Covariance of order statistics I'm a researcher in social science and I have encountered the following math formulation of a problem in my field. Note that I'm relatively new to stack exchange and I have already posted this on math.stackexchange as well as mathoverflow, before being asked to post it here. Please let me know if I should delete the double post from the sites (otherwise I will definitely update all sites when I receive an answer). Thank you!
Let $x_1,x_2,...,x_n,x_{n+1}$ be $n+1$ i.i.d. random variable with non-negative support and strictly positive probability mass around zero. Let $$z_k\equiv \min\{x_1,...,x_k\}.$$ 
In simulations, I find $Cov(z_{n+1},z_n)$ to be very close to $Var(z_{n+1})$ and very different from $Var(z_{n})$, as long as $n>>1$. I have tried this for many distributions (with non-negative support): $Cov(z_{n+1},z_n) / Var(z_{n+1}) \approx 1$ and the approximation gets better as $n$ increases, even for small $n$ such as $n=5$. On the other hand, $Cov(z_{n+1},z_n)/Var(z_n)$ is very far from 1.
How can I formalize this? That is, I'm looking for some kind of bounds on how the approximation improves with $n$. Of course, as $n\to\infty$, $Var(z_n)=Var(z_{n+1})$, so I'm looking for results either for $n$ finite, or an asymptotic result that takes $n$ to $\infty$ on $m\equiv floor(c\cdot n)$ for constant $c>1$ such that $Cov(z_n,z_m)$ is closer to $Var(z_m)$ than $Var(z_n)$.
Note that the approximation only works for X with non-negative support and has a positive probability mass in a neighborhood around zero. I believe that from results in extreme value theory, such distributions have exponential distribution in the limit. I don't know if this is important.
 A: 
Let $(X_1,\dots,X_n,X_{n+1})$ denote a random sample of size $(n+1)$ drawn on $X$, and let $$Z_n = \min\{X_1,...,X_n\} \quad \text{and} \quad Z_{n+1} = \min\{X_1,...,X_n,X_{n+1}\}$$ 

By including the extra $X_{n+1}$ term, there are only 2 possibilities:



*

*EITHER CASE A $\rightarrow$ with probability $\frac{n}{n+1}$


$\quad \quad \text{The extra term } X_{n+1}$ does NOT change the sample minimum i.e. $z_{n+1} = z_n$. Then:
$$\text{Cov}(Z_n, Z_{n+1})\big|_\text{Case A} \; = \; \text{Cov}(Z_{n+1}, Z_{n+1}) \; = \; \text{Var}(Z_{n+1})$$
Since Event A occurs with probability $\frac{n}{n+1}$, this immediately explains why your observed unconditional covariance $\text{Cov}(Z_n, Z_{n+1})$ is well approximated by $\text{Var}(Z_{n+1})$, as $n$ increases.



*

*OR CASE B $\rightarrow$ with probability $\frac{1}{n+1}$


$\quad \quad \text{The extra term } X_{n+1}$ DOES change the sample minimum i.e. $Z_{n+1} < Z_n$. Then $Z_{n+1}$ and $Z_n$ must be the $1^{\text{st}}$ and $2^{\text{nd}}$ order statistics from a sample of size $n+1$ i.e.
$$\text{Cov}(Z_n, Z_{n+1})\big|_\text{Case B} \; = \; \text{Cov}\big(X_{(1)}, X_{(2)}\big) \text{ in a sample of size: } n+1$$

In summary:
\begin{align*}\displaystyle  \text{Cov}(Z_n, Z_{n+1}) \; &= \frac{n}{n+1}\text{Cov}(Z_n, Z_{n+1})\big|_\text{Case A} \quad  + \quad  \frac{1}{n+1}\text{Cov}(Z_n, Z_{n+1})\big|_\text{Case B} \\ 
&= \frac{n}{n+1} \text{Var}(Z_{n+1})   \quad  + \quad  \frac{1}{n+1}  \text{Cov}\big(X_{(1)}, X_{(2)}\big)_{\text{sample size } = n+1} \\ &   
\end{align*}
This makes it easy to see why the result is similar to  $\text{Var}(Z_{n+1})$: because Case A dominates with probability $\frac{n}{n+1}$

Example and Check: Uniform Parent
In the case of $X \sim \text{Uniform}(0,1)$ parent:


*

*Case A:  $\text{Var}(Z_{n+1}) = \text{Var}(X_{(1)})_{\text{sample size } = n+1} = \frac{n+1}{(n+2)^2 (n+3)}$

*Case B: $\text{Cov}\big(X_{(1)}, X_{(2)}\big)_{\text{sample size } = n+1} = \frac{n}{(n+2)^2 (n+3)}$

*Then: $\text{Cov}(Z_n, Z_{n+1})  = \frac{n}{(n+1) (n+2) (n+3)}$
The following diagram compares:


*

*this exact theoretical solution for $\text{Cov}(Z_n, Z_{n+1})$, as $n$ increases from 1 to 30  $\rightarrow$ the red curve

*to a Monte Carlo calculation of $\text{Cov}(Z_n, Z_{n+1})$  $\rightarrow$  the blue dots

Looks fine.

The following diagram compares the exact theoretical solution for $\text{Cov}(Z_n, Z_{n+1})$, $\text{Var}(Z_n)$ and $\text{Var}(Z_{n+1})$:  as the OP reports, by the time $n = 5$, $\text{Cov}(Z_n, Z_{n+1})$ is well approximated by $\text{Var}(Z_{n+1})$:

A: This seems to imply that Var(zn)/Var(zn+1) if far from 1 as well. If zi is really the minimum of a sample of size i, then for standard computable examples, like the exponential distribution, the ratio of variances is close to 1 for n reasonably large, for example ((n+1)/n)^2 for the exponential. Perhaps you're simulating the second smallest order statistic?
