Question :
Let us define the cumulative sum (Brownian motion): $$x_k = \sum_{i=1}^k y_i$$ and the running average : $$ \overline{x_k} =\frac{1}{W}\sum_{i=k-W+1}^k x_i$$ for $ k>W $, $W$ is a positive integer. Knowing that $ y_i \equiv \mathcal{N}(\mu,\sigma^2)$ (at convenience $\mu=0$) and independent; what is the variance of $(\overline{x_{k+2K}}-\overline{x_{k+K}})-(\overline{x_{k+K}}-\overline{x_{k}})$ ? $$ Var\left((\overline{x_{k+2K}}-\overline{x_{k+K}})-(\overline{x_{k+K}}-\overline{x_{k}})\right)$$ $K$ is a positive integer (at convenience over $W$).
This variance is inspired from Allan Variance. It's more efficient in the sens no preliminary calculation of $\mu$ is needed.
Attempt :
I was successful on calculating $Var\left(\overline{x_{k+K}}-\overline{x_{k}}\right)$ for $\mu=0$ and $W<K$.
$$ \overline{x_k} =\frac{1}{W}\sum_{i=k-W+1}^k \sum_{j=1}^i y_j$$
$$=\frac{1}{W}\left(W\sum_{i=1}^{k-W+1} y_i + \sum_{i=1}^{W-1} iy_{k-i+1}\right)$$
$$=\frac{1}{W}\left(W\sum_{i=1}^{k-W+1} y_i + \sum_{i=k-W+2}^{k} (k-i+1)y_{i}\right)\tag{1}$$
So: $$ \overline{x_{k+K}}- \overline{x_k}=\frac{1}{W}\left(W\sum_{i=1}^{k+K-W+1} y_i + \sum_{i=k+K-W+2}^{k+K} (k+K-i+1)y_{i}-W\sum_{i=1}^{k-W+1} y_i - \sum_{i=k-W+2}^{k} (k-i+1)y_{i}\right)$$ $$=\frac{1}{W}\left(\sum_{i=k-W+2}^{k} (W-(k-i+1))y_i + \sum_{i=k+1}^{k+K-W+1} Wy_{i}+\sum_{i=k+K-W+2}^{k+K} (k+K-i+1)y_i \right)\tag{2}$$
Since the sums don't overlap anymore and $y_i$ are independent, we can calculate the variance of each terms. The variance of the sum is the sum of the variance for independent variables:
$$Var\left(\sum_{i=k+1}^{k+K-W+1} Wy_{i}\right)=(K-W+1)W^2\sigma^2$$ $$Var\left(\sum_{i=k-W+2}^{k} (W-(k-i+1))y_i\right)=\sigma^2\sum_{i=k-W+2}^{k} (W-(k-i+1))^2$$ $$=\sigma^2\sum_{i=1}^{W-1} i^2=\sigma^2\frac{W(W-1)(2W-1)}{6}\tag{3}$$ $$Var\left(\sum_{i=k+K-W+2}^{k+K} (k+K-j+1)y_i \right)=\sigma^2\sum_{i=k+K-W+2}^{k+K} (k+K-j+1)^2$$ $$=\sigma^2\sum_{i=1}^{W-1} i^2=\sigma^2\frac{W(W-1)(2W-1)}{6}$$
So we have for the sum of the three terms: $$Var\left(\overline{x_{k+K}}-\overline{x_{k}}\right) = \frac{\sigma^2}{W^2}\left(\frac{W(W-1)(2W-1)}{3}+(K-W+1)W^2\right)$$ $$=K\sigma^2\left(1+\frac{1}{3K}\left(\frac{1}{W}-W\right)\right)\tag{4}$$
I checked this solution and it works; in the graph at the end, the blue line overlaps the blue dots. And we can see it is consistent since when $W=1$, the result is just $K\sigma^2$ and when $K$ is large, the bias $\frac{1}{3K}\left(\frac{1}{W}-W\right)$ becomes negligible.
But I didn't succeed in applying this method for the calculation of $ Var\left((\overline{x_{k+2K}}-\overline{x_{k+K}})-(\overline{x_{k+K}}-\overline{x_{k}})\right)$.
