The standard deviation of the subsample of correlated random data. Let's assume to observe N random variable x, which has some standard deviation sigma and mean=0. N data points correlate with some typical correlation length, let's say ~ 5.
This means we have N data points
$\{x_1, x_2, x_3, x_4, x_5, x_6, ... x_N\}$
and 5 adjacent observed values correlate.
Now, we randomly take the M sample from the $N$ data points. Given the correlation length of 5, if $M$ is much less than $N/5$, the $M$ sample can be thought of as roughly independent (no correlation). In this case, are the expected standard deviation of the $M$ sample and that of the original N sample the same? Does this change if $M$ becomes close to $N$.
I am happy to hear intuitive explanations, proof by rigorous mathematics or numerical experiments. I would appreciate it if you could edit my question or let me know if you find any inappropriate use of the mathematical term.
 A: The expected sample standard deviations would not be the exactly the same even if the points were independent, because you have different sample sizes.  With a Bessel correction the expected sample variances would then be the same but you lose that equality when taking the square-root.  So you might instead analyse variances with a Bessel correction to avoid this point.
If you do that, then I do not a reason to believe the expectation of the subsample variance and the expectation of the full sample variance would be different, despite the correlation issue, and this does not depend on the relative sizes of the two samples (so long as $N \ge M \ge 2$).
As an illustration, here are some simulations using R, where $K$ is what you call the correlation length.
twovariances <- function(N, M=5, K=1){
   randomvalues <- rnorm(N+K)
   mat <- matrix(rep(randomvalues, K)[1:((N+K-1)*K)], ncol=K)
   bigsample <- rowSums(mat)[K:(N+K-1)] / sqrt(K)
   subsample <- sample(bigsample, M)
   c(var(bigsample), var(subsample))
   } 

set.seed(2022)

Showing the effect of the square root on the average sample standard deviation, even with no correlation:
# compare variances and standard deviations
sims <- replicate(10^5, twovariances(N=100, M=4, K=1))
rownames(sims) = c("bigsample", "subsample")
rowMeans(sims)        # variances with Bessel correction
# bigsample subsample 
# 0.9998452 0.9996055         
rowMeans(sqrt(sims))  # standard deviations
# bigsample subsample 
# 0.9973976 0.9215901 

Showing that with large $N/M$ the average sample variances are similar, given simulation approximations:
# large N/M
sims <- replicate(10^5, twovariances(N=100, M=4, K=5))
rownames(sims) = c("bigsample", "subsample")
rowMeans(sims) 
# bigsample subsample 
# 0.9593476 0.9577393 

Showing that with small $N/M$ the average sample variances are still similar:
# small N/M
sims <- replicate(10^5, twovariances(N=100, M=84, K=5))
rownames(sims) = c("bigsample", "subsample")
rowMeans(sims)   
# bigsample subsample 
# 0.9604660 0.9605773      

Showing that with large $M$ and $K$ the average sample variances are still similar (they are both smaller due to the high correlation in the big sample)
# large M and K
sims <- replicate(10^5, twovariances(N=100, M=84, K=95))
rownames(sims) = c("bigsample", "subsample")
rowMeans(sims)    
# bigsample subsample 
# 0.3561925 0.3562596 

Showing that with small $N$ the average sample variances are still similar (they are both smaller due to the high correlation in the big sample)
# small N 
sims <- replicate(10^5, twovariances(N=4, M=2, K=5))
rownames(sims) = c("bigsample", "subsample")
rowMeans(sims)  
# bigsample subsample 
# 0.3344043 0.3359496 

