.

Guesttt
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    As $X$ and $Y$ are independent Gamma variables , it can be shown that $V$ is a Beta variable. – StubbornAtom Mar 12 '18 at 10:09
  • Ok. As I know names of those distribution I found https://math.stackexchange.com/questions/190670/how-exactly-are-the-beta-and-gamma-distributions-related

    which is what I was searching for. Thank you

     – Guesttt Mar 12 '18 at 10:23

1 Answers1

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So to answer your question...

With $\text{G}(\alpha,p)$ denoting a Gamma distribution having pdf $\displaystyle f(u)=\frac{\alpha^p e^{-\alpha u}u^{p-1}}{\Gamma(p)}\mathbf1_{u>0}$ where $\alpha$ and $p$ are positive, you have $X\sim\text{G}(2,3)$ and $Y\sim\text{G}(2,4)$ independent of each other.

So, $\displaystyle V=\frac{X}{X+Y}\sim\text{Beta}(3,4)$, the beta distribution of the first kind.

Hence $\displaystyle\mathrm{Var}(V)=\frac{3\times4}{(3+4)^2(3+4+1)}$

StubbornAtom
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