Sum of Discrete Uniform Variables times extraction index

Question

Let $X_i$ be the $i-th$ extraction from an urn with $N$ balls numbered from $1$ to $N$. Let's make $N$ extractions without replacement, so that that the urn is left without balls.

Let $Y_i = X_i \cdot i$ basically multiplying the number written on the extracted ball by the index of the extraction.

Obviously $\Bbb E(Y_i) = i \cdot \frac{N+1}{2}$

You can get $Var(Y_i) = \frac{i^2(N^2-1)}{12}$

My question is

$Var(\sum_{i=1}^N Y_i)$

Expected value of the sum is quite easy to find so let's assume we have it, any idea guys?

Answer 1

Typically such calculations lead to simple polynomials of $N$ which can be found by brute force, such as listing all the possible permutations and doing the calculations. Here looking at $N=1$ to $8$ is more than enough, with the means being $1, \frac92, 12,25,45,\frac{147}{2},112,162$ and the variances being $0, \frac14, 2, \frac{25}{3}, 25, \frac{245}{4}, \frac{392}{3}, 252$. This leads to the empirical results

• $\mathbb E\left[\sum\limits_{i=1}^N Y_i\right] = \frac14 N(N+1)^2$ as can also be found easily using linearity of expectation
• $\mathrm{Var}\left[\sum\limits_{i=1}^N Y_i\right] = \frac1{144} (N-1)N^2(N+1)^2$

and it looks as if, when you use these results to standardise $\sum\limits_{i=1}^N Y_i$, that it converges in distribution to a Gaussian as $N$ increases, with the cumulative distributions when $N=8$ looking like