36Alternatively one can take ⟨𝒪⟩ = ⟨𝒪 ⟩ = (1∕n)∑n 𝒪(t′) 0 t t′=0 without noticeable difference for t ≪ n . The choice in (13.48) results in a more accurate calculation and smaller finite sample effects. The choice (13.48) , instead of ∑n −1−t ′ ′ ρ𝒪 (t) ∝ (1∕(n− t)) t′=0 𝒪(t)𝒪 (t + t)− ⟨𝒪 ⟩0⟨𝒪 ⟩t , has smaller roundoff errors.