Since $({x_{k+2K}}-{x_{k+K}}) $ has a mean value of $K\mu$ and $-({x_{k+K}}-{x_{k}})$ a mean value of $-K\mu$ then I figured that $\left(({x_{k+2K}}-{x_{k+K}})-({x_{k+K}}-{x_{k}})\right)$ has a mean value of $0$. The mean value of $\left((\overline{x_{k+2K}}-\overline{x_{k+K}})-(\overline{x_{k+K}}-\overline{x_{k}})\right)$ should be pretty much the same : $0$. With this argument, I considered I could just take $\mu=0$ for this calculation. From the result (1) and with $W<K$:
$$(\overline{x_{k+2K}}-\overline{x_{k+K}})-(\overline{x_{k+K}}-\overline{x_{k}}) =\overline{x_{k+2K}}-2\overline{x_{k+K}}+\overline{x_{k}} =\frac{1}{W}\left(W\sum_{i=1}^{k+2K-W+1} y_i+\sum_{i=k+2K-W+2}^{k+2K} (k+2K-i+1)y_i-2W\sum_{i=1}^{k+K-W+1} y_i-2\sum_{i=k+K-W+2}^{k+K} (k+K-i+1)y_i +W\sum_{i=1}^{k-W+1} y_i+\sum_{i=k-W+2}^{k} (k-i+1)y_i\right)$$
$$=\frac{1}{W}\left(\sum_{i=k-W+2}^{k} ((k-i+1)-W)y_i-\sum_{i=k+1}^{k+K-W+1} Wy_i+ \sum_{i=k+K-W+2}^{k+K} (W-2(k+K-i+1))y_i+\sum_{i=k+K+1}^{k+2K-W+1} Wy_i+\sum_{i=k+2K-W+2}^{k+2K} (k+2K-i+1)y_i\right)$$$$\tag{5}$$
With this modification the sum don't overlap anymore and the variance of the sum is the sum of the variance. Lets compute the variance of each term :
$$ Var\left(\sum_{i=k-W+2}^{k} ((k-i+1)-W)y_i\right)=\sigma^2\sum_{i=k-W+2}^{k} ((k-i+1)-W)^2=\sigma^2\sum_{i=1}^{W-1} i^2=\sigma^2\frac{W(W-1)(2W-1)}{6}$$ $$ Var\left(\sum_{i=k+1}^{k+K-W+1} Wy_i\right)=\sigma^2(K-W+1)W^2 $$ $$ Var\left(\sum_{i=k+K-W+2}^{k+K} (W-2(k+K-i+1))y_i\right) = \sigma^2\sum_{i=k+K-W+2}^{k+K} (W-2(k+K-i+1))^2 = \sigma^2\sum_{i=-W+2}^{W-2} i^2= 2\sigma^2\sum_{i=1}^{W-2} i^2=\sigma^2\frac{(W-2)(W-1)(2W-3)}{3}$$ $$Var\left(\sum_{i=k+K+1}^{k+2K-W+1} Wy_i\right)=\sigma^2(K-W+1)W^2\tag{6}$$ $$ Var\left(\sum_{i=k+2K-W+2}^{k+2K} (k+2K-i+1)y_i\right)=\sigma^2\sum_{i=k+2K-W+2}^{k+2K} (k+2K-i+1)^2=\sigma^2\sum_{i=1}^{W-1} i^2=\sigma^2\frac{W(W-1)(2W-1)}{6}$$
So the sum of these 5 terms makes : $$Var(\overline{x_{k+2K}}-2\overline{x_{k+K}}+\overline{x_{k}})=\frac{1}{W^2}\left(\sigma^2\frac{W(W-1)(2W-1)}{3}+2\sigma^2(K-W+1)W^2+\sigma^2\frac{(W-2)(W-1)(2W-3)}{3}\right)$$ $$=2K\sigma^2\left(1-\frac{1}{K}\left(1-\frac{7}{3W}+\frac{W}{3}+\frac{1}{W^2}\right)\right)\tag{7}$$ But I checked this formula with statistical test and it doesn't fit (green line doesn't overlap the green dots).
The Matlab code for this verification is available here. Thank you for any help, it would be absolutely great.
There is an error before formula $(6)$ the correct calculation is $$ Var\left(\sum_{i=k+K-W+2}^{k+K} (W-2(k+K-i+1))y_i\right) = \sigma^2\sum_{i=k+K-W+2}^{k+K} (W-2(k+K-i+1))^2 = \sigma^2\sum_{i=-W+2}^0(W+2i-2)^2=\sigma^2\frac{W(W-1)(W-2)}{3} $$