I have a long range dependence process which is defined in such a way: $$σ^2_{μ,X}=∫_T∫_{R^2}Cov_X(t,u,v) μ(du) μ(dv) dt. $$ We can use derived formula for covariation using distribution functions here: $$Cov_X (t,u,v)=F_{X_0,X_t} (u,v)-F_{X_0} (u) F_{X_0} (v)$$ Then, I got $$∫_T∫_{R^2}(F_{X_0,X_t} (u,v)-F_{X_0} (u) F_{X_0} (v)) μ(du) μ(dv) dt$$ This result was estimated by empirical distribution functions $$\hat{F}_{X_0} (u)=P ̂(X_0≤u)=1/n ∑_{i=1}^nI_{X_i≤u}, \space \hat{F}_{X_0} (v)=P ̂(X_0≤v)=1/n ∑_{i=1}^nI_{X_i≤v}, \space \hat {F}_{(X_0,X_t)} (u,v)=\frac{1}{(n-t)} ∑_{i=1}^{n-t}I_{X_i≤u} I_{X_{i+t}≤v}.$$ After inserting these estimators into the previous equation and skipping some steps I received such result: $$σ^2_{μ,X} = ∫_T∫_{R^2}\frac{1}{(n-t)} ∑_{i=1}^{n-t}I_{X_i≤u} * I_{X_{i+t}≤v} \space μ(du) μ(dv) dt - ∫_T∫_{R^2}\frac{1}{n} ∑_{i=1}^nI_{X_i≤u}* \space\frac{1}{n} ∑_{i=1}^nI_{X_i≤v} \space μ(du) μ(dv) dt.$$ Now I need to compute order statistics from these integrals, but I have no idea how to proceed, maybe someone can help me with that.

  • $\begingroup$ Great question. Keep it up. Welcome to Math Stack Exchange. :-D $\endgroup$ – The Great Duck Nov 25 '18 at 23:58
  • $\begingroup$ @TheGreatDuck thanks, hope to find help here =) $\endgroup$ – Dmytro Bihun Nov 26 '18 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.