Expectation and variance without replacement

Question

Let $X_{N_1},\cdots,X_{N_n}$ is a sample without replacement from the set $\{1,2,\cdots,N\}$, and let $\bar X_n=\sum_{i=1}^n X_{N_{i}}/n$. Then, how one can find $E(\bar X_n)$, $Var(\bar X_n)$, $\max_{1\le i\le N} (X_i-\bar X_n)^2$, and $\sum_{i=1}^N(X_i-\bar X_n)^2$?

Answer 1

1. Expectation is linear if the expectations exist:

$$E[\bar X_n]=E[\sum_{i=1}^n X_{N_{i}}/n] = \sum_{i=1}^n E[X_{N_{i}}]/n$$

2. Variance is sort of linear for independent random variables:

$$Var[\bar X_n]=Var[\sum_{i=1}^n X_{N_{i}}/n] = \sum_{i=1}^n Var[X_{N_{i}}]/n^2$$

If the random variables are not independent (from Wiki):

3. $$\max_{1\le i\le N} (X_i-\bar X_n)^2$$

Statistically speaking, compute the values

$$(X_1-\bar X_n)^2, (X_2-\bar X_n)^2, ..., (X_n-\bar X_n)^2$$

then pick the highest.

Alternatively, we note that:

$$\max_{1\le i\le N} (X_i-\bar X_n)^2 = (\max_{1\le i\le N} |X_i-\bar X_n|)^2$$

so we can instead compute

$$|X_1-\bar X_n|, |X_2-\bar X_n|, ..., |X_n-\bar X_n|$$

then pick the highest and square it.

Probabilistically speaking, $\max_{1\le i\le N} (X_i-\bar X_n)^2$ is a random variable equal to

$(X_1-\bar X_n)^2$ if $(X_1-\bar X_n)^2 > (X_2-\bar X_n)^2, ..., (X_N-\bar X_n)^2$

$(X_2-\bar X_n)^2$ if $(X_2-\bar X_n)^2 > (X_1-\bar X_n)^2, (X_3-\bar X_n)^2, ..., (X_N-\bar X_n)^2$

$$\vdots$$

$(X_N-\bar X_n)^2$ if $(X_N-\bar X_n)^2 > (X_1-\bar X_n)^2, (X_2-\bar X_n)^2, ..., (X_{N-1}-\bar X_n)^2$

4. $$\sum_{i=1}^N(X_i-\bar X_n)^2 = \sum_{i=1}^N(X_i-\sum_{i=1}^n X_{N_{i}}/n)^2$$