# independence of uniform random variables1

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 $c$

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But $\sum_{j=1}^n Y_j = 1$. Are you seeking to prove independence of certain subset of $\{Y_j\}_{j=1}^n$? – Sasha Jan 21 '13 at 16:27
@Sasha ,no i want to show the independence of $Y_1 ,Y_2,...,Y_n$ – yalda Jan 21 '13 at 16:30
@yalda The point Sasha is making is that the $Y_i$ are clearly not independent; what you're trying to show is false. – Jonathan Christensen Jan 21 '13 at 16:40
@JonathanChristensen I don't know why they are not independence!! – yalda Jan 21 '13 at 16:44
@yalda Consider the case $n=2$. Then $Y_2 = 1-Y_1$: once we know the value of $Y_1$, we also know the value of $Y_2$. They aren't independent for higher $n$, either. – Jonathan Christensen Jan 21 '13 at 16:45

$\newcommand{\var}{\operatorname{var}}$ $\newcommand{\cov}{\operatorname{cov}}$
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 $c$.)