Variance of $\frac{x_i}{\sum_{i=1}^nx_i}$ when $x_1,\ldots,x_n$ are iid random variables

Suppose $$x_1,\ldots,x_n$$ are iid random variables with finite variances. Is the variance of $$z_i=\frac{x_i}{\sum_{i=1}^nx_i}$$ a function of $$n$$ only? I am interested in two cases. Case 1: the variance of $$x_i$$ is zero. Case 2: the variance of $$x_i$$ is nonzero.

I know that the expectation is $$\frac{1}{n}$$ because $$\sum_{i=1}^nz_i=1$$ so that both $$\sum_{i=1}^nE(z_i)=nE(z_i)$$ and $$\sum_{i=1}^nE(z_i)=E(\sum_{i=1}^nz_i)=1$$. To find the variance $$V(z_i)$$ we would want to find $$E(z_i^2)$$, and we could do so by finding $$E(\sum_{i=1}^nz_i^2)=E(\frac{\sum_{i=1}^nx_i^2}{(\sum_{i=1}^nx_i)^2})$$.

• The best I could do was to consider $<z_1\sum z> =<z_1>$ to get ${1\over n}= <z^2>+(n-1)<zz'>$ Feb 18, 2020 at 9:09
• I don't quite understand the question. Certainly we can find the value of the variance from the definition. Are you asking whether we can find it given some particular set of data? E.g. given the means and variances of the $x_i$? Feb 18, 2020 at 12:37
• @joriki Can we find its value in terms of known constants ($n$ and real numbers). Or is the variance distribution-dependent? Its expectation is not distribution-dependent but always equal to $1/n$. Feb 18, 2020 at 16:01

As clarified in the comments, the question is whether the variance is a function of $$n$$ only or depends on the distribution.
It depends on the distribution. The variance of $$z_i$$ is zero or not according as the variance of the $$x_i$$ is zero or not.
For a slightly more interesting case, consider $$x_i$$ uniformly distributed over $$\{a,b\}$$. Then $$z_i$$ takes the value $$\frac12$$ with probability $$\frac12$$ and the values $$\frac a{a+b}$$ and $$\frac b{a+b}$$ with probabilities $$\frac14$$ each, so the variance is
$$\frac14\left(0+0+\left(\frac a{a+b}-\frac12\right)^2+\left(\frac b{a+b}-\frac12\right)^2\right)=\frac18-\frac{ab}{2(a+b)^2}\;,$$
which depends on $$a$$ and $$b$$ and thus on the distribution.