I want to figure out how the fraction of the biggest sample of the sum of some iid samples from an arbitrary distribution varies.
For example, I have a random variable $X \sim f(x)$, and I sampled $n$ iid samples, denoted as $\{x_1, x_2, \cdots, x_n\}$. Clearly when $n=1$, the fraction $\left(\max\{x_i\}_{i=1}^n \right) / \left(\sum_{i=1}^n x_i\right)$ is simply $1$. When the variance is infinite, as $n$ grows, the fraction may reach $1$. However, if the variance is finite, it should slowly go to $\frac{1}{n}$.
It is a hunch that how the fraction converges to where is related to some properties of $f(x)$. Is there any way that I can describe this fraction as some function of $f$ and $n$, maybe an upper/lower bound?
Thanks a lot.
If we assume that $X$ is almost surely bounded, then we can give a pretty straightforward answer :
Let, $X, x_1,\ldots,x_n$ be i.i.d. and denote by $x_{(n)}:=\max\{x_i\}_{i=1}^n$ the maximum order statistic, and assume that $\|X\|_\infty = M$ for some $M>0$, where $\|\cdot\|_\infty$ denotes the essential supremum.
Then it can be shown that $x_{(n)}\to M$ in probability (in fact almost surely, by Borel-Cantelli).
On the other hand, the law of large numbers yields $\frac1n \sum_{i=1}^n x_i \to \mathbb E[X] $ in probability as well.
Therefore, provided that $\mathbb E[X]\ne 0$, continuous mapping theorem yields $$n\cdot \frac{x_{(n)}}{\sum_{i=1}^n x_i} = \frac{x_{(n)}}{\frac 1 n\sum_{i=1}^n x_i} \to \frac{M}{\mathbb E[X]} $$ in probability (or almost surely...) as $n\to\infty$. Hence for large $n$, the ratio $\frac{x_{(n)}}{\sum_{i=1}^n x_i}$ behaves like $\mathcal O(1/n)$, as you observed in your simulations.
The case unbounded $X$ or infinite variance is much more delicate to handle.
One thing you can do is note that the CDF of $x_{n}$ is given by $\mathbb P(x_{(n)}\le t) = F_X(t)^n $ where $F_X$ is the CDF of $X$. Hence if the sequence $F_X^n$ converges pointwise to some limit CDF $F$, then you can apply Slutsky's theorem to find a limiting distribution of $\frac{x_{(n)}}{\sum_{i=1}^n x_i}$.