The decomposition and the amount of time it takes to perform it depends on whether you are using
the Tau Overlap or the Maximal Overlap.Maximal Overlap Allan Variance
Given $N$ equally spaced samples with averaging time $tau = n*tau_0$,
where $n$ is an integer such that $1<= n="" <="N/2$." therefore,="" $n$="" is="" able="" to="" be="" selected from="" ${n|n<="" floor(log2(n))}$="" then,="" $m="N" -="" 2n$="" samples="" exist.="" the="" maximal-overlap="" estimator="" given="" by:="" $\frac{1}{{2\left(="" {n="" 2k="" +="" 1}="" \right)}}\sum\limits_{t="2k}^N" {{{\left[="" {{{\bar="" y}_t}\left(="" k="" \right)="" {{\bar="" y}_{t="" k}}\left(="" \right)}="" \right]}^2}}="" $<="" p="">
where $ {{\bar y}_t}\left( \tau \right) = \frac{1}{\tau }\sum\limits_{i = 0}^{\tau - 1} {{{\bar y}_{t - i}}} $.
Tau-Overlap Allan Variance
Given $N$ equally spaced samples with averaging time $tau = n*tau_0$,
where $n$ is an integer such that $1<= n="" <="N/2$." therefore,="" $n$="" is="" able="" to="" be="" selected from="" ${n|n<="" floor(log2(n))}$="" then,="" a="" sampling="" of="" $m="\left\lfloor" {\frac{{n="" -="" 1}}{n}}="" \right\rfloor="" 1$="" samples="" exist.="" the="" tau-overlap="" estimator="" given="" by:<="" p="">
where $ {{\bar y}_t}\left( \tau \right) = \frac{1}{\tau }\sum\limits_{i = 0}^{\tau - 1} {{{\bar y}_{t - i}}} $.
=>=>