independence of uniform random variables1

Question

let $X_j \sim U(0,1)$

if $$Y_j=\frac{X_j}{X_1+X_2+\cdots+X_n}$$

I want to show that:

• $Y_j $are independent

• $\operatorname{Var}(Y_1)=\dfrac{c}{n^2} +o\left(\dfrac{1}{n^2}\right)$ then calculate $c$

Answer 1

$\newcommand{\var}{\operatorname{var}}$ $\newcommand{\cov}{\operatorname{cov}}$

Since $Y_1+\cdots+Y_n$ is constrained to be $1$, we have $\var(Y_1+\cdots+Y_n)=0$. But $$ \var(Y_1+\cdots+Y_n)= \var(Y_1)+\cdots+\var(Y_n) + \underbrace{2\cov(Y_1,Y_2)+\cdots}_{\binom n 2 \text{ terms}} $$ By symmetry, all of the variances are equal to each other and all of the covariances are equal to each other. Thus you have $$ n\var + 2\binom n 2 \cov = 0. $$ Thus $$ \var = \frac{-2\binom n 2}{n}\cov = \frac{-\cov}{n}. $$ So it seems $c/n$ rather than $c/n^2$ is what you need. (And they can't be independent since $\var>0$, so $\cov<0$, and there's still the problem of finding $c$.)