Error of a quantity computed from a time series I start with a presumably non-stationary time series. By some criterium $k$ I extract $n_k$ pairs of times $(t_{k,i,2},t_{k,i,1})$.
Now I compute the following quantity: $\bar{\tau} = \frac{1}{K} \sum_k \frac{1}{n_k} \sum_i (t_{k,i,2}-t_{k,i,1})$
What I'm looking for is the error of that quantity. I first began by summing up the variances of the "inner" mean, but I guess thats far off because of the high correlations.
So I guess, somehow the autocorrelation function has to be incorporated. But as I'm not too familiar with time series, I don't even really know what to look for.
Updated with info from comments:
Here is the example of how the pairs of times are chosen:

$A(t)$ is some observed quantity. $A_k$ is an upper threshold and $A_0$ some fixed lower one $(0\notin k)$. In physics we call the $y_{ki}=t_{k,i,2}-t_{k,i,1}$ first passage times.
Edit:
I've been thinking on this for a while and I've got some ideas, but I'm not sure. First I came up with this (acv is the autocovariance function): $\sigma^2(\bar\tau) = \frac{1}{K^2}\sum_{\forall t_{k_1,i_1,j_1}, \forall t_{k_2,i_2,j_2}} a(j_1,j_2) \frac{1}{n_{k_1} n_{k_2}} acv(|t_{k_1,i_1,j_1}-t_{k_2,i_2,j_2}|)$. $a(j_1, j_2)$ is supposed to be some function that is -1 if either of the arguments is 1 (meaning that the corresponding $t$ is subtracted in the formula in my original post).
But I don't see that this reduces to $\sigma^2(\bar\tau) = \frac{1}{K^2}\sum_k\frac{1}{n_k^2}\sigma_k^2$, which was my original guess assuming no correlation. $\sigma_k^2$ denotes the variance of the $k$th "inner mean". So finally I thought about something like this: $\sigma^2(\bar\tau) = \frac{1}{K^2}\sum_k\frac{1}{n_k^2}\sigma_k^2 + \frac{1}{K^2}\sum_{\forall t_{k_1,i_1,j_1}, \forall t_{k_2,i_2,j_2}, t_{k_1,i_1,j_1}\neq t_{k_2,i_2,j_2}} a(j_1, j_2) \frac{1}{n_{k_1} n_{k_2}} acv(|t_{k_1,i_1,j_1}-t_{k_2,i_2,j_2}|)$
Can you help me? Am I thinking at least in the right direction? Thanks, Julius
 A: First let us introduce notation $y_{ki}=t_{k,i,2}-t_{k,i,1}$. Then $\bar\tau=\frac{1}{K}\sum_k\frac{1}{n_k}\sum_iy_{ki}$. The variance then is
\begin{align}
Var(\bar\tau)&=E(\bar\tau-E\bar\tau)^2=E\left(\frac{1}{K}\sum_k\frac{1}{n_k}\sum_i(y_{ki}-Ey_{ki})\right)^2\\
&=E\frac{1}{K^2}\sum_k\sum_l\frac{1}{n_kn_l}\sum_i\sum_jcov(y_{ki},y_{lj})
\end{align}
Now the question is what can you say about $cov(y_{ki},y_{lj})$ which converted to original notation is $cov(t_{k,i,2}-t_{k,i,1},t_{l,j,2}-t_{l,j,1})$. If they are not correlated when $k\neq l$ or $i\neq j$ then the variance is simply
\begin{align}
Var(\bar\tau)=\frac{1}{K^2}\sum_k\frac{1}{n_k^2}\sum_i\sigma_{ki}^2.
\end{align}
I would be careful to employ stationarity. Your process has 2 indexes, when stationarity is usually defined for single indexed processes. When you have single index there is natural order of the observations $t_1<t_2<...$ and stationarity means that if you fix times $t_1<...<t_k$ then vectors $(X_{t_1},...,X_{t_k})$ and $(X_{t_1+h},...,X_{t_k+h})$ behave similarly for every shift $h$. If the index is two-dimensional there is no natural order, also shifting is now two-dimensional so there is no one clear way how to define stationarity.
So what process exactly you observe and how $t_{k,i}$ is chosen from it?
