Why is the covariance of the hypergeometric distribution not dependent i and j?

Question

Suppose we have $\{1,\ldots,20\}$ and $X_i$ will be the drawn element number $i$ without placing it back (hypergeometric distribution)

My question is why is the covariance $\textbf{cov}=[X_i,X_j]$ for ($1\le i\neq j \le 20$) not dependent on $i$ and $j$?

I also don't understand covariance at all, expectation and variance is understable

Answer 1

Note: You are missing some information; I have added the necessary details in bold.

Suppose we select $n$ elements from $\{1,\ldots,20\}$, and $X_i$ will be the indicator random variable that we have drawn element number $i$ without placing it back (hypergeometric distribution.)

My question is why is the covariance $\textsf{Cov}[X_i,X_j]$ for ($1\le i\neq j \le 20$) not dependent on $i$ and $j$?

From the definition of covariance, and the fact that these are indicator random variables.

$$\begin{split}\mathsf {Cov}(X_i,X_j) &= \mathsf E(X_iX_j)-\mathsf E(X_i)\mathsf E(X_j)\\ &= \mathsf P(X_i=1,X_j=1)-\mathsf P(X_i=1)\;\mathsf P(X_j=1) \\ &= \frac 1{n(n-1)}-\frac 1n\cdot\frac 1 n & \text{ if } 0\leqslant i\neq j\leqslant 20\\ &= \frac 1{n^2(n-1)} \end{split}$$

The probabilities that the numbers are selected do not depend on the values of the numbers, only that they are different (and from the set, of course).   Ergo, neither does the covariance.

Answer 2

If $X_i$ denotes the element drawn at $i$-th draw then for distinct elements $r,s$ of $\{1,\dots,20\}$:$$P(X_i=r, X_j=s)=P(X_i=r)P(X_j=s\mid X_i=r)=\frac1{20}\frac1{19}$$So the distribution (and consequently covariance ) do not depend on $i, j$